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METHOD OF COMPLETING THE SQUARE

If we try to solve the equation 2k2 – 6k + 3 = 0 using the factorization method explained in the previous pages, we will realize that there are no factors.

The factorization method, though simple, cannot solve all quadratic equation questions, hence the need for some other methods that can solve such equations.

The general form of a quadratic equation ax2 ± bx ± c = 0 still remains the same.

            a is the coefficient of x2

            b is the coefficient of x

            c is the constant

the above terms a, b and c will be used in the next examples so master them.

 

 

Example

Solve the equation 2k2 – 6k + 3 = 0 using completing the square.

 

Solution

2k2 – 6k + 3 = 0

 

Step1: re-arrange the equation such that c is away from others.

 

            2k2 – 6k = –3

 

Step2: make sure a is 1. If a has value other than 1, divide through by such value.

In the case of the equation above, divide through by 2 

          

          \dpi{100} \frac{2k^2}{2} - \frac{6k}{2}=\frac{-3}{2} 

          k2 – 3k = –3/2 

 

Step3: take the square of half of b and add to both sides of the equation.

            b = -3

            half of b = -3/2

            square of half of b = (-3/2)2

                                          = 9/4

            Add the 9/4 to both sides of the equation.

 

                        k2 – 3k + 9/4 = -3/2 + 9/4

 

Step4: at the left side of the equation, put k and – 3/2 (half of b) in a bracket and square the bracket. At the right side, add -3/2 and 9/4

                 

                        (k – 3/2)2 =  \dpi{100} \frac{-6+9}{4}

                        (k – 3/2)2 =  3/4

Step5: cancel the square at the left side by taking the square root of both sides

                      \dpi{100} \sqrt{\left ( k-\frac{3}{2} \right )^2} = \sqrt{\frac{3}{4}} \\ \\ \\ \left ( k-\frac{3}{2} \right ) = \pm \frac{\sqrt{3}}{2}

Note: the square root of a number will always take ±. i.e.\dpi{100} \sqrt{9} = \pm 3, \ \sqrt{2} = \pm 1.414 \ \ etc.   

 

Step6: make k the subject of the equation. Place the 3/2 immediately after the = sign.

   

              \dpi{100} k = \frac{3}{2} \pm \frac{\sqrt{3}}{2}\ \ \ \ \ \ \ \ \ lcm \ is \ 2 \\ \\ k = \frac{3\pm \sqrt{3}}{2}

 

Step7: at this point split the equation in to two. The first with + and the second with – 

          \dpi{100} k = \frac{3+\sqrt{3}}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{3-\sqrt{3}}{2}  

 

**you can stop here, or proceed to find the square root of 3 so that your answer will be in decimal. It depends on where the question wants you to stop. if the question does not specify where to stop, then proceed.

 

                  \dpi{100} \fn_cm k = \frac{3+1.732}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{3-1.732}{2} \\ \\ \ \ k = \frac{4.732}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{1.268}{2} \\ \\

               k = 2.366           or         k = 0.634

 

Example 2 

Solve the equation d2 + 5d = 2 

 

Solution 

              d2 + 5d = 2

** Step 1:  c is already at the right side of the equation, so we proceed to step 2 

                    d2 + 5d = 2

** Step 2: There is no need to divide both sides by anything since a is already 1. So we move to step 3

                  d2 + 5d = 2

** Step 3:  take the square of half of b and add to both sides of the equation. 

                 b = 5 

                 half of b = 5/2

                square of half of b = 25/4

       add the 25/4 above to both sides of the equation 

                     d2 + 5d = 2

                      d^2 + 5d + \frac{25}{4} = 2 + \frac{25}{4}  

Step4: at the left side of the equation, put d and + 5/2 (half of d) in a bracket and square the bracket. At the right side, add 2 and 25/4

                  \left ( d + \frac{5}{2} \right )^2 = 2 + \frac{25}{4} \\ \\ \left ( d + \frac{5}{2} \right )^2 = \frac{8 + 25}{4}

                 (d + 5/2)2 = 33/4 

 

Step5: cancel the square at the left side by taking the square root of both sides  

                        \sqrt{\left ( d + \frac{5}{2} \right )^2} = \sqrt{\frac{33}{4}} \ \\ \\ \\ d + \frac{5}{2} = \pm \frac{\sqrt{33}}{2}

 

Step 6: make k the subject of the equation. Place the 3/2 immediately after the = sign.

            d = - \frac{5}{2} \pm \frac{\sqrt{33}}{2} \\ \\ lcm \ is \ 2 \\ \\ d= \frac{- 5 \pm \sqrt{33}}{2}

 

Step7: at this point split the equation in to two. The first with + and the second with –  

                d= \frac{- 5 + \sqrt{33}}{2} \ \ \ \ or \ \ \ \ d= \frac{- 5 - \sqrt{33}} {2} \\ \\ d= \frac{- 5 + 5.745}{2} \ \ \ \ or \ \ \ \ d= \frac{- 5 - 5.745} {2} \\ \\ d= \frac{0.745}{2} \ \ \ \ or \ \ \ \ d= \frac{- 10.745} {2}

                d= 0.372           or     d = -5.572  

 

                       Example 3

     Solve the equation 70v = 81v2 - 37 

                 

                 Solution 

** Re-arrange 

                    37 =  81v2 - 70v 

                   81v2 - 70v = 37 

 

** divide through by a  

                 v^{2} - \frac{70v}{81} = \frac{37}{81}  

** Add the square of half of b 

              b = 70/81

               half of b = 35/81

               square of half of b = 1225/6561

         v^{2} - \frac{70v}{81} + \frac{1225}{6561} \ \ \ = \ \ \ \frac{37}{81} + \frac{1225}{6561} \\ \\ \left ( v- \frac{35}{81} \right )^2 \ \ \ = \ \ \ \frac{2997+1225}{6561} \\ \\ \left ( v- \frac{35}{81} \right )^2 \ \ \ \ = \ \ \ \ \frac{4222}{6561} \\ \\ Take \ square \ root \ of \ both \ sides \\ \\ \sqrt{ \left ( v- \frac{35}{81} \right )^2} \ \ \ \ = \ \ \ \sqrt { \frac{4222}{6561}} \\ \\ v- \frac{35}{81} \ \ \ \ = \ \ \ \ \pm \frac{64.977}{81} \\ \\ v \ \ = \ \ \ \frac{35}{81} \ \ \ \ \pm \ \ \ \ \frac{64.977}{81} \\ \\ v \ \ = \ \ \ \frac{35 \pm 64.977}{81} \\ \\ Split \ the \ equation \\      

 

         v \ \ = \ \ \ \frac{35 + 64.977}{81} \ \ \ \ or \ \ \ \ v \ \ = \ \ \ \frac{35 - 64.977}{81} \\ \\

        v = 99.977/81    or     v = - 29.977/81  

         v = 1.234          or       v = - 0.3701

 

EXERCISE 1

 

Solve the equations below using completing the square method
1.        x2 + 8x + 3 = 0
2.        p2 - 3p = 12
3.        a2  - 2a - 9 = 0
4.        0 = e2  + 3e + 1
5.        c2  = 6c - 4
6.        2m2  + 5m + 1 = 0
7.        3b2  - 12b - 8 = 0
8.        7 = 4y + 4y2 
9.        - 2n2  = 7n + 2
10.       2 + 5d2  = 9d
11.       6 - z = 3z2 
12.       x2 = 3 - x
13.       23j = 16j2 + 8
14.       9q2  + 13q + 3 = 0
15.      12 - 11r - 2r2 
16.      6 - 3u - 5u2 = 0
17.      1 + 2f = 5f2 
18.      2 = 5m + 6m2 
19.      11g2 - 12g + 3 = 0
20.       2h2  = 3h + 7
21.       0 = 2a2  - 5a - 9
22.       25e2 - 42e + 17 = 0 

 

 

NEXT: QUADRATIC FORMULA METHOD 

PRE: FACTORIZATION METHOD  

 

PRE

NEXT

                       Other problems

 

1.  (a) (i)  Use identities sin (A + B) = sinA cosB + sinB cosA and cos (A + B) = cosA  cosB – sinA       sinB to express cos 2θ and cos 3θ in terms of cosθ.

(ii)   2 cos3θ cos2θ as a sum of two cosines and hence, or otherwise, obtain an expression for cos 5θ in terms of cosθ.

(b)   solve sin 2x = sin 3x for values of x in the interval 0° ≤ x ≥ 360°. 

