Month: June 2017

  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.

 

NEXT: QUADRANTS AND ANGLES

PREVIOUS: INTRODUCTION

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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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                        COMPOUND ANGLES

Below are the list of compound angles formula needed for this topic

1.       sin (A+B) = sin A cos B + sin B cos A

2.       sin (A – B) = sin A cos B – sin B cos A

3.       cos(A+B) = cos A cos B – sin A sin B

4.       cos(A – B) = cos A cos B + sin A sin B

5.       tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

6.       tan (A – B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

Master them, as we will use them to solve the questions below.

 

                           Example

If cos P = 3/5 and sin Q = 5/13, evaluate

 

1 . sin (P+Q)                         2 . cos (P – Q)                      3 .   tan (P – Q)

                         Solution

We need to get sin P and cos Q first.

** we will get sin P from cos P = 3/5 using

sin2P + cos2P = 1

sin2P + (3/5)2 = 1

sin2P + 9/25 = 1

sin2P = 1 – 9/25

sin2P = 16/25

sin P = √(16/25)

sinP = 4/5

** we also get cos Q from sin Q

sin2Q + cos2Q = 1

(5/13)2 + cos2Q = 1

25/169 + cos2Q = 1

cos2Q = 1 – 25/169

cos2Q = 144/169

cosQ = √(144/169)

cos Q = 12/13

we can now proceed to evaluate

 

1 .  sin (P+Q) = sin P cos Q + sin Q cos P

= 4/5 x 12/13 + 5/13 x 3/5

= 48/65 + 15/65

= 63/65

 

2 .   cos (P – Q) = cos P cos Q + sin P sin Q

= 3/5 x 12/13 + 4/5 x 5/13

= 36/65 + 20/65

= 56/65

 

3 .  tan (P – Q) =

** Since the value of either tan P or tan Q is not given, we calculate it using

Tan P = sin P/ cos P                        or                    tan2P + 1 = sec2P

tan P = (4/5) ÷ (3/5)                                              tan2P + 1 = 1/cos2P

tan P = (4/5) x (5/3)                                               tan2P + 1 = \frac{1}{9/25}

5 cancels 5                                                               tan2P + 1 = 25/9

tan P = 4/3                                                               tan2P = \frac{25}{9} – 1

tan2P = \frac{25 - 9}{9}

tan2P = 16/9

tan P = √(16/9)

tan P = 4/3

** for tan Q, we use

Tan Q = sin Q / cos Q                                 or        1 + cot2Q = cosec2Q

tan Q = (5/13) ÷ (12/13)                                        1 + \frac{1}{\tan^2 Q} = \frac{1}{\sin^2 Q}

tan Q = (5/13) x (13/12)                                         1 + \frac{1}{\tan^2 Q} = \frac{1}{25/169}

tan Q = 5/12                                                           1 + \frac{1}{\tan^2 Q} = \frac{169}{25}

\frac{1}{\tan^2 Q} = \frac{169}{25} - 1

\frac{1}{\tan^2 Q} = \frac{169}{25} - \frac{1}{1}

\frac{1}{\tan^2 Q} = \frac{169 - 25}{25} \\ \\ \frac{1}{\tan^2 Q} = \frac{144}{25} \\ \\ \tan^2 Q = \frac{25}{144} \\ \\ \tan Q = \frac{5}{12}

now that we have our tan P and tan Q, we can now evaluate

tan (P - Q) = \frac{\tan P - \tan Q}{1 + \tan P \tan Q} \\ \\

= \frac{\frac{4}{3} - \frac{5}{12}}{1 + \frac{4}{3} \times \frac{5}{12}} \\ \\ \\ = \frac{\frac{16 - 5}{12}}{1 + \frac{1}{3} \times \frac{5}{3}} \\ \\ \\ = \frac{\frac{16 - 5}{12}}{1 + \frac{1}{3} \times \frac{5}{3}} \\ \\ \\ = \frac{\frac{11}{12}}{1 + \frac{5}{9}} \\ \\ \\

= \frac{\frac{11}{12}}{\frac{14}{9}} \\ \\ = \frac{11}{12} \div \frac{14}{9} \\ \\ = \frac{11}{12} \times \frac{9}{14} \\ \\ = \frac{11}{12} \times \frac{9}{14} \\ \\ = \frac{11}{4} \times \frac{3}{14} \\ \\ = \frac{33}{56}

 

Exercise

1.     If sinα = 7/25 and cos β = 35/37, evaluate the following

a.     sin (α – β)        b.     cos (α + β)           c.  tan (α + β)

2.    Given that cos θ = 8/17 and sin α = 9/41, evaluate

a.    sin (α + θ)         b.     cos (α – θ)             c.    tan (α – θ)

3.    if sin x = 0.8 and sin y = 0.3, find the value of

a.  cos (x – y)           b.   sin (x – y)            c.   tan (x + y)

 

Example 2

Without using tables evaluate the following

1.    cos 75°               2.   Sin 105°              3.   Tan 15°

 

Solution

Without using tables means we should use our special angle values.

I hope you have mastered how to draw the triangles (30° by 60°) and 45°?

 

1.      cos 75°

75° is not a special angle, so we split it to form two special angles of 45° and 30°.

