© 2014, John Bird

706

CHAPTER 42 TRIGONOMETRIC IDENTITIES AND EQUATIONS

EXERCISE 175 Page 477

1. Prove the identity: sin x cot x = cos x

L.H.S. = sin x cot x = sin x 1 cossintan sin

xxx x

=

= cos x = R.H.S.

2. Prove the identity: 2

1(1 cos )θ−

= cosec θ

L.H.S. = ( ) 22

1 1sin1 cos θθ

=−

(since 2 2sin cos 1θ θ+ = )

= 1sinθ

= cosec θ = R.H.S.

3. Prove the identity: 2 cos2 A – 1 = cos2 A – sin2 A R.H.S. = ( )2 2 2 2cos sin cos 1 cosA A A A− = − − since 2 2cos sin 1A A+ =

= 2 2 2cos 1 cos 2cos 1A A A− + = − = L.H.S.

4. Prove the identity: 3cos cos

sinx x

x− = sin x cos x

L.H.S. = 3 2 2cos cos cos (1 cos ) cos sin

sin sin sinx x x x x x

x x x− −

= = = cos x sin x = sin x cos x = R.H.S.

5. Prove the identity: (1 + cot θ)2 + (1 – cot θ)2 = 2 cosec2θ L.H.S. = ( ) ( )2 2 2 21 cot 1 cot 1 2cot cot 1 2cot cotθ θ θ θ θ θ+ + − = + + + − +

= ( )2 22 2cot 2 2 cosec 1θ θ+ = + −

= 2 22 2cos 2 2cosecec θ θ+ − = = R.H.S.

© 2014, John Bird

707

6. Prove the identity: 2sin (sec cosec )

cos tanx x x

x x+ = 1 + tan x

L.H.S. = ( )2 2

21 1 sin cossin sinsin sec cosec cos sin cos sin

sincos tan sincoscos

x xx xx x x x x x xxx x xxx

+ + + = =

= sin cos sin cossincos sin cos

x x x xxx x x+ + =

= sin cos tan 1 1 tancos cos

x x x xx x+ = + = +

= R.H.S.

© 2014, John Bird

708

EXERCISE 176 Page 479

1. Solve for angles between 0° and 360°: 4 – 7 sin θ = 0

Since 4 – 7 sin θ = 0 then 4 = 7 sin θ and sin θ = 47

from which, θ = 1 4sin7

−

= 34.85°

Since sine is positive, the angle 34.85° occurs in the 1st and 2nd quadrants as shown in the diagram

below

Hence, the two angles for θ between 0° and 360° whose sine is 47

are:

34.85° and 180° – 34.85° = 145.15° 2. Solve for angles between 0° and 360°: 3 cosec A + 5.5 = 0

Since 3 cosec A + 5.5 = 0 then 3 cosec A = –5.5 and cosec A = 5.53

−

i.e. 1 5.5sin 3A

= − or sin A = 35.5

−

from which, A = 1 3sin5.5

− −

= –33.06°

Since sine is negative, the angle 33.06° occurs in the 3rd and 4th quadrants as shown in the diagram

below

© 2014, John Bird

709

Hence, the two angles for A between 0° and 360° whose sine is 3

5.5− are:

180° + 33.06° = 213.06° and 360° – 33.06° = 326.94° 3. Solve for angles between 0° and 360°: 4(2.32 – 5.4 cot t) = 0

Since 4(2.32 – 5.4 cot t) = 0 then 2.32 – 5.4 cot t = 0 and 2.32 = 5.4 cot t

i.e. cot t = 2.325.4

from which, tan t = 5.42.32

Hence, t = 1 5.4tan 66.752.32

− = °

Since tan is positive, the angle 66.75° occurs in the 1st and 3rd quadrants as shown in the diagram

below

Hence, the two angles for t between 0° and 360° whose tan is 5.4

2.32 are:

66.75° and 180° + 66.75° = 246.75° 4. Solve for θ in the range 0 360θ≤ ≤ : sec 2θ =

Since sec 2θ = then cos θ = 1 0.52=

© 2014, John Bird

710

Hence, θ = 1cos 0.5− = 60° or 300° since cosine is positive in the 1st and 4th

quadrants (check CAST)

5. Solve for θ in the range 0 360θ≤ ≤ : cot 0.6θ =

Since cot 0.6θ = then tan θ = 1 1.666670.6

=

Hence, θ = 1tan 1.66667− = 59° or 239° since tangent is positive in the 1st and 3rd

quadrants (check CAST)

6. Solve for θ in the range 0 360θ° ≤ ≤ ° : cosec θ = 1.5

Since cosec θ = 1.5 then sin θ = 1 0.666671.5

=

Hence, θ = 1sin 0.66667− = 41.81° or 138.19° since sine is positive in the 1st and 2nd

quadrants (check CAST)

7. Solve for x in the range 180 180x− ° ≤ ≤ ° : sec 1.5x = −

Since sec 1.5x = − then cos x = 1 0.666671.5

= −−

Hence, θ = 1cos 0.66667− − = 131.81° or 228.81° since cosine is negative in the 1st and 2nd

quadrants (check CAST)

Hence, in the range 180 180x− ° ≤ ≤ ° , θ = 131.81° or – 131.81° i.e. θ = ±131.81°

8. Solve for x in the range 180 180x− ° ≤ ≤ ° : cot 1.2x =

Since cot 1.2x = then tan x = 11.2

Hence, x = 1