 

                            Solution

    1.   (a)  i.  Cos 2θ

          cos (θ + θ)

          = cos θ cos θ – sin θ sin θ

          = cos2 θ – sin2 θ

From sin2 θ + cos2 θ = 1,

sin2 θ =1 – cos2 θ

                   = cos2θ – (1 – cos2θ)

                   = cos2θ – 1 + cos2θ

                   = 2cos2 θ – 1

 

          cos 3θ

          = cos (2θ + θ)

          = cos 2θ cos θ – sin2 θ sin θ

          = (cos2 θ – sin2 θ) cos θ – (2 sin θ cos θ) sin θ

          = (cos2 θ – (1 - cos2 θ)) cos θ – 2 sin2θ cos θ                              

          = (cos2 θ – 1 + cos2 θ) cos θ – 2 (1 – cos2 θ) cos θ

          = (2cos2 θ – 1) cos θ – 2cos θ (1 – cos2 θ)

          = 2 cos3 θ - cos θ – 2 cos θ + 2 cos3 θ

          = 2 cos3 θ + 2 cos3 θ – cos θ – 2 cos θ

          = 4 cos3 θ – 3 cos θ

 

                   ii  

2 cos 3θ cos 2θ

cos X + cos Y = 2 cos (X + Y)/2 cos (X – Y)/2

since our original question is having 2, there is no need to divide both sides by 2.

Equate the right hand of the equation to the original question.

2 cos (X + Y)/2 cos (X – Y)/2 = 2 cos 3θ cos 2θ

          (X + Y)/2 = 3θ                                   (X – Y)/2 = 2θ

          X + Y = 6θ ---------- (1)                      X – Y = 4θ ------------(2)  

Solving (1) and (2) simultaneously

          2X = 10θ

          X = 5θ

          From X + Y = 6θ  

                   5θ + Y = 6θ

                   Y = 6θ – 5θ

                   Y = θ

 

Cos X + cos Y

= cos5θ + cosθ.

Expressing 2 cos 3θ cos 2θ as the sum of two cosines = cos 5θ + cos θ

 

∴cos5θ + cosθ = 2 cos3θ cos2θ

 

          from the above equation, make  cos5θ from the above.

cos 5θ + cos θ = 2 cos 3θ cos 2θ

cos 5θ = 2 cos 3θ cos 2θ – cos θ

cos 5θ = 2 (4 cos3 θ – 3 cos θ) (2cos2 θ – 1) – cos θ

cos 5θ = (8 cos3 θ – 6 cos θ) (2cos2 θ – 1) – cos θ

cos 5θ = 16 cos5θ – 8 cos3θ – 12 cos3 θ + 6 cos θ – cos θ

cos 5θ = 16 cos5θ – 20 cos3 θ + 5 cos θ 

 

 

                   (b)

sin 2x = sin 3x

2 sinx cosx = sin (2x + x)

2 sinx cosx = sin 2x cos x + sin x cos 2x

2 sinx cosx = (2 sinx cosx) cosx + sinx (cos2x – sin2x)

2 sinx cosx = 2 sinx cos2x + sinx (cos2x – (1 – sin2x))

0 = 2 sinx cos2x + sinx (2cos2x – 1) – 2 sinx cosx

2 sinx cos2x + sinx (2cos2x – 1) – 2 sinx cosx = 0

          Factorize sinx

Sinx [2cos2x + (2cos2x – 1) – 2 cosx] = 0

Sinx [4cos2x – 2cosx – 1] = 0

Sinx = 0       or       4cos2x – 2cosx – 1 = 0

          Keep sinx = 0 first and Solve 4cos2x – 2cosx – 1 = 0 using quadratic formula

4cos2x – 2cosx – 1 = 0

a = 4            b = -2           c = -1 

\dpi{100} \small cos x = \frac{-b \pm \sqrt{b ^2 -4ac}}{2a} \\ cos x = \frac{-(-2) \pm \sqrt{(-2) ^2 -4\times 4\times -1}}{2 \times 4} \\ cos x = \frac{2 \pm \sqrt{4 + 16}}{8} \\ cos x = \frac{2 \pm \sqrt{20}}{8} \\ cos x = \frac{2 \pm 4.472}{8} \\ cos x = \frac{2 + 4.472}{8}\ \ \ \ \ \ or \ \ cos x = \frac{2 - 4.472}{8} \\ \\cos x = \frac{6.472}{8}\ \ \ \ \ \ or \ \ \ \cos x = \frac{-2.472}{8} \\

cos x = 0.809        or      cos x = - 0.309

 

note that we kept sin x = 0

 

sin x = 0                       or                          cos x = 0.809                   or                      cos x = - 0.309

**      When sin x = 0     (first and second quadrant)

x = sin-1 0     

x = 0°/360° (0° is same as 360°)

          or

x = 180 – 0

x = 180°

 

**      when cos x = 0.809         (cos is +ve, so x will be in 1st and 4th quadrant)

          x = cos-10.809

          x = 36°        (1st quadrant)

                   or

          x = 360 – 36

          x = 324°      (4th quadrant)

 

**      When cos x = - 0.309      (notice that cos is –ve, x will be in 2nd and 3rd quadrant)

          x = cos-1 0.309

          x = 72°

          x = 180 - 72°

          x = 108°

                   or

          x = 180 + 72

          x = 252°

 

values of x so far are 0°/360°, 180°, 36°, 324°, 108°, 252°

x = 0°, 36°, 108°, 252°, 180°, 324°, 360°

 

 

2.  Given that cos (A + B) = cosA  cosB – sinA sinB and sin (A + B) = sinA cosB + cosA sinB prove that cos 3A = 4 cos3A – 3 cosA, hence or otherwise, express the equation 2 cos 3x – 3cos 2x cos x + cos x = 0 in terms of cos x, and find all the values of x such that 0° ≤ x ≥ 360°, which satisfy the equation.

   

Solution

cos 3A = 4cos3A – 3cosA 

cos (2A + A)

= cos 2A cos A – sin 2A sin A

= (cos2A – sin2A)cos A – (2sinA cosA) sinA

= (cos2A – (1 – cos2A)) cosA – 2sin2A cosA

= (cos2A – 1 + cos2A) cos A – 2 (1 - cos2A) cos A

= (2cos2A – 1) cos A – 2cos A (1 – cos2A)

= 2cos3A – cosA – 2cosA + 2cos3A

= 2cos3A + 2cos3A  – cosA – 2cosA

= 4cos3A – 3cosA

Therefore, cos 3A = 4cos3A – 3cosA as required. 

 

Now to the second part of the question     

 

2cos3x – 3cos2x cosx + cosx = 0

            We will use the proof for cos 3A above for cos 3x, just replace A with x.

2(4cos3x – 3cosx) – 3(2cos2x – 1)cosx + cosx = 0

2(4cos3x – 3cosx) – 3cosx(2cos2x – 1) + cosx = 0

8cos3x – 6cosx – 6cos3x + 3cosx + cosx = 0

8cos3x – 6cos3x – 6cosx + 3cosx + cosx = 0

2cos3x – 2cosx = 0

             2cosx is common, so factorize it.

2cosx (cos2x – 1) = 0

2cosx = 0                                        or                                cos2x – 1 = 0

   We will keep cos2x – 1 = 0 first and solve for 2cosx = 0

 

cosx = 0/2

cosx = 0

x = cos-10

       we will take 0 as positive, so x will be in 1st and 4th quadrant.

x = 90° (1st quadrant value)

     or

x = 360 – 90

x = 270° (4th quadrant value)

  

  Also, if we decide to take 0 as a negative number, we will still get the same thing.

  Taking 0 as negative, x will be in 2nd and 3rd quadrant.

 

cosx = 0

x = cos-10

x = 90

x = 180 – 90

x = 90° (2nd quadrant value)

     or

x = 180 + 90

x = 270°  (3rd quadrant value)

 

   We are not through yet. Remember! We kept cos2x – 1 = 0

cos2x – 1 = 0

cos2x – 12 = 0

           Apply difference of two squares

(cosx – 1) (cosx + 1) = 0

cosx – 1 = 0                                   or                               cosx + 1 = 0

cosx = 0 + 1                                   or                            cosx = 0 – 1

cosx = 1 (1st and 4th)                    or                            cosx = – 1 (2nd and 3rd)

x = cos-11                                       or                            x = cos-11

x = 0° (1st quadrant value)            or                             x = 0°

        or

x = 360° - 0°                                                              x = 180° – 0°

x = 360° (4th quad value)                                           x = 180° (2nd quad value)

                                                                                      or

                                                                                   x = 180° + 0°

                                                                                   x = 180°

the required values of x are the ones in bold.

 

Notice that the values of x are same for 1st and 4th quadrant (0° is same as 360°), same with 2nd and 3rd. So we will pick just one of these same values.

 

Values of x so far are 90°, 270°, 0°/360°, 180°

 

x = 0°, 90°, 180°, 270°, 360°

 

I will keep updating this page as I see more challenging questions. 

 

Drop your comments and questions in the comment box below

 

 

PREVIOUS: Multiple Angles

PRE

SOLVING QUADRATIC EQUATION

The introduction of = sign to the expressions above will change them to equation, and hence we will be required to solve the equations for the given values of the unknown. Because quadratic equation is of power 2, we will get two values of the unknown. In order to solve quadratic equation, we will first factorize as done in the previous section, then we later proceed to solve.