= cos (45° + 30°)

= cos 45° cos 30° – sin 45° sin 30°

= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} \\ \\ = \frac{\sqrt{3}}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}} \\ \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ \\ rationalize \\ \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \times \frac{2 \sqrt{2}}{2 \sqrt{2}} \\ \\ = \frac{2 \sqrt{6} - 2 \sqrt{2}}{4 \times 2}

= \frac {2 (\sqrt{6} - \sqrt{2})}{8} \\ \\ = \frac {\sqrt{6} - \sqrt{2}}{4}

 

2.   Sin 105°

= sin (60 + 45)

= sin 60 cos 45 + sin 45 cos 60

= √3/2 x 1/√2 + 1/√2 x ½

= √3/2√2 + 1/2√2

= (√3 + 1)/2√2

rationalize

= \frac{\sqrt{3} + 1}{2\sqrt{2}} \times \frac{2\sqrt{2}}{2\sqrt{2}} \\ \\ = \frac{2 \sqrt{6} + 2\sqrt{2}}{8} \\ \\ \\ = \frac{\sqrt{6} + \sqrt{2}}{4}

 

3 . tan 15°

= tan (45 – 30)

\small = \frac{\tan 45 - \tan 30}{1 + \tan 45 \tan 30} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \times \frac{1}{\sqrt{3}}} \\ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \\ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \\ = \frac{\sqrt{3} - 1}{\sqrt{3}} \div \frac{\sqrt{3} + 1}{\sqrt{3}} \\ = \frac{\sqrt{3} - 1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3} + 1} \\

\small = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \\ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{3 - 1} \\ = \frac{4 - 2\sqrt{3}}{2} \\ = \frac{2(2 - \sqrt{3})}{2} \\ = 2 -\sqrt{3}

 

                                    Exercise

1 .     Evaluate the following without using tables

a .  sin 15°   b . cos 105°    c . tan 75°   d .  cos 195°

2 .    If tan A = 2/3 and tan B = ¼, find the values of

a . tan (A – B)      b .   sin (A + B)              c . cos (A – B)

 

PRE: TRIGONOMETRICAL EQUATION

NEXT: TRIGONOMETRY PRODUCT FORMULA

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TRIGONOMETRY EQUATIONS

             Example

Solve the equations below for values of y between 0° and 360° 

 

        1.   5sin y + 1 = 3 

                 Solution

 5sin y + 1 = 3

Collect like terms

5sin y = 3 – 1

5sin y = 2

Divide both sides by 5

Sin y = 2/5

Sin y = 0.4

y = sin – 1 0.4 (sin is positive)

y = 23.579°              or        y = 180 – 23.579°

y = 156.421°.

 

        2.   cos2y = 1/25

                        Solution

Cos2y = 1/25 take the square root of both sides

Cos y = ±√1/25

Cos y = ± 1/5

Cos y = ± 0.2

Split the equation

Cos y = 0.2                            or                    cos y = - 0.2

y = cos – 1 0.2                                               y = cos – 1 0.2

y = 78.463°                                                  y = 78.463°

cos is +ve                                                      cos is –ve

y = 360 – 78.463                                        y = 180 – 78.463; y = 101.537°

y = 281.537°                                                 y = 180 + 78.463; y = 258.463°

∴ y = 78.463°, 281.537°, 101.537°, 258.463°

 

        3.    6 + 25tan2x = 27

               Solution

6 + 25tan2x = 27

25tan2x = 27 – 6

25tan2x = 21

tan2x = 21/25

tan2x = 0.84

tan x = ±√0.84

tan x = ± 0.9185

tan x = 0.9185                     or                    tan x = - 0.9185

x = tan – 1 0.9185                                                                                                               

x = 42. 5057°                                                          

tan is positive                                              tan is negative

x = 180 + 42. 5057                                     x = 180 - 42.5057 or x = 360 – 42. 5057

x = 222.5057°                                             x = 137.4943°      or    x = 317.4943°

∴ x = 42. 5057°, 222.5057°, 137.4943°, 317.4943°

 

        4.    5 sin y + 2 = 6 + 11 sin y

                        Solution

5 sin y + 2 = 6 + 11 sin y

Collect like terms

5 sin y – 11 sin y = 6 – 2

– 6 sin y = 4

Sin y = – 4/6

Sin y = – 0.6667

y = sin – 1 0.6667

y = 41.8103°

sin is negative (3rd and 4th quadrant)

y = 180 + 41.8103              or                    y = 360 – 41.8103

y = 221.8103°                                              y = 318.1897°

∴ y = 221.8103°, 318.1897°.

 

            Exercise

Solve the equations below for the values of the angles between 0 and 360°

        1.   6 sin x = 5

        2.   2 + 7cos p = 5

        3.   12 tan2k = 7

        4.   5 sin x + 8 = 7

        5.   8 cos2 y = 3

        6.   7 + 15 cos x = 5

        7.   6 tan2k – 7 = 5

        8.   13 – 5 cos2 p = 9

        9.   28 = 35 – 13 sin2 θ

       10.   8 – 3 cos x = 13 – 10 cos x

       11.   6 tan2 Q – 9 = 3 – 5 tan2 Q

 

              Example 2 

Solve the equations below

        1.    2 tan2P – tan P – 10 = 0

        2.    8 sin k = 3 – 4 sin2k

            Solution

The two equations above are quadratic equations, so we solve them using the various method for solving quadratic equation.

 

    1.    2 tan2P – tan P – 10 = 0                  (see it as if you want to solve 2p2 – p – 10 = 0).

            Factorize

            2tan2p + 4tan p – 5tan p – 10 = 0

            (2tan2p + 4tan p) – (5tan p + 10) = 0

            2 tan p (tan p + 2) – 5 (tan p + 2) = 0

            (tan p + 2)( 2tan p – 5) = 0

            Tan p + 2 = 0                                                or                    2tan – 5 = 0

            Tan p = 0 – 2                                                                        2 tan p = 0 + 5

            tan p = - 2                                                                             2 tan p = 5

            p = tan – 1 2                                                                           tan p = 5/2

            p = 63.4349                                                                         tan p = 2.5

            tan is negative                                                                    p = tan – 1 2.5  (tan is +ve)

            p = 180 – 63. 4349 or 360 – 63. 4349                          p = 68.1986°           

p = 116.5651° or 296.5651°                                                      or p = 180 + 68.1986

                                                                                                            p = 248.1986°

 

∴ p = 116.5651°, 296.5651°, 68.1986°, 248.1986°.

