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

 

 

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