Note that a quadratic equation has the general form ax2 + bx + c = 0

          Example

Solve the quadratic equations below:

       1.    x2 + 8x – 20 = 0

       2.    2a2 – 7a = 15

       3.    36p2 – 25 = 0

       4.    7d = 2d2 – 28

       5.    81 = 25z2

       6.    0 = 24e – 36e2

            Solution

 

1.       x2 + 8x – 20 = 0  

          Factorize the left side of the equation.

 

I believe by now you have mastered the factorization steps, so I will insert the factors into the equations directly. The long process of using product and sum to bring out the factors will be skipped.

 

            x2 – 2x + 10x – 20 = 0

            (x2 – 2x) + (10x – 20) = 0

            x(x – 2) + 10(x – 2) = 0

            (x – 2) (x + 10) = 0

 

Now that factorization is complete, we will now equate each expression in the bracket to zero and separating them with or

            (x – 2) =0                   or                   (x + 10) = 0

            x = 0 + 2                     or                    x = 0 – 10

            x = 2                           or                    x = - 10

 

            x = 2 or – 10

 

the two values of x (2 and –10) above are the solutions to the equation and are called the roots of the equation.

 

 

2.       2a2 – 7a = 15

          Re-arrange the equation so that one side is 0

            2a2 – 7a – 15 = 0

            2a2 + 3a – 10a – 15 = 0

            (2a2 + 3a) – (10a + 15) = 0

            a(2a + 3) – 5(2a + 3) = 0

            (2a + 3)(a – 5) = 0

            2a + 3 = 0                  or                    a – 5 = 0

            2a = 0 – 3                  or                    a = 0 + 5

            2a = –3                      or                    a = 5

            a = –3/2

 

            a = - 3/2 or 5

           

 

3.         36p2 – 25 = 0

            62p2 – 52 = 0

            (6p)2 - 52 = 0

            (6p – 5)(6p + 5) = 0

            6p – 5 = 0                  or                    6p + 5 = 0

            6p = 0 + 5                  or                    6p = 0 – 5

            6p = 5                        or                    6p = - 5

            p = 5/6                       or                    p = - 5/6

 

            p = – 5/6 or 5/6

 

 4.        d = 2d2 – 28

            Re-arrange

            0 = 2d2 – d – 28

            0 = 2d2 – d – 28 

            0 = 2d2 + 7d – 8d – 28

            0 = (2d2 + 7d) – (8d + 28)

            0 = d(2d + 7) – 4(2d + 7)

            0 = (2d + 7)(d – 4)

            0 = 2d + 7                  or                    0 = d – 4

            0 – 7 = 2d                  or                    0 + 4 = d

            - 7 = 2d                      or                    4 = d

            -7/2 = d

            d = -7/2                     or                    d = 4

 

            d = 4 or – 7/2

 

 5.         81 = 25z2

            0 = 81 – 25z2

            0 = 92 – 52z2

            0 = (9 – 5z)( 9 + 5z)

            0 = 9 – 5z                  or                    0 = 9 + 5z

            0 + 5z = 9                  or                    0 – 9 = 5z

            5z = 9                         or                    - 9 = 5z

            z = 9/5                       or                    -9/5 = z

                       

z = 9/5 or – 9/5

 

Note: for equations involving difference of two squares, we will always get the same value but the signs will be different.

           

   6.      0 = 24e – 36e2

            0 = 12e (e – 3)

            0 = 12e                      or                    0 = e – 3

            0/12 = e                     or                    0 + 3 = e

            0 = e                           or                    3 = e

 

                        e = 3 or 0

 

                       

                        Exercise J

Find the roots of the equations below.

            1.        3v2 + 19v + 20 = 0

            2.         3r2 = 11r – 6

            3.        14 – 15t = 9t2

            4.        16 = 81z2

            5.         28d2 + 15d = 0

            6.        4e – 21 =  – e2 

            7.        k2 – 3k – 10 = 0

            8.        62 – 2k = 5k2

            9.        49p2 = 225

            10.      32 – 4d = 32

            11.      2x2 – 5x + 3 = 0

            12.      t2 = 9t – 8

            13.      1 + 2c = c2

            14.      y2 = 3y

            15.      n2 + 7n = 30

            16.      48 + 8d = d2

            17.      8y = y2 + 16 

            18.      18 – 5m + 2m2 = 0

            19.      81 = 169r2

            20.      e2 – 25e + 136 = 0

            21.      2a2 = 19a – 9

            22.      0 = 9n2 – 21

            23.      81 = 121k

            24.      9p + 52 = p2

            25.      2y2 + 15 = 31y

            26.      30 – c = c2

            27.      8j2 – 44j = 0

            28.      6d2 = 4 + 5d

            29.      50 = 72p2

            30.      m2 + 18m + 77 = 0

            31.      8 = 5g2 – 6g

 

                              Forming quadratic equations from the given roots

Here, the roots of the quadratic equation will be given, we will then use the roots to get the equation. This is just the reverse of the previous section.

 

            Example

Find the quadratic equation with the following roots:

1.        2 and 3

2.        ½ and ¾

3.         – 1 and – 7

4.        4/5 and – 5 

5.        \dpi{100} \small - 2\frac{1}{3} \ \ and \ \ 1\frac{5}{6} 

6.         1.25 and 1.6 

7.        0 and –3

8.        –7 and 7 

 

            Solution       

1.        2 and 3

** Since 2 and 3 are the roots, that means x = 2 or x =3

            x = 2               or                    x = 3

** Move 2 and 3 to the left side to meet x and leave the right side with 0

Note: the signs will change on crossing the equality sign   

x – 2 = 0        or                    x – 3 = 0

** since the two equations are equal to zero, we can put them together without or.

            (x – 2)(x – 3) = 0

Expand

x2 – 3x – 2x + 6 = 0

x2 – 5x + 6 = 0

 

If we solve the above equation, we will get our initial roots.

 

2.         ½ and ¾

            p = ½                          or                                p = ¾

            p – ½ = 0                   or                                p – ¾ = 0

            (p – ½)(p – ¾) = 0  

              \dpi{100} \small 8 ×p^2 - 8 \times \frac{3p}{4}- 8 \times \frac{p}{2}+ 8 \times \frac{3}{8}= 8 \times 0         

**4 divides 8 to give 2, 2 divides 8 to give 4, 8 divides 8 to give 1

            8p2 – 2 x 3p – 4 x p + 1 x 3 = 0

            8p2 – 6p – 4p + 3 = 0

            8p2 – 10p + 3 = 0

 

 

3.         – 1 and – 7

            e = -1                          or                    e = -7

** move -1 and – 7 to join e at the left side, - will change to +

            e + 1 = 0                    or                    e + 7 = 0

            (e + 1)(e + 7) = 0

            e2 + 7e + e + 7 = 0

            e2 + 8e + 7 = 0  

 

4.        \dpi{100} \small \frac{4}{5} and –5 

         

            \dpi{100} \small k = \frac{4}{5}                        or                    k = –5

           \dpi{100} \small k - \frac{4}{5} = 0                   or                    k + 5 = 0

           (k – \dpi{100} \small \frac{4}{5})( k + 5) = 0  

            k2 + 5k – \dpi{100} \small \frac{4k}{5} – 4 = 0 

           

** multiply through with 5

            5 x k2 + 5 x 5k – \dpi{100} \small 5 \times \frac{4k}{5} – 5 x 4 = 0

            5k2 + 25k – 4k – 20 = 0

            5k2 + 21k – 20 = 0

5.    

      \dpi{100} \small -2\frac{1}{3} \ and \ 1\frac{5}{6} \\ x = -2\frac{1}{3} \ \ \ \ \ or \ \ \ \ \ \ \ x = 1\frac{5}{6} \\ x + 2\frac{1}{3} = 0 \ \ \ or \ \ \ \ x - 1\frac{5}{6} = 0 \\ \left ( x + 2\frac{1}{3}\right )\left ( x - 1\frac{5}{6} \right ) = 0 \\ x^2 - \frac{11x}{6} + \frac{7x}{3} - \frac{77}{18} = 0 \\ ** clear fraction \\ \\ 18 \times x^2 - 18 \times \frac{11x}{6} + 18 \times \frac{7x}{3} - 18 \times \frac{77}{18} = 18 \times 0 \\ \\ 18x^2 - 3 \times 11x + 6 \times 7x - 1 \times 77 = 0 \\ 18x^2 - 33x + 42x - 77 = 0 \\ \\ 18x^2 - 9x - 77 = 0 \\

      

6.        