 

 

        2.   8 sin k = 3 – 4 sin2k

            Rearrange

4 sin2k + 8 sin k – 3 = 0                                 (see it as if you want to solve 4k2 + 8 k – 3 = 0)

We cannot solve the above equation using factorization method, hence we apply quadratic formula.

a = 4, b = 8, c = - 3.

 \sin k = \frac{- b \pm \sqrt{b^2 - 4 ac}}{2a} \\ = \frac{- 8 \pm \sqrt{8^2 - 4 \times 4\times -3}}{2\times 4} \\ \\ = \frac{- 8 \pm \sqrt{64 + 48}}{8} \\ \\ = \frac{- 8 \pm \sqrt{112}}{8} \\ \\ = \frac{- 8 \pm 10.583}{8} \\ \\

sink = (- 8 + 10.583)/8                         or                           sink = (- 8 - 10.583)/8

sink = 2.583/8                                                                     sink = - 18.583/8

sink = 0.3229                                                                      sink = - 2.3229

k = sin – 10.3229                                                                  k = sin – 12.3229

k = 18.8384°                                                                        k = 󠇯ꝏ

k = 180 – 18.8384°

k = 161.1616°

 ∴ k = 18.8384°, 161.1616°

 

Exercise

1 . 12cos2y = cos y + 2

2 . 2sin2α + 3sinα = 2

3 . 5sinβ + 6 = 2 sin2β

4 . 3tan2p + 2tanp = 0

5 . 2 + coty = cot2y

6 . 0 = 3sin2α – 7sinα – 5

7 . 15cosx + 2cos2x = – 7

 

            Example 3

1 .  3cosx = cotx

2 . 5tanx = 3sinx

3 . 7secx = 12tanx

 

            Solution

1 .       3cosx = cotx

            \frac {3cosx}{1} = \frac {cosx}{sinx}

Cross multiply

            3 cosx sinx – cos x = 0

            Cosx (3sinx – 1) = 0

            Cos x = 0                                                       or                    3sinx – 1 = 0

            X = cos – 10                                                                            3sinx = 1

            x = 90°                                                                                   sinx = 1/3

                        or                                                                                sinx = 0.333     (sin is ­­­+ve)

             x = 360 – 90                                                                        x = sin – 1 0.333

            x = 270°                                                                                x = 19. 4692°

                                                                                                            or

                                                                                                            x = 180 – 19.4692

                                                                                                            x = 160.5308°

            ∴ x = 90°, 270°, 19. 4692°, 160.5308°

 

2 .       5tanx = 3sinx 

        \frac{5 sin x}{cos x} = \frac {3sinx}{1} 

Cross multiply

            3cosx sinx = 5 sinx

            3cosx sinx – 5 sinx = 0

            Sin x (3cosx – 5) = 0

Sin x = 0                                 or                                            3cosx – 5 = 0

X = sin – 1 0                                                                            3cos x = 5

x = 0                                                                                       cos x = 5/3

or                                                                                            cos x = 1.6667

x = 0 + 180                                                                            x = cos – 1 1.6667

x = 180                                                                                  x = ꝏ

 

∴ x = 0°, 180°

 

3 .       7secx = 12tanx

7(1/cosx) = 12 (sinx/cosx)

7/cosx = 12sinx/cosx

Cross multiply

12 sinx cosx  = 7 cosx

12 sinx cosx – 7 cosx = 0

cosx (12 sinx – 7) = 0

cos x = 0                                            or                                12 sinx – 7 = 0

x = cos – 1 0                                                                           12 sinx = 7

x = 90 or 360 – 90                                                              sinx = 7/12

x = 270                                                                                  sinx = 0.5833

                                                                                                x = sin – 1 0.5833

                                                                                                x = 35.683

                                                                                                x = 144.317

            ∴ x = 90°, 270°, 35.683°, 144.317°

 

Exercise

Solve the equations below for the given angles between 0 and 360

1 .       4 cosec p = 7 cot p

2 .       13 tan x = 2 sec x

3 .       3cos2x = 2 cosx

4 .       8cot2k = 11 cos2k

 

            Example 4

Solve the equations below

1 .       cos2x + sin x + 1 = 0

2 .       2 cot2x – 7 cosec x + 8 = 0

 

            Solution

1 .       cos2x + sin x + 1 = 0

From sin2x + cos2x = 1       (make cos2x the subject)

            cos2x = 1 – sin2x      (substitute 1 – sin2x for cos2x)

1 – sin2x + sin x + 1 = 0

2 – sin2x + sin x = 0

– sin2x + sin x + 2 = 0         (clear the –)

sin2x – sin x – 2 = 0

sin2x + sinx – 2sinx – 2 = 0

(sin2x + sinx) – (2sinx + 2) = 0

Sinx (sinx + 1) – 2(sin x + 1) = 0

(sinx + 1)(sin x – 2 ) = 0

sin x + 1 = 0                                                  or                                sin x – 2 = 0

sin x = – 1                                                                                         sin x = 2

x = sin – 1 – 1                                                                                     x = sin – 1 2

x = 90                                                                                                 x = ꝏ

sin is negative ( 2nd and 3rd quadrants)

x = 180 + 90 or 360 - 90

x = 270°

 

∴ x = 270°

 