6,        1.25 and 1.6

** change decimals to fraction  

            1.25                                        1.6

            125/100                               16/10

            5/4                                         8/5

            t = 5/4                        or                    t = 8/5

            t – 5/4 = 0                 or                    t – 8/5 = 0   

            \dpi{100} \left ( t - \frac{5}{4} \right )\left ( t-\frac{8}{5} \right ) = 0 \\ \\ t^2 - \frac{8t}{5} - \frac{5t}{4} + 2 = 0 \\ \\ ** clear \ fraction \ with \ 20 \\ \\ 20 \times t^2 - 20 \times \frac{8t}{5} - 20\times \frac{5t}{4} + 20 \times 2 = 20 \times 0 

             20t2 – 4 x 8t – 5 x 5t + 40 = 0

             20t2 – 32t – 25t + 40 = 0 

           20t2 – 57t + 40 = 0  

 

7,        0 and –3

          n = 0               or                    n = –3

            n = 0               or                    n + 3 = 0

            n(n+3) = 0

            n2 + 3n = 0

 

8,        –7 and 7

            g = –7             or                    g = 7

            g + 7 = 0        or                    g – 7 = 0

            (g + 7)(g – 7) = 0

            g2 – 7g + 7g – 49 = 0

                        note that –7g + 7g = 0

            g2 – 49 = 0

 

Exercise

Find the quadratic equations with the following roots

1.    3 and – 4

2.    –2/3 and 5/6

3.    2 ½  and –4⅓

4.    - 8 and –4

5.    5 and 1

6.    0 and 8

7.    –4/9 and 5/6

8.    –2⅖ and 0

9.     +5 and –5

10.   4 and –1

11.   0.3 and 0.1

12.    1.25 and –0.5

13.    –2.2 and –1.3 

14.    1.1 and 0.6

15.    +12 and –12

16.   –4/5 and 4/5   

 

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Case 3: when the sum is positive and product is negative

 

In this case, we look for the two pairs that we can subtract to get the sum. The smaller of the two factors will take a negative sign.

 

Example

Factorize the following

     1.   y2 + y - 20

     2.   6c2 + 13c – 8

     3.   15 + 4m – 3m2

 

Solution

      1.   y2 + y – 20

            a = 1 b = 1 c = -20

            product = - 20

            sum = 1

20 = 1, 2, 4, 5, 10, 20      

(1 x 20)        (2 x 10)        (4 x 5)

 

4 and 5 will subtract to give 1. 4 will take the negative sign

 

factors are – 4 and 5.

 

y2 + y – 20

y2 – 4y + 5y – 20

(y2 – 4y) + (5y – 20)

y(y – 4) + 5(y – 4)

(y – 4)( y + 5)

 

      2.   6c2 + 13c – 8

            a = 6 b = 13           c = - 8

            product = - 48

            sum = 13

48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

(1 x 48)        (2 x 24)        (3 x 16)        (4 x 12)        (6 x 8)

 

Factors are – 3 and 16

6c2 + 13c – 8

6c2 – 3c + 16c – 8

(6c2 – 3c) + (16c – 8)

3c(2c – 1) + 8(2c – 1)

(2c – 1)(3c + 8)

 

      3.  15 + 4m – 3m2

            Product = - 45

            Sum = 4

45 = 1, 3, 5, 9, 15, 45

(1 x 45)        (3 x 15)        (5 x 9)

 

Factors are – 5 and 9  

 

15 + 4m – 3m2

15 – 5m + 9m – 3m2

(15 – 5m) + (9m – 3m2)

5(3 – m) + 3m(3 – m)

(3 – m)(5 + 3m)

 

            Exercise C

Factorize the following

      1.   y2 + 2y – 35

      2.   5p2 + 9p – 18

      3.   3k2 + 14k – 24

      4.   55 + 6t – t2

      5.   6 + 17p – 3p2

      6.   6d2 + 5d – 25

      7.   2a2 + 15a – 19

      8.   m2 + 2m – 8

 

Case 4: when both the sum and the product are negative.

This is just like case 3, only that the negative sign goes to the bigger factor.

 

            Example

Factorize the following

      1.   k2 – k – 6

      2.   5α2 – 8α – 21

      3.   12 – 5x – 3x2

 

Solution

      1.   k2 – k – 6

 

Product = - 6

Sum = -1

6 = 1, 2, 3, 6

(1 x 6)           (2 x 3)

 

2 and 3 will subtract to give 1. And 3 (the larger factor) will take the negative sign

 

Factors are 2 and – 3

k2 – k – 6

k2 +2k – 3k – 6

(k2 +2k) – (3k + 6)

k(k + 2) – 3(k + 2)

(k + 2)(k – 3)

 

      2.  2 – 8α – 21

Solution

Product = -105

Sum = -8

105 = 1, 2, 3, 5, 7, 15, 21, 35, 75, 105

(1 x 105)     (2 x 75)        (3 x 35)        (5 x 21)        (7 x 15)

Factors are 7 and – 15

2 – 8α – 21

2 + 7α – 15α – 21

(5α2 + 7α) – (15α + 21)

α(5α + 7) – 3(5α + 7)

(5α + 7)( α – 3)

 

      3.   12 – 5x – 3x2

 

Solution

Product = - 36

Sum = - 5

36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

(1 x 36)        (2 x 18)        (3 x 12)        (4 x 9)          

 

Factors are 4 and – 9

12 – 5x – 3x2

12 + 4x – 9x – 3x2

(12 + 4x) – (9x + 3x2)

4(3 + x) – 3x(3 + x)

(3 + x)(4 – 3x)

           

 

            Exercise D

Factorize the following

      1.   2m2 – 15m + 7

      2.   5y2 – 11y – 12

      3.   p2 – p – 6

      4.   20 + x - x2

      5.   3y2 – 4y – 15

      6.   k2 – 3k – 18

      7.   3t2 – 7t – 12

      8.   2d2 – d – 6

 

 

Case 5: when there are two unknowns in the expression as in the examples below.

 

Factorize the following

      1.   3a2 – 11ab + 6b2

      2.   8c2 – 7cd – d2

 

    Solution

 

      1.   3a2 – 11ab + 6b2

            a = 3              b = - 11       c = 6

            product = 18

            sum = - 11

18 = 1, 2, 3, 6, 9, 18

(1 x 18)        (2 x 9)           (3 x 6)

            Factors are – 2 and – 9

3a2 – 2ab – 9ab + 6b2

(3a2 – 2ab) – (9ab – 6b2)

a(3a – 2b) – 3b(3a – 2b)

(3a – 2b)(a – 3b)

 

The process involved is still the same as the ones we have done so far.

 

 

      2.   8c2 – 7cd – d2

a = - 1                       b = -7            c = 8

product = - 8

sum = - 7

8 = 1, 2, 4, 8

(1 x 8)           (2 x 4)

Factors are 1 and – 8

8c2 – 7cd – d2

8c2 + cd – 8cd – d2
(8c2 + cd) – (8cd + d2)
c(8c + d) – d(8c + d)

(8c + d)(c – d)

 

            Exercise E

Factorize the following

      1.   x2 – 2xy + y2

      2.   5p2 – 23pq + 12q2

      3.   2m2 – 11mn + 15n2

      4.   j2 – jk – 6k

           

Case 6: When there is a common factor to all the terms in the expression.

            We first factorize the common factor before proceeding to factorize.

 

Factorize the following:

1.        6x2 – 9x – 27

2.        60 + 20v – 5v2

 

            Solution

1.       6x2 – 9x – 27

          3 is a common factor

            3(2x2 – 3x - 9)

                        Factorize 2x2 – 3x – 9

                                    a = 2              b = -3            c = -9

                                    product = - 18

                                    sum = -3

                                    18 = 1, 2, 3, 6, 9, 18

                                    (1 x18)         (2x9)            (3 x 6)

                                    Factors are 3 and – 6

                        2x2 – 3x - 9

                        2x2 + 3x – 6x – 9

                        (2x2 + 3x) – (6x + 9)

                        x(2x + 3) – 3(2x +3)

(2x +3)(x – 3)

Now place the 3 in front of the factorized expression

            3(2x +3)(x – 3)

 

2.      60 + 20v – 5v2 

          5 is a common factor

            5 (12 + 4v – v2)

            Factorize 12 + 4v – v2

                        a = -1            b = 4             c = 12

                        product = - 12

                        sum = 4

                        12 = 1, 2, 3, 4, 6, 12

                        (1 x 12)        (2 x 6)           (3 x 4)

                        Factors are – 2 and 6

                        12 + 4v – v2

                        12 – 2v + 6v – v2 

                        (12 – 2v) + (6v – v2)

                        2(6 – v) + v (6 – v)

                         (6 – v)(2 +v)

            5(6 – v)(2 +v)

 

                   Exercise F

Factorize

1.        9t2 – 15t – 6

2.        10a – 8a2 +12

3.        45 + 12d – 9d2

4.        8x2 – 4x – 60

 

Case 7: When c is absent.

            We factorize directly by looking for what is common.

 

Factorize the following:

      1.   5r2 – 35r

      2.   3p – 4p2

      3.   16t – 12t2

 

              Solution

1.      5r2 – 35r

            5r is common

            5r (r – 7)

 

2.      3p – 4p2

          p (3 – 4p)

 

3.       16t – 12t2

          4t (4 – 3t)

           

          Exercise G

Factorize the following

1.        k2 – 3k

2.        72y – 48y2

3.        6e2 – 15e

4.         18y2 – 40y

5.        12k2 – 21k

6.         7u – 8u2

 

 

Case 8: When b is absent and c is a -ve number.

             In this case use difference of two squares to factorize the expressions.

 

            Difference of two squares

Difference of two squares is used to factorize expressions such as a2 – b2, x2 – 25, 25p2 – 49 etc. Take note of the minus between the two expressions.

To factorize such expressions e.g. a2 – b2 we apply difference of two squares. This will give (a – b)(a + b).

x2 – y2 = (x – y)(x + y)

k2 – 4

= k2 – 22

= (k – 2)(k + 2)

 

c2 – 1

= c2 – 12

= (c – 1)(c + 1)

 

Note: we can only apply difference of two squares when c is a negative number. we can not apply difference of two squares to expressions such as a2 + b2, x2 + 25, 25p2 + 49 etc.

 

            Examples

Factorize the following

      1.   k2 – 36

      2.   81a2b2 – 400c2

      3.   36p2 – 16q2

 

                     Solution     

      1.   k2 – 36

            k2 – 62

            (k – 6)(k + 6)

 

      2.   81a2b2 – 400c2

            92 a2b2 – 202c2

            (9ab)2 – (20c)2

            (9ab – 20c)(9ab – 20c)

 

      3.   3c2 – 12

            3 is common, so factorize first

            3(c2 – 4)

            3(c2 – 22)

            3(c – 2)(c + 2)

 

         4.   36p2 – 16q2

Always factorize when there are common factors before applying difference of two squares.

            4(9p2 – 4q2)

            4(32p2 – 22q2)

            4(3p – 2q)(3p – 2q)

 

                        Exercise H

Factorize the following

      1.   x2 – 49

      2.   m2 – 169

      3.   25p2 – 64

      4.   5 – 75a2b2

      5.   28f2 – 36k2

 

                        Exercise I

Factorize the following

      1.   15y2 – y – 6

      2.   7e2 + 24e – 14

      3.   9d2 – 5d + 6

      4.   5m2 + 18m + 9

      5.   100j2k2 – 441r2

      6.   12 + 4x – x2

      7.   15α2 – αβ – β2

      8.   b2 – 121

      9.   12 – 5x – 3x2

      10.   7 + 8u + u2

      11.   12 – 5x – 3x2

      12.   16e2 – 225

      13.   θ2 – 16θ + 15

      14.   y2 – 2yz – 8z2

      15.   24 + 7c – 5c2

      16.   b2 – 10b + 25

      17.   3g2 + 38g – 13

      18.   20 – t – t2

      19.   3 + 2f – f2

      20.   17 + 16r – r2

      21.   18 – 3d - d2

      22.   c2 – 21c + 20

      23.   27 – 12n – 4n2

      24.   7a2 – 18a + 8

      25.   5 + 8j – 4j2

      26.   8 – 7q – q2

      27.   256x2y2z2 – 144

      28.   s2 + 30s + 225

      29.   4n2 – 12n – 27

      30.   9p2 – 23p + 10

      31.   6q2 – 4q

      32.   1 – 2500t2    

      33.   15a2 – 16a – 7

      34.   12p2 – 4p – 8

 

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QUADRATIC EQUATION

Quadratic equations are equations in this form, x2 + 6x + 9 = 0; x2  – y2 = 0; 2p2 – 11pq + 15q2; 36 – y2 etc., the highest power of the unknowns in these equations is 2. Generally, a quadratic expression is in the form ax2 + bx + c. Notice there is no equality sign while a quadratic equation has the form ax2 + bx + c = 0.

            The coefficient of the unknown with power 2 is a while that one with power 1 is b and the constant is c. When the expression is in this form:

ap2bpq + cq2

ax2bxy + cy2

Here, we have two unknown p and q. The coefficient of the first p2 is a, that of pq is b and q2 is c.

 

                        FACTORIZING QUADRATIC EXPRESSIONS

 

Factorizing quadratic equation involves looking for two numbers which add up to give the sum and multiply to give the product.

The sum is b while the product is a x c. The signs of the factors depend on the signs of the product and sum. 

We will now look at the different forms of a quadratic equation and how to factorize them

        

Case 1: When both the sum and the product are positive.

 

Factorize the quadratic expressions below:

      1.     t2 + 9t + 20

      2.     3p2 + 14p + 8

      3.     2y2 + 13y + 20

      4.     18 + 13d + 2d2

 

Solution

Follow the steps below to factorize 

 

Step 1

Write the product and sum for the equation. The product is a x c while the sum is b

Product = a x c

Sum = b

 

From t2 + 9t + 20,

a = 1 (coefficient of t2)

b = +9 (coefficient of t)

c = +20 (the constant)

 

Product = a x c

              = 1 x 20

              = 20

Sum = +9

 

Step 2

        Look for the factors of the product.

       20 = 1, 2, 4, 5, 10, 20

       pair the factors in two so that they multiply to give 20.

            ** 1 and 20

            ** 2 and 10

            ** 4 and 5

                        There are three pairs formed

Step 3

            Pick the pairs that add up to give our sum (+9). Only 4 and 5 satisfy this. Therefore, we write:

                      Factors are 4 and 5    

We will now replace the 9t in the question with 4t and 5t          

 

            t2 + 9t + 20                               replace 9t with 4t and 5t

            t2 + 4t + 5t + 20                       bracket in twos

            (t2 + 4t) + (5t + 20)                  factorize each bracket

            t(t+4) + 5(t+4)            

(t+4) is common to the two expressions, we factorize again.

            (t+4)(t+5).

The above is the result obtained when t2 + 9t + 20 is factorized.

 

Once you grab the above steps, you will be able to factorize easily without passing through the long steps above.

 

        2.    3p2 + 14p + 8

 

           Solution

a = 3, b = 14, c = 8

product = ac = 3 x 8 = 24

sum = 14

 

** factors of 24

24 = 1, 2, 3, 4, 6, 8, 12, 24

** pair the factors

            (1 x 24)            (2 x 12)            (3 x 8)             (4 x 6)

** Of the four pairs above, only 2 and 12 add to give 14

 

            Therefore, factors are 2 and 12

** we will now replace 14p in the original expression with 2p and 12p

           

            3p2 + 14p + 8

            3p2 + 2p + 12p + 8

** bracket them in pair of twos

            (3p2 + 2p) + (12p + 8)

** factorize each bracket

            p(3p + 2) + 4(3p + 2)

** factorize again

            (3p + 2)(p + 4)

The above is the result obtained when 3p2 + 14p + 8 is factorized.

 

       3.    2y2 + 13y + 20

     Solution

 

    Product = 2 x 20 = 40

    Sum = 13

1, 2, 5, 8, 20, 40         

(1 x 40)            (2 x 20)            (5 x 8)

 

Factors are 5 and 8

            2y2 + 13y + 20

            2y2 + 5y + 8y + 20

            (2y2 + 5y) + (8y + 20)

y is common to the first bracket and 4 to the second

            y(2y + 5) + 4(2y + 5)

factorize again

            (2y + 5)(y+4)

 

        4.    18 + 13d + 2d2

don't be confuse with c coming before a and b.

 

Solution

a = 2, b = 13, c = 18

Product = 2 x 18 = 36

Sum = 13

 

36 = 1, 2, 3, 4, 9, 12, 18,36

(1 x 36)            (2 x 18)            (3x12)              (4 x 9)

 

Factors are 4 and 9

 

18 + 13d + 2d2

18 + 4d + 9d + 2d2

(18 + 4d) + (9d + 2d2)

2(9 + 2d) + d(9 + 2d)

(9 + 2d)(2 + d)

 

            Exercise A

Factorize the following

      1.      t2 + 16t + 63

      2.      3m2 + 11m + 6

      3.      e2 + 10e + 25

      4.      4x2 + 13m + 10

      5.      6 + 5f + f2

      6.      8 + 10v + 3v2

      7.      12 + 20y + 7y2

 

Case2: when the sum is negative and the product is positive.

            This is just like case 1, only that we now add minus (–) to the two factors

 

Factorize the following

    1.   n2 – 7n + 6

    2.   3e2 – 20e + 28

    3.   12a2 – 19a + 5

    4.   p2 – 10p – 25

 

Solution

     1.   n2 – 7n + 6

a = 1, b = - 7, c = 6

product = 6

sum = - 7

factors = 1, 2, 3, 6

            (1 x 6)              (2 x 3)             

1 and 6 will add to give 7.