2 .       2cot2x – 7cosec x + 8 = 0

            From 1 + cot2x = cosec2x              (make cot2x the subject)

            cot2x = cosec2x – 1

            2 (cosec2x – 1) – 7cosec x + 8 = 0

            2cosec2x – 2 – 7cosec x + 8 = 0

            2cosec2x – 7cosec x + 6 = 0

            2cosec2x – 4cosec x – 3cosec x + 6 = 0

            (2cosec2x – 4cosec x) – (3cosec x – 6) = 0

            2 cosec x (cosec x – 2) – 3(cosec x – 2) = 0

            (cosec x – 2)(2 cosec x – 3) = 0

cosec x – 2 = 0                                 or                                2 cosec x – 3 = 0

cosec x = 0 + 2                                                                     2 cosec x = 3

cosec x = 2                                                                            cosec x = 3/2

since you can’t press cosec directly from your calculator, change it to sin

\frac{1}{\sin x}= \frac{2}{1}                                           or                                    \frac{1}{\sin x} = \frac{3}{2}

Take the reciprocal of both sides

Sin x = ½                                            or                                sin x = 2/3

Sin x = 0.5                                                                             sin x = 0.6667

x = 30°                                                                                               x = 41.81°

x = 150°                                                                                           x = 138.19°

 

            Exercise

Solve the equations below for values of p between 0 and 360°

1 .   sec2p – 3tanp + 1 = 0

2 .   2 cot p + tan p – 3 = 0

3 .   3 sin p = 2 – 2sin2p

4 .   cot p = tan p

5 .   5 cos p = 3

6 .   cot p = 3 – tan p

7 .   sin p = 2cosp

8 .   sec2p + tanp = 3

9 .   5 cos2p + 5sinp = 1

10 .  3cosec p = 4 sin p – 4

11 . 18 cosec p = 3 – 5 cot2p

12 .  5/cotp = 2sec2p – 4

13 .  \cos p + \sin p = \frac{1}{\cos p - \sin p}

 

PRE: SPECIAL ANGLES

NEXT: COMPOUND ANGLES

PRE

NEXT

SPECIAL ANGLES

Special angles are angles whose trigonometry function can be got without using tables or calculator. Example of such angles are 30°, 60° and 45°.

The triangle below will help you to get the sin, cos and tan of these angles. 

From fig 1, if we consider angle 30°, opp =1, adj = √3 and hyp = 2

Sin θ = opp/hyp

Sin 30 = ½

Cos 30 = adj/hyp =  \frac{\sqrt{3}}{2}

Tan 30 = 1/√3.

 

Considering angle 60, opp is √3, adj is 1 and hyp is 2

Sin 60 =  \frac{\sqrt{3}}{2}

Cos 60 = ½

Tan 60 = \frac{\sqrt{3}}{1} 

            = √3.

 

Considering the angle 45 in fig 2, opposite =1, adjacent =1 and hypotenuse = √2.

Sin 45 = 1/√2

Cos 45 = 1/√2

Tan 45 = 1/1

            = 1.

 

Master the two diagrams (fig 1 and fig 2) and how to put the values to each side correctly. This is very important when solving special angles questions.

 

    Example

Without using tables or calculator, find the value of the following.

   1. Cos 60°

   2. Sin 45°

   3. Tan 135°

   4. Cos 240°

   5. Cos 315°

   6. Sin 210°

 

         Solution

Once you see special angles questions telling you not to use tables, then draw your two triangles (fig1 and fig2 above.)

        1. Cos 60 = ½

 

        2. Sin 45 = 1/√2

 

        3. Tan 135

135 is not a special angle, so we will get its second value from the second quad. formula (180 – θ).

135 = (180 – 45)

 The 45 we subtracted to get 135 is the second value and it’s a special angle.

We will therefore look for the value of tan 45. from our fig 2,

Tan 45 = 1.

 

Our original angle 135 is on the second quadrant, tan is negative here. It means our answer will also take –.

∴ tan 135 = – 1.

 

** our final answer above is tan 135 and not 45 since the question is asking for tan 135 and not tan 45. We only used tan 45 to get our answer.

     

          4.  Cos 240°.

240 is a 3rd quadrant angle (180 + θ) and cos is negative here.

= – Cos (180 + 60)

= – cos 60

= – ½.

∴ cos 240 = – ½.

     

          5.  Cos 315°.

= cos (360 – 45)         cos is positive here

= cos 45

= 1/√2

∴ Cos 315 = 1/√2.

 

      6.  Sin 210°

= sin (180 + 30)       3rd quad. Sin is negative

= – sin 30

= – ½

∴ sin 240 = – ½.

 

              Exercise

Without using tables, find the value of the following

   1. Tan 120

   2. Cos 300

   3. Sin 225

   4. Sin 150

   5. Cos 210

   6. Tan 330

   7. Sin 315 + Cos 120

   8. 1 – 2 cos 15

   9. 2 sin 45 cos 60

  10. Tan 210 – sin 210

  11. \frac{\tan 300 + \tan 210}{1 - \tan 300 \tan210 }

   12. Cos2225 – tan2225 

   13. \frac{\sin 60 + \cos 30}{1 - \sin 30 \ tan 45} 

   14. \frac{1}{\cos 315} 

    15. Sin (– 150)

 

            Example 2

Without using table, evaluate the following.

        1. Sin (–240)

        2. Tan (–330)

        3. Cos (-120)

        4.   \frac{\sin ^2 60 + \tan ^2 315}{1 + 2 \sin 135\cos 300} leaving your answer in surd form

           

             SOLUTION

1     Sin (–240)

              = – Sin 240

              = – Sin (180 + 60)

              = – (– Sin 60)

The first – is the initial one from the question, the second – is because 240 belongs to the 3rd quad. And sin is negative here.

             = – (– \frac{\sqrt{3}}{2})

             =   \frac{\sqrt{3}}{2}

 

          2.  Tan (–330)

           = – tan 330

           = – tan (360 – 30)

           = – (– tan 30) 

330 belongs to 4th quad. And tan is negative here, this means

           = – (– 1/√3)

           = 1/√3.