 

∴ Factors are – 1 and – 6.

n2 – 7n + 6

n2 – n – 6n + 6

(n2 – n) – (6n – 6)

 

Notice that + 6 has changed to – 6. This is because of the – outside the bracket.

Anytime you open a bracket with – outside the bracket, the signs in the bracket will change, + will become – and – will change to +.

 

n(n – 1) – 6(n – 1)

(n – 1)(n – 6)

 

    2.   3e2 – 20e + 28

            Solution

a = 3    b = -20             c = 28

 

product = 84

sum = - 20

 

84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

(1 x 84)            (2 x 42)            (3 x 28)            (4 x 21)            (6 x 14)            (7 x 12)

 

6 and 14 add to give 20

 

∴ factors are – 6 and – 20

3e2 – 20e + 28

3e2 – 6e – 14e + 28

(3e2 – 6e) – (14e – 28)

3e(e – 2) – 14(e – 2)

(e – 2)(3e – 14)

 

    3.   12a2 – 19a + 5

            Solution

a = 12              b = – 19           c = 5

product = 60

sum = – 16

60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

(1 x 60)            (2 x 30)            (3 x 20)            (4 x 15)            (5 x 12)            (6 x 10)

 

Factors are – 4 and – 15

12a2 – 19a + 5

12a2 – 4a – 15a + 5

(12a2 – 4a) – (15a – 5)

4a(3a – 1) – 5(3a – 1)

(3a – 1)(4a – 5)

 

    4.    p2 – 10p + 25

                Solution

            a = 1                b = -10             c = -25

            product = 1 x 25 = 25

            sum = - 10

            factors of 25 = 1, 5, 25

(1 x 25)             (5 x 5)

            Factors are – 5 and – 5

            p2 – 5p – 5p + 25

            (p2 – 5p) – (5p – 25)   

            p(p – 5) – 5(p – 5)

            (p – 5)(p – 5) 

 

            Exercise B

Factorize the following

   1.   3p2 – 17p + 10

   2.   5y2 – 18y + 9

   3.   m2 – 6m + 9

   4.   9k2 – 18k + 8

   5.   7x2 – 19x + 10

   6.   2α2 - 13α + 18

   7.   t2 – 18t +45  

 

NEXT: Factorizing quadratic expressions continuation

 

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MULTIPLE ANGLES

If sin (A + B) = sinA cosB + sinB cosA

Then,

sin (A + A) = sinA cosA + sinA cosA

Notice that we have replaced the B with A  

sin (A + A) = 2 sinA cosA

          A + A = 2A

sin 2A = 2 sinA cosA

 

also, from

cos (A + B) = cosA cosB – sinA sinB

cos (A + A) = cosA cosA – sinA sinA

cos (A + A) = cos2A – sin2A

cos 2A = cos2A – sin2A ---------------------------------------- (equation J)

          but

sin2A + cos2a = 1

sin2A = 1 – cos2a

            substitute 1 – cos2A for sin2A in (J

cos 2A = cos2A – sin2A

cos 2A = cos2A – (1 – cos2A)

cos 2A = cos2A – 1 + cos2A

cos 2A = cos2A + cos2A – 1

cos2A = 2cos2A – 1

 

we can also make cos2a the subject of sin2A + cos2a = 1 to give cos2a = 1 – sin2A.

 

substitute 1 – sin2A for cos2a in (J)

 

cos 2A = cos2A – sin2A

cos 2A = 1 – sin2A – sin2A

cos 2A = 1 – 2sin2A

 

In summary,

          sin 2A = 2 sinA cosA

          cos 2A = cos2A – sin2A

          cos 2A = 2cos2A – 1

          cos 2A = 1 – 2sin2A

 

Same goes for tan.

      if \ tan (A+B) = \frac{tan A + tan B}{1 - tanA \ tanB} \ then \\ \\ tan (A+A) = \frac{tan A + tan A}{1 - tanA \ tanA} \\ \\ tan (2A) = \frac{2tan A}{1 - tan^2A} \\ \\

the above proofs also applies to half angles 

sin (A/2 + A/2) = 2 sin (A/2) cos (A/2) 

sin A = 2 sin (A/2) cos (A/2) 

             Note that: A/2 + A/2 = A 

cos A = cos2(A/2) – sin2(A/2)          or       cosA = cos2(A/2) – 1          or        cosA = 1 – 2sin2(A/2) 

 

                           Note:   sinA x sinA = sin2A

                                      sin2A x sinA = sin3A

                                      sin2A x sin3A = sin5A

                                      sin4A x sin2A = sin6A

                      Add the powers.

 

    Examples

1.   Find the values of the following  (a) sin 3A    (b) cos 4A  

 

Solution

a.         Sin 3A

= sin (2A + A)

= sin 2A cos A + sin A cos 2A

            Sin 2A = 2 sin A cos A           and      cos 2A = cos2A – sin2A

= (2sin A cos A) cos A + sin A (cos2A – sin2A)

= 2 sinA cos2A + sin A (cos2A – sin2A)

            From sin2A + cos2A = 1,

             cos2A = 1 - sin2A

= 2 sin A (1 - sin2A) + sin A (1 - sin2A – sin2A)

=2 sin A (1 - sin2A) + sin A (1 – 2sin2A)

= 2 sin A – 2 sin3A + sin A – 2 sin3A

= 2 sin A + sin A – 2 sin3A – 2 sin3A

= 3 sin A – 4 sin3A

 

b.         cos 4A

= cos (2A + 2A)

= cos2 2A – sin2 2A

= (cos 2A)2 – (sin 2A)2

= (2 cos2A – 1)2 – (2 sin A cos A) 2

= (2 cos2A – 1) (2 cos2A – 1) – (2 sin A cos A) (2 sin A cos A)

= 4cos4A – 4cos2A + 1 – 4 sin2A cos2A

= 4cos4A – 4cos2A + 1 – 4 (1 – cos2A) cos2A

= 4cos4A – 4cos2A + 1 – 4 cos2A (1 – cos2A)

= 4cos4A – 4cos2A + 1 – 4 cos2A + 4 cos4A

= 4cos4A + 4 cos4A – 4cos2A – 4cos2A + 1

= 8 cos4A – 8cos2A + 1

 

2.         Find without using tables the value of sin 2P when sin P = 3/5.

 

            Solution

Sin 2P = 2 sin P cos P

First, We need to get cos p from sin P = 3/5

sin2 P + cos2 P = 1

(3/5) 2 + cos2 P = 1

9/25 + cos2 P = 1

cos2 P = 1 – 9/25

cos2 P = 16/25

cos P = 4/5

sin 2P = 2 sin P cos P

            = 2 x (3/5) x (4/5)

            = 24/25.  

 

3.         Find without using tables the value of cos 2θ when tan θ = 2/3.

           

Solution

cos 2θ = 2cos2θ – 1

          from tan2θ = 2/3, get cos2θ.

          sin2θ + cos2θ = 1

          dividing through by cos2θ

          tan2θ + 1 = 1/cos2θ

          (2/3)2 + 1 = 1/cos2θ

          4/9 + 1 = 1/cos2θ

          13/9 = 1/cos2θ

          9/13 = cos2θ

cos 2θ = 2cos2θ – 1

          = 2 (9/13) – 1

          = (18/13) – 1

          = 5/13.

 

4.       If cos 2α = 9/49, find without using tables the value of

(a) sin α       (b) cos α      (c) tan α

 

Solution

a          cos 2α = 1 – 2sin2α

            9/49 = 1 – 2sin2α

            2sin2α = 1 – 9/49

            2sin2α = 40/49

            Divide through by 2

            sin2α = 20/49

            \sin \alpha = \sqrt{\frac{20}{49}} \ \ \ \ = \ \frac{\sqrt{20}}{7} \ or \ \frac{2\sqrt{5}}{7}

            sinα = √20/7        or        2√5/7

 

b          cos α

            cos 2α = 9/49

            2cos2 α – 1 = 9/49

          2cos2 α = 9/49 + 1

          2cos2 α = 58/49

          cos2 α = 29/49 

         \cos \alpha = \sqrt{\frac{29}{49}} \ \ = \ \frac{\sqrt{29}}{7}

         cos α = 5.385/7

         cos α = 0.769   

 

c          tan α

            the easiest way to get tan α is to divide sin α above by cos α

            i.e.

            tan α = sinα/cosα

            tan α = sinα ÷ cosα 

            \tan \alpha = \frac{\sqrt{20}}{7} \div \frac{\sqrt{29}}{7} \\ \\ \tan \alpha = \frac{\sqrt{20}}{7} \times \frac{7}{\sqrt{29}} \\ \\ \tan \alpha = \frac{\sqrt{20}}{\sqrt{29}} \ rationalize \\ \\ \tan \alpha = \frac{\sqrt{20}}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} \\ \\ \tan \alpha = \frac{\sqrt{580}}{29}

           

When the values of sin α and cos α are not initially known, we have to find the tan direct instead of getting it from sin and cos.

 

cos 2α = 9/49

            from sin2α + cos2α = 1

            divide through by cos2α

            tan2 α + 1 = 1/cos2α

                we will now get cos2α from the original question and substitute it in the equation above.

cos 2α = 9/49

2cos2 α – 1 = 9/49

2cos2 α = 9/49 + 1

2cos2 α = 58/49

cos2 α = 29/49

1/cos2 α = 49/29

          The 49/29 above will be substituted for 1/cos2 α in tan2 α + 1 = 1/cos2α.

tan2 α + 1 = 1/cos2α

tan2 α + 1 = 49/29

tan2 α = (49/29) – 1

tan2 α = (49 – 29)/29

tan2 α = 20/29

tanα = √(20/29)

 

          our answer is the same as the first we got from sin and cos. 

 

                                  Exercise

 

1.       Find the value of the following

(a)      cos 3A        (b)     sin 4A

 

2.       Find without using tables the value of

(i) sin 2θ when 

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(ii) cos 2θ when

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(iii) tan 2θ when

      (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

(iv) sin 3θ when 

     (a) sin θ = 3/5  (b) sin θ = √3/2  (c) cos θ = 5/13 (d) tan θ = 2/3

 

3.       If cos 2θ = 18/25, find without using tables the values of

          (a) tan θ       (b) sin θ       (c) cos θ      (d) tan 2θ     (e) sin 2θ 

 

NEXT: Other trigonometry problems

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Example 2

In this example, we will look at how to express product as sum of two trigonometry ratios.

 

Write the following as the sum of two trigonometry ratios.

1.       sin 19x cos 13x         2.    sin11α sin21α          3.    Cos 6p sin 4p          4.    Cos 13α cos 11α 5.    Sin 4x cos 9x

 

Solution

 

  1.  sin 19x cos 13x

From our product formula, sin and cos are connected by either sin X + sin Y or sin X – sin Y. The two when solved correctly will give us the same answer.

 

Using sin X + sin Y

 

sin X + sin Y = 2 sin (X + Y)/2 cos (X – Y)/2

divide both sides by 2

½ (sin X + sin Y) = sin (X + Y)/2 cos (X – Y)/2

 

We will relate our bolded part with the original question.

sin 19x cos 13x = sin (X + Y)/2 cos (X – Y)/2

 

equating sin

 

19x = (X + Y)/2

38x = X + Y --------------------------------- (equation 1)

Equating cos

 

13x = (X – Y)/2

26x = X – Y -------------------------- (equation 2)

 

From (1) and (2) above, eliminate y

X + Y = 38x

X – Y = 26x

2X      = 64x

X        = 64x/2

X        = 32x

 

Substitute 32x for X in (1)

X + Y = 38x

32x + Y = 38x

Y = 38x – 32x

Y = 6x

 

We will now substitute 32x and 6x for X and Y respectively in ½ (sin X + sin Y)

= ½ (sin X + sin Y)

= ½ (sin 32x + sin 6x)

∴ sin 19x cos 13x = ½ (sin 32x + sin 6x)

 

Also, if we decide to use sin X – sin Y

 

            Sin X – sin Y = 2 sin (X – Y)/2 cos (X + Y)/2

Divide both sides by 2

            ½ ( sinX – sinY) = sin (X – Y)/2 cos (X + Y)/2

            Equate the bolded to the original question

            sin (X – Y)/2 cos (X + Y)/2 = sin 19x cos 13x

 

            equate sin

            (X – Y)/2 = 19x

            X – Y = 38x --------------------------------------------------- (1)

             

            Equate cos

            (X + Y)/2 = 13x

            X + Y = 26x --------------------------------------------------- (2)

 

            Eliminate Y from (1) and (2)

            2X   =    64x

            X = 32x

            Substitute 32x for X in (2)

            X + Y = 26x

            32x + Y = 26x

            Y = 26x – 32x

            Y = - 6x

 

Substitute 32x and – 6x for X and Y respectively in ½ (sinX – sinY)

= ½ (sin X – sin Y)

= ½ (sin 32x – sin (-6x))

= ½ (sin 32x + sin 6x)

 

∴ sin 19x cos 13x = ½ (sin 32x + sin 6x)

Our answer is still the same as the first one above.

 

         2.     sin11α sin21α

Product of two sin can be obtained from cos X – cos Y

cos X – cos Y = -2 sin (X+Y)/2 sin (X – Y)/2

Divide both sides by – 2

            - ½ (cos X – cos Y) = sin (X+Y)/2 sin (X – Y)/2

            ½ (cos Y – cos X) = sin (X+Y)/2 sin (X – Y)/2

            Notice how the removal of minus affects the expression in the bracket.

            sin (X+Y)/2 sin (X – Y)/2 = sin11α sin21α

                        from the first sin

            (X+Y)/2 = 11α

            X + Y = 22α ------------------------------------- (1)

                        From second sin

            (X – Y)/2 = 21α

            X – Y = 42α ------------------------------------- (2)

 

            From (1) and (2) eliminate Y

            2X = 64α

            X = 32α

 

            From (1) sub 32α for X

            X + Y = 22α

            32α + Y = 22α

            Y = 22α - 32α

            Y = -10α

 

            Substitute for X and Y in ½ (cos Y – cos X)

            = ½ (cos Y – cos X)

            = ½ (cos (-10x) – cos 32x)

            = ½ (cos 10x – cos 32x)

 

            Note: cos(–x) is same as cos x

 

         3.     Cos 6p sin 4p

As we already know that the product of sin and cos can be obtained from either sin X + sin Y or sin X – sin Y, so I will use the latter to solve this question.

You can also use the former, we will get the same answer.

 

Sin X – sin Y = 2 sin (X – Y)/2 cos (X + Y)/2

Divide both sides by 2

            ½ ( sinX – sinY) = sin (X – Y)/2 cos (X + Y)/2

            Equate the bolded to the original question

            sin (X – Y)/2 cos (X + Y)/2 = Cos 6p sin 4p

          sin (X – Y)/2 cos (X + Y)/2 = sin 4p cos 6p

                   (X – Y)/2 = 4p

                        X – Y = 8p ------------------------------------- (1)

 

                        (X + Y)/2 = 6p

                        X + Y = 12p ------------------------------------ (2)

From (1) and (2) eliminate Y

 

                        2X = 20p

                        X = 10p

Substitute 10p for X in (2)

           

                        X + Y = 12p

                        10p + Y = 12p

                        Y = 12p – 10p

                        Y = 2p

Substitute 10p and 2p for X and Y respectively in ½ ( sinX – sinY)

= ½ (sinX – sinY)

= ½ (sin 10p – sin 2p)

 

∴ Cos 6p sin 4p = ½ (sin 10p – sin 2p)

 

        4.   cos 13α cos 11α

cos X + cos Y = 2 cos (X + Y)/2cos (X – Y)/2

½ (cos X + cos Y) = cos (X + Y)/2cos (X – Y)/2

cos (X + Y)/2 cos (X – Y)/2 = cos 13α cos 11α

 

(X + Y)/2 = 13α

X + Y = 26α ---------------------------------------------------- (1)

 

(X – Y)/2 = 11α

X – Y = 22α ---------------------------------------------------- (2)

 

From (1) and (2) eliminate Y

 

2X = 48α

X = 24α

 

From (1)  

X + Y = 26α

24α + Y = 26α

Y = 26α - 24α

Y = 2α

 

Substitute 24α and 2α for X and Y respectively in ½ (cos X + cos Y)

= ½ (cos X + cos Y)

= ½ (cos 24α + cos 2α)

 

∴ cos 13α cos 11α = ½ (cos 24α + cos 2α)  

              

      Exercise

Express the following as the sum of two trigonometric ratios.

1.  cos 3x cos 12x

2.  sin 13p cos5p

3.  sin 15α sin21α

4.  sin 9x cos 4x

5.  cos 36x sin 20x

6.  sin 27x sin 3x

7.  cos 5a cos 17a

8.  sin 8f cos 15f

 

PREVOUS: Product formula

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  Answer: 

           1.    sinα = 12/13, tanα = 12/5, cosecα = 13/12

we will now solve question 2 together.