 

          3. Cos (-120)

Unlike in sin and tan, cos of a negative number is same as that of a positive number i.e. cos (-210) = cos 210 and cos (- 150) = cos 150 etc.

               = cos 120

              = cos (180 – 60)        Notice there is no – in front of cos like in sin and tan.

             = – cos 60        the new here is because 120 is 2nd quad.

  

            = – ½

∴ cos 120 = – ½

     

      4.    \frac{\sin ^2 60 + \tan ^2 315}{1 + 2 \sin 135\cos 300} 

        first change all the angles (315, 135, 300) that are not special angles to special angles

            ** NOTE: tan2 315 means (tan 315)2  and sin2 60 means (sin 60)2 

             = \frac{(\sin 60) ^2 + (\tan 360 - 45)^2}{1 + 2 \sin (180 -45) \cos (360 - 60) } \\ \\ \\ = \ \ \ \frac{(\sin 60) ^2 + (- \tan 45)^2}{1 + 2 \sin 45 \cos 60 } \\

note: tan is negative in 4th quad. hence the –

= \frac{(\sin 60) ^2 + (- \tan 45)^2}{1 + 2 \sin 45 \cos 60 } \\ \\ \\ = \frac{(\frac{\sqrt{3}}{2}) ^2 + (-1)^2}{1 + 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} } \\

= \frac{\frac{3}{4} + 1}{1 + \frac{1}{\sqrt{2}} } \\

= \frac{\frac{3 + 4}{4}}{\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\ = \frac{\frac{7}{4}}{\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\ =\frac{7}{4} \div {\frac{\sqrt{2} + 1}{\sqrt{2}} } \\ \\

=\frac{7}{4} \times {\frac{\sqrt{2}}{\sqrt{2} + 1} } \\ \\ =\frac{7\sqrt{2}}{4\sqrt{2} + 4}

rationalise the denominator

= \frac{7\sqrt{2}}{4\sqrt{2} + 4} \times \frac{4 \sqrt{2} - 4}{4\sqrt{2} - 4} \\ \\ = \frac{7\sqrt{2} \times 4\sqrt{2} - 7\sqrt{2} \times 4 }{4\sqrt{2} \times 4\sqrt{2} - 4 \times 4} \\ \\ \\ = \frac{28 \times 2 -28\sqrt{2}}{16 \times 2 - 4 \times 4}

=\frac {28 (2 - \sqrt 2)}{16} \\ \\ =\frac {7 (2 - \sqrt 2)}{4}

 

          Exercise 

Evaluate the following without using tables

      1.   sin (–330)

      2.   tan (–315)

      3.   2 sin (–240)

      4.   cos (– 225)

      5.   tan2 (–300)

      6.   5 sin2 (–210) cos2(–300)

 

PRE: COMPLEMENTARY ANGLES

NEXT: TRIGONOMETRICAL EQUATIONS

PRE

NEXT

COMPLEMENTARY ANGLES

Two angles are complementary if their sum is 90°. Consider the right – angled triangle below,

Since one angle is 90°, the other two will be complementary. i.e. if one is θ the other will be 90 – θ.

** if one is 40° the other will be 50°

** if one is 85° the other will be 05°

** if one is 27° the other will be ……..

From the triangle above, sin θ = a/c . Also from the other angle (90 – θ), cos (90 – θ) = a/c. Since the two are equal, we say sin θ = cos (90 – θ)

Also, cos θ =  and sin (90 – θ) = , i.e. cos θ = sin (90 – θ)

 

      Example

1.   Given that cos 2θ = sin θ, find the value of θ.

 

Solution 

cos 2θ = sin θ

but Sin θ = cos (90 – θ). Substitute cos (90 – θ) for Sin θ.

∴ cos 2θ = cos (90 – θ)

Cos will cancel cos

 

We now have 2θ = 90 – θ

Collect like terms

2θ + θ = 90

3θ = 90

Divide both sides by 3,

θ = 30°.

                         OR

cos 2θ = sin θ

sin (90 - 2θ) = sin θ

90 - 2θ = θ

90 = 2θ + θ

90 = 3θ

30 = θ

θ = 30

2.  Find the value of K if cos K = sin 7K.

Solution

Cos K = sin 7K

Sin (90 – K) = sin 7K

90 – K = 7K

90 = 7K + K

90 = 8K

Divide both sides by 8

90/8 = k

11.25 = K

∴ K = 11.25°.

 

3.   Find the value of P if cot P = tan 14P

 

Solution 

Cot P = tan 14P

Substitute tan (90 – P) for Cot P

Tan (90 – P) = tan 14P

90 – P = 14P

90 = 14P + P

90 = 15P

Divide through by 15

90/15 = P

6 = P

∴ P = 6°.

 

4. Solve for y in the equation below sin3y = cos 2y

Solution

sin3y = cos2y

sin3y = sin (90 - 2y)

sin cancel sin

3y = 90 - 2y

3y + 2y = 90 

5y = 90

y = 18°

 

 

Exercise E 

Find the value of the angles below given that:

1. Cos Q = sin 5Q

2. Tan 7P = cot P

3. Cos 3X = sin X

4. Sin 7K = cos K

5. Tan B = cot 5B

6. Sin 10Q = cos 5Q

7. cos 7t = sin 3t

8. sin 2p = cos 5p

9. cos 6.5x = sin 2.3x

10. cot 8a = tan 5a

 

PRE: ANGLES AND QUADRANTS

NEXT: SPECIAL ANGLES

 

PRE

NEXT

                 Angles and Quadrants
 

From the quadrant diagram above.

1st quadrant is from 0 to 90°, on this quadrant Sin, Cos and Tan are all positive.

2nd quadrant is 90 to 180°, only Sin is positive here while others are negative.