 

Solution to number 2

tan B = 2/3

tan2B = 4/9

tan2B + 1 = sec2B

\frac{4}{9} + 1 = sec2B

\frac{4}{9} + \frac{1}{1} = sec2B

lcm is 9

\frac{4 + 9}{9} = sec^2B \\ \frac{13}{9} = sec^2B \\ \\ \sqrt{\frac{13}{9}} = secB

root 9 is 3

\frac{\sqrt{13}}{3} = secB   but cosB = 1/secB

cosB = \frac{1}{\sqrt{13}/3}

\therefore cosB = \frac{3}{\sqrt{13}}

since we already know tanB and cosB, we can easily get sinB using

tanB = sinB/cosB

tanB = \frac{sinB}{cosB} \\ \frac{2}{3} = \frac{sinB}{3/\sqrt{13}} \\ \frac{2}{3} = sinB \div \frac{3}{\sqrt{13}}\\ \frac{2}{3} = \frac{sinB}{1} \div \frac{3}{\sqrt{13}}\\ \\ \frac{2}{3} = \frac{sinB}{1} \times \frac{\sqrt{13}}{3}

\frac{2}{3} = \frac{\sqrt{13} \ sinB}{3}

3 cancel 3 at the denominator

2 = \sqrt{13} \ sinB

divide both sides by \sqrt{13}

\therefore \frac{2}{\sqrt{13}} = sinB.

 

EXAMPLE 

If cosA = 0.8, find the value of sinA and cotA.

Solution

Sin2A + cos2A = 1

Sin2A + (0.8)2 = 1

Sin2A + 0.64 = 1 (collect like terms)

Sin2A = 1 – 0.64

Sin2A = 0.36 (take the square root of both sides)

\sqrt{Sin^2A} \ = \sqrt{0.36}

sinA = 0.6 

cotA = \frac{cosA}{sinA}

cot A = \frac{0.8}{0.6}

cot A = 1.33

we can also get our cot a from

1 + cot2A = cosec2A

1 + cot2A = \frac{1}{sin^2A}           (cosec2A  = \frac{1}{sin^2A} )

1 + cot2A = \frac{1}{0.36}                  (Sin2A = 0.36 before taking the roots of both sides from above)

1 + cot2A = 2.7778

collect like terms

cot2A = 2.7778 - 1

cot2A = 1.7778

take the square root of both sides

\sqrt{cot^2A} = \sqrt{1.7778}

cotA = 1.333 (same as our answer above)

 

               Exercise B

Using trigonometric formulas only to solve these problems

           1.   If cos A = \frac{5}{13}, find tan A and cosec2A

           2.   Given that tan B = \frac{8}{15}, find the value of cos B and cosec B

           3.   What is the value of tan C and cos C given that sin C = \frac{12}{13}

           4.   Given that sec2P = 1/9, find the value of cosP, sin P and cot2P

          5.   If cot x = 3.7, evaluate sin x, cos x and sec x.

          6.   What is the value of tan2 Y and sec Y when Sin Y = 2/5

          7.   If cot θ = 7, find the value of tan θ , sin θ and sec θ.

          8.   Find in terms of k sin θ and cos θ given that tan θ = k

          9.   If cosec α = \frac{1}{\beta } find in terms of sec β and cot β.

        10.   if sec x = \sqrt{5}, what is the value of tan x and cosec x.

 

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                       Product Formula

 

Take a look at the compound angles formulas below:

      sin (A+B) = sinAcosB + sinBcosA

     sin (A - B) = sinAcosB - sinBcosA

     cos (A+B) = cosAcosB - sinAsinB

    cos (A - B) = cosAcosB + sinAsinB

 

we are going to prove some formulas from the above.

 

let A + B = X --------------------- (equation 1)

A - B = Y ------------------------ (equation 2)

 If we add (1) and (2)

 

            A + B + A - B = X + Y

           2A = X + Y

           A = \frac{X + Y}{2}

Also if we subtract (1) - (2)

           A + B - (A - B) = X - Y

           A + B - A + B = X - Y

          2B = X - Y

         B = \frac{X - Y}{2}

 

From equation (1)

             X = A + B

            sin X = sin (A+B)

            sin X = sinAcosB + sinBcosA

  but A = (X+Y)/2 and B = (X-Y)/2 

 

\sin X = sin \frac{X + Y}{2} \cos \frac{X - Y}{2} + \sin \frac{X - Y}{2} \cos \frac{X + Y}{2}

 

Also from (2) 

        Y = A - B 

        sin Y = sin (A - B)

        sin Y = sinAcosB - sinBcosA

        \sin Y = \sin \frac{X+Y}{2} \cos \frac{X-Y}{2} - \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}

 And 

      cos X = cos(A+B)

     cos X = cosAcosB - sinAsinB

     \cos X = \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} - \sin \frac{X+Y}{2} \sin\frac{X-Y}{2}

and lastly

          cos Y = cos (A-B)

         cos Y = cosAcosB + sinAsinB

        \cos Y = \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} + \sin \frac{X+Y}{2} \sin\frac{X-Y}{2} 

 

from the above proofs,

sinX + sinY = [\sin \frac{X + Y}{2} \cos \frac{X - Y}{2} + \sin \frac{X - Y}{2} \cos \frac{X + Y}{2}] + [

\sin \frac{X+Y}{2} \cos \frac{X-Y}{2} - \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}]

 

                                    sinX +sinY = 2sin\frac{X+Y}{2}cos\frac{X-Y}{2} 

 

sinX - sinY = [\sin \frac{X + Y}{2} \cos \frac{X - Y}{2} + \sin \frac{X - Y}{2} \cos \frac{X + Y}{2}] - [

\sin \frac{X+Y}{2} \cos \frac{X-Y}{2} - \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}]

 

sinX - sinY = \sin \frac{X + Y}{2} \cos \frac{X - Y}{2} + \sin \frac{X - Y}{2} \cos \frac{X + Y}{2}

- \sin \frac{X+Y}{2} \cos \frac{X-Y}{2} + \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}

 

                  sinX - sinY = 2sin\frac{X-Y}{2}cos\frac{X+Y}{2}

 

cosX + cosY = \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} - \sin \frac{X+Y}{2} \sin\frac{X-Y}{2} + 

\cos \frac{X+Y}{2} \cos\frac{X-Y}{2} + \sin \frac{X+Y}{2} \sin\frac{X-Y}{2}

 

                cosX + cosY = 2 cos \frac{X+Y}{2} cos \frac{X-Y}{2}

 

cosX - cosY = \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} - \sin \frac{X+Y}{2} \sin\frac{X-Y}{2} - [

\cos \frac{X+Y}{2} \cos\frac{X-Y}{2} + \sin \frac{X+Y}{2} \sin\frac{X-Y}{2}]

 

cos X - cos Y = \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} - \sin \frac{X+Y}{2} \sin\frac{X-Y}{2}

- \cos \frac{X+Y}{2} \cos\frac{X-Y}{2} - \sin \frac{X+Y}{2} \sin\frac{X-Y}{2}

 

                             cosX - cosY = -2 sin \frac{X+Y}{2} sin\frac{X-Y}{2}

 

 

          Example

Express the following as product of trigonometrical ratios

 

           1.  sin 8x + sin 2x

           2.  cos 12x - cos 6x

          3.  sin 10x - sin 18x

          4.  cos 4x + cos 14x

 

        Solution

        1.     sin 8x + sin 2x = 2 sin \frac{X+Y}{2} cos \frac{X-Y}{2}

                                        = 2 sin \frac{8x+2x}{2} cos \frac{8x-2x}{2}

                                        = 2 sin \frac{10x}{2} cos \frac{6x}{2}

                                        = 2 sin 5x cos 3x

 

      2.    cos 12x - cos 6x = - 2 sin (X+Y)/2 sin (X-Y)/2

                                      = -2 sin (12x + 6x)/2  sin (12x - 6x)/2

                                      = -2 sin (18x)/2 sin (6x)/2

                                      = -2 sin 9x sin 3x

 

    3.    sin 10x - sin 18x = 2 sin (X-Y)/2 cos (X+Y)/2

                                     = 2 sin (10x - 18x)/2 cos (10x + 18x)/2

                                     = 2 sin (-8x)/2 cos (28x)/2

                                     = 2 sin (-4x) cos 14x

                                     = - 2 sin 4x cos 14x  

 

4.    cos 4x + cos 14x = 2 cos \frac{X+Y}{2} cos \frac{X-Y}{2}

                                  = 2 cos (4x + 14x)/2 cos (4x - 14x)/2

                                  = 2 cos 18x/2 cos -10x/2

                                 = 2 cos 9x cos -5x   

                                 = 2 cos 9x cos 5x

 

NOTE: unlike sin and tan, the cos of a negative number is same as the positive.. i.e.cos (-p) = cos p, cos (-2x) = cos 2x etc.

 

                    EXERCISE

Write the following as product of trigonometry ratios

 

      1.   sin 21t + sin 15t 

      2.   sin 18θ - sin 12θ 

     3.   cos 10α - cos 4α 

     4.   cos 6x - cos 2x

     5.   sin 3x + sin 5x

     6.   cos 28y - cos 36y

     7.   cos 5x + cos 9x 

     8.   sin 2B - sin 12B

     9.   sin 3p - sin 7p

    10.  cos 2θ - cos 8θ

 

NEXT: Product formula continuation

PRE: Compound angles 

 

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