3rd quadrant is 180° to 270°, only tan is positive here.

4th quadrant is 270° to 360°, only cos is positive.

 
 
 

The diagram above shows that angles are measured with respect to the x – axis.

** 1st quad – angle remains the same as it is.

** 2nd quad – subtract the given angle from 180°.

** 3rd – add the given angle to 180°

** 4th – subtract from 360°.

Confused? Relax, you will see the application of all these in the examples below.

 

           Example

 

Find the values of θ between 0° and 360° for which

      1.      Sin θ = 0.345

 

Note:

      1.   0° ≤ x ≤ 360° is same as between 0° and 360°.

      2.   While 0° ≤ x ≤ 180° means between 0° and 180°. Values of x greater than 180 will be discarded here.

 

Solution

1.   sin θ = 0.345

θ = sin1 0.345

Pressing shift + sin 0.345 or 2nd function + sin 0.345 on our calculator gives

θ = 20.182°.

 

Take a look at the question again

 

sin θ = 0.345  

 

0.345 is a positive number, which means sin is positive in that quadrant. In which quadrant is sin positive? 1st and 2nd quadrant. i.e. θ and 180 – θ.

 

 

** θ = 20.182°

 

** 180 – θ

      = 180 – 20.182

       = 159.818°.

 

A quick check. Pick your calculator and press sin 20.182° and sin 159.818°.

 

 

What did you observe? The same answer.

 

∴ if sin θ = 0.345, then θ = 20.182° or 159.818°. 

 

    2.   Find the values of P between 0 and 360° for which

tan P = 1.28

P = tan–1 1.28

P = 52.00°

since tan P is positive at the 1st (P) and 3rd quad (180 + P)

= 180 + 52

= 232°.

∴ P = 52° or 232°.

 

      3.   Find the values of A between 0 and 360° for which

cos A = – 0.8329

Ignore the – sign. It only means cos is negative in that quadrant.

cos A = 0.8329

A = cos – 1 0.8329

A = 33.6°

Since Cos A is negative, then A belongs to 2nd and 3rd quadrants.

2nd quadrant = 180 – A

= 180 – 33.6

= 146.4°. 

 

3rd quadrant = 180 + A

= 180 + 33.6

= 213.6°.

∴ A = 146.4° and 213.6°.

 

You can confirm the above answers on your calculator by pressing cos 146.4° and cos 213.6°. you will get – 0.8329 in both cases.

 

       4. Find the values of θ between 0 and 360° for which

sin θ = – 0.749

sin θ = 0.749

θ = sin – 1 0.749

θ = 48.5°

sin is negative in 3rd and 4th quadrants. 180 + θ and 360 – θ.

180 + 48.5 = 228.5

360 – 48.5 = 311.5

∴ θ = 228.5° and 311.5°

 

         Exercise C 

Find the values of P between 0 and 360 for which

      1. Cos P = 0.377

      2. Tan P = – 1.5527

      3. Sin P = 0.982

      4. Cos P = – 0.848

      5. Tan P = 4

      6. Sin P = – 0.7

Find the values between 0 and 360 of the angles given below.

      7. Tan y = 1.22

      8. Cos k = 0.332

      9. Sin t = - 0.08

     10. Cos q = - 0.0952

 

                Example 2 

      1.  If sin β = - 0.4629 and cos β is negative, what quadrant does β belongs, hence find β.

 

Solution 

From our question above, sin and cos are negative. That is second quadrant.

Sin β = 0.4629

β = sin-1 0.4629

β = 25.574°.

sin and cos are negative in 2nd quad. only, i.e. 180 + β

= 180 + 25.574

= 205.574°.

∴ β = 205.574°.

 

       2.   Find t if tan t = - 2.6 and sin t is negative.

Solution

Tan t = 2.6

t = tan-1 2.6

t = 68.963°. since both tan and sin are –ve, then it is 4th quadrant.

= 360 – t

= 360 – 68.963

= 291.037°.

∴ t = 291.037°.

 

       3.   If cos q = 0.4882 and tan q is positive, find q

Solution

cos q = 0.4882

q = cos-1 0.4882

q = 60.778°.

both cos and tan are positive only on the 1st quadrant, on this quadrant, our angle remains as it is.

∴ q = 60.778°.

 

                       Exercise D

Find the angles below given the conditions stated in front.

      1. A, if cos A = -0.588 and sin A is positive

      2. B, if tan B = 1.7592 and cos B is negative

      3. C, if tan C is negative and sin C = -0.6

      4. D, if cos B = -0.83 and tan B is negative.

      5. K, if tan K = 0.69 and sin K is positive. 

 

PRE: INTRO CONTINUATION

NEXT: COMPLEMENTARY ANGLES

 

PRE   

NEXT

     consider the diagram above.

We all know that

\sin \theta = \frac{opposite}{hypotenuse} \ \ \cos \theta = \frac{adjacent}{hypotenuse} \ \ \ and \ \ \tan \theta = \frac{opposite}{adjacent}  

we will now introduce

- cosec/csc (the reciprocal of sin)

- sec (the reciprocal of cos) 

- cot (the reciprocal of tan)

 

This means cosec \ \theta = \frac{Hypotenuse}{Opposite} \ \ sec \ \theta = \frac{Hypotenuse}{Adjacent} \ \ cot \ \theta = \frac{Adjacent}{Opposite}

 

from our diagram above, \sin \theta = \frac{b}{c} , \csc \theta = \frac{c}{b}       note that csc is the short form of cosec

\cos \theta = \frac{a}{c}, \ \ \sec \theta = \frac{c}{a} \\ \tan \theta = \frac{b}{a}, \ \ \cot \theta = \frac{a}{b}

from the above, we realized that sin and cosec are reciprocal of each other, same with cos and sec, also tan and cot.

 

PROOF 1: To show that sin and cosec are reciprocal of each other. i.e     \sin \theta = \frac{1}{\csc \theta }

Solution

from our above triangle, \ sin \theta = \frac{b}{c} \ \ while\ \csc \theta = \frac{c}{b}

\sin \theta = \frac{1}{\csc \theta }

\frac{b}{c} = \frac{1}{\frac{c}{b}} \\ \frac{b}{c} = 1 \div \frac{c}{b}

at the right side of the equation, if we change divide to X, then c/b will become b/c

\frac{b}{c} = 1 \times \frac{b}{c}\\ \frac{b}{c} = \frac{b}{c}\\ \therefore \sin \theta = \frac{1}{\csc \theta } \ as \ required.

 

Note: Once the expression on the right side of the equation is the same with that of the left, then the proof is correct.

the above proof can also be done for cos and sec, tan and cot.

also if   \sin \Theta = \frac{1}{\csc \Theta }, then

sin ^{2}\Theta = \frac{1}{\csc^{2} \Theta }

\cos^{2} \Theta = \frac{1}{sec^{2}\Theta} \\ \\ \tan^{2} \Theta = \frac{1}{\cot^2 \Theta}

Students can verify the last three proofs on their own.

 

PROOF 2: To show that \tan \Theta = \frac{\sin \Theta }{\cos \Theta }

Solution

We will still make use of our triangle above. where   \sin \Theta = \frac{b}{c}, \cos \Theta = \frac{a}{c}, \tan \Theta = \frac{b}{a}

\tan \Theta = \frac{\sin \Theta }{\cos \Theta }

\frac{b}{a} = \frac{\frac{b}{c}}{\frac{a}{c}} \\ \frac{b}{a} = \frac{b}{c} \div \frac{a}{c} \\ \\ \frac{b}{a} = \frac{b}{c} \times \frac{c}{a} \\ \\ at \ the \ left \ side, \ c \ cancel \ c \\ \\ \frac{b}{a} = \frac{b}{1} \times \frac{1}{a} \\ \\ \frac{b}{a} = \frac{b}{a} \\ \\ \ since \ left \ side \ of \ the \ equation = \ right \ side \\

\therefore \tan \Theta = \frac{\sin \Theta }{\cos \Theta }

 

PROOF 3: Show that   \tan ^2 \alpha = \frac{1}{\cot ^2\alpha } 

Solution

NOTE: we can use β, α, γ, δ, π, ρ, σ, letters A..Z in place of θ

.\tan \alpha = \frac{b}{a}, \ \cot \alpha = \frac{a}{b}

\tan ^2 \alpha = \frac{1}{\cot ^2\alpha } , \tan ^2 \alpha \ means \left ( \tan \alpha \right )^2 = \left ( \frac{b}{a} \right )^2, \ while \cot ^2\alpha = \left ( \frac{a}{b} \right )^2

\left ( \frac{b}{a} \right )^2 = \frac{1}{\left ( \frac{a}{b}\right )^2} \\ \\ \\ \left ( \frac{b}{a} \right )^2 = \left ( \frac{b}{a}\right )^2

Since left side (LS) equals right side (RS) then \tan ^2 \alpha = \frac{1}{\cot ^2\alpha } .

Note: \frac{1}{\left ( \frac{a}{b} \right )^2} = \left ( \frac{b}{a} \right )^2 reciprocal of fractions can be obtained easily by removing the 1 and swapping the numerator and denominator.

** Notice that b initially at the denominator is now at the numerator and a is now at the denominator.

 

                                 Exercise A

Show that:

(1)  \cot \Theta = \frac{\cos \Theta }{\sin \Theta }     

(2)   \sec ^2 \Theta = \frac{1}{\cos ^2 \Theta }   

(3)   \tan^2 \Theta = \frac{\sin^2 \Theta }{\cos^2 \Theta }   

(4)   \cot^2 \Theta = \frac{\cos^2 \Theta }{\sin^2 \Theta }

 

PROOF 4: To show that Sin2θ + Cos2θ = 1

Solution

from our triangle at the top of this page, Sinθ = \frac{b}{c}  and Cosθ = \frac{a}{c}

Sin2θ + Cos2θ = 1

\left ( \frac{b}{c} \right )^{2} + \left ( \frac{a}{c} \right )^{2} = 1 \\ \frac{b^2}{c^2} + \frac{a^2}{c^2} = 1 \\

the lcm is C2  

\frac{b^2 + a^2}{c^2} = 1, applying Pythagoras' theorem to our above right-angled triangle, we have C2 = a2 + b2   

we will substitute, C2 for a2 + b2 in the above equation.

\frac{c^2}{c^2} = 1

C2 will cancel C2 to give 1.

1 = 1.

 

therefore, Sin2θ + Cos2θ = 1.

From the above proof, if we divide both sides by Sin2θ, we have

\frac{\sin ^2\theta }{\sin ^2\theta } \ + \frac{\cos^2\Theta }{\sin ^2\theta } = \frac{1}{\sin ^2\theta }

 

1 + cot2θ = cosec2θ. 

Similarly, if we divide through by  Cos2θ, then

\frac{\sin ^2\Theta }{\cos ^2\Theta } \ + \frac{\cos ^2\Theta }{\cos ^2\Theta } = \frac{1}{\cos ^2\Theta }

 

Tan2 θ + 1 = sec2 θ.

Take note of the above proofs, as we will use them later.

 

Sin2θ + Cos2θ = 1

1 + cot2θ = cosec2 θ.

Tan2 θ + 1 = sec2 θ. 

 

                            EXAMPLES

           1.   Given that sinθ = 4/5, find the value of cosθ and tanθ.

Solution

sinθ = 4/5, cosθ = ?

Sin2θ + Cos2θ = 1

\left ( \frac{4}{5} \right )^2 + \cos ^{2}\Theta = 1 \\ \frac{16}{25} + \cos ^{2}\Theta = 1

make Cos2θ the subject

\cos ^{2}\Theta = 1 - \frac{16}{25}

\cos ^{2}\Theta = \frac{1}{1} - \frac{16}{25}

L.C.M is 25

\cos ^2\Theta = \frac{25 - 16}{25} \\ \cos ^2\Theta = \frac{9}{25} \\ \cos\Theta = \sqrt{\frac{9}{25}} \\ \cos\Theta = \frac{3}{5}

After getting cosθ, we can now get tanθ.

sinθ = 4/5, cosθ = 3/5, tanθ = ?

using tan\Theta = \frac{sin\Theta}{cos\Theta } \\ tan\Theta = \frac{4/5}{3/5} \\ tan\Theta = \frac{4}{5} \div \frac{3}{5} \\ tan\Theta = \frac{4}{5} \times \frac{5}{3}

5 will cancel 5

tan\Theta = \frac{4}{1} \times \frac{1}{3} \\ \therefore tan \Theta = \frac{4}{3}

also we can get tanθ from Tan2 θ + 1 = sec2 θ. 

sec2 θ = 1/Cos2θ

from our above calculation, cos2θ = 9/25

sec2 θ = \frac{1}{9/25}

therefore 25/9.

tan2 θ + 1 = sec2 θ

tan2 θ + 1 = 25/9

tan2 θ  = \frac{25}{9} - 1

tan2 θ = tan^2 \Theta = \frac{25}{9} - \frac{1}{1} \ lcm \ is \ 9 \\ \\ tan^2 \Theta = \frac{25 - 9}{9} \\ \\ tan^2 \Theta = \frac{16}{9} \\ \\ tan \Theta = \sqrt{\frac{16}{9}} \\ \\ tan \Theta = \frac{4}{3}

Our answer is still the same as before, so use any method you like.

try these on your own and check the answer in the next page.

 

        1.    Given that cos α = 5/13, find the value of sin α, tan α and cosec α  

        2.    If tan B = 2/3, find the value of sin B, cos B and cosec B.

                                                                                                               NEXT

               L.C.M and H.C.F of fractions

Just like we found for whole numbers, we can also calculate the L.C.M and H.C.F of fractions. Such as ½, ¾, 0.3, ¼, 2.1 etc.

 

L.C.M = \frac{L.C.M \ of \ numerator \ numbers}{H.C.F \ of \ denominator \ numbers}

H.C.F = \frac{H.C.F \ of \ numerator \ numbers}{L.C.M \ of \ denominator \ numbers}

 

When decimal numbers are involved, change the decimal numbers to fractions and proceed to find the lcm and hcf with the above formulas.

 

            Example

       1.   Find the lcm of 5/8, 3/4 and 9/20.

Solution

Lcm = lcm of numerator numbers/hcf of denominator numbers.

**Numerator numbers are 5, 3, and 9

Their l.c.m = 45.

**denominator numbers are 8, 4 and 20

Their hcf = 4

Lcm = 45/4 or 11.25

 

      2.    Find the hcf of 0.8, 2/9, 0.72 and 8/9

Solution

First change all the decimal numbers to fractions.

0.8 = 8/10 = 4/5 when reduced.

0.72 = 72/100 = 18/25.

Our fractions will be

4/5, 2/9, 18/25 and 8/9.

h.c.f = hcf of numerator numbers/lcm of denominator numbers

            = hcf of 4, 2, 18 and 8 / lcm of 5, 9, 25 and 9

            = 2/225.

 

        EXERCISE

Find the lcm and hcf of the following numbers

        1.  20/27 and 4/9

        2.  225 and 0.6

        3.  18/25, 21/40 and 12/25

        4.  16/25, 0.4 and 12/15.

 

Other problems

           1.    If two alarm clocks are set such that they ring every 12 minutes and 20 minutes. After how many minutes will they ring together?

Solution

The lcm of 12 and 20 will give us when they will ring together.

2

12

20

2

6

10

3

3

5

5

1

5

 

1

1

Lcm = 2 x 2 x 3 x 5

    = 60

Therefore, they will ring together after 60 minutes.

 

            2.    Smart, Joy and Ken usually visit Mama T's cafe every 25 minutes, 30 minutes and 50 minutes respectively. If the three friends met at the cafe at 11:20 am, what time will they meet again?

            Solution

Find the lcm of 25, 30 and 50

Lcm = 150

That means they will meet every 150 minutes, to get the required time, convert the 150 minutes to hours

= 150/60

= 2 hours 30 minutes.

Add 2 hours 30 mins to 11:20 am

            = 13:50

Which is 01:50 pm.

 

        3.   The l.c.m of 12 and another number is 60. If their h.c.f is 4, find the number.

             Solution

Product of the two numbers = product of their l.c.m and h.c.f

i.e. if the two numbers are a and b, then

            a x b = l.c.m x h.c.f

in the above question, a = 12, b = ?, l.c.m = 60, h.c.f = 4.

            12 x b = 60 x 4

            12b = 240

            b = 240/12

            b = 20.

 

                  Practice questions

          1.     After a budget planning consultation Mr. A, B and C decided that they will change their car after every 16 months, 20 months and 30 months respectively. If they all bought their cars together in January 10, 2017, when next will the buy together?

         2.     The hcf of 18 and another number is 6, if their l.c.m is 72, find the number.

         3.     The lcm of 6 and another number is 12. if their h.c.f is 2, find the number.

         4.     Three vigilante men usually blow their whistles every 20, 15 and 25 minutes. If they blow together at 10:00pm, when next will they blow together?

 

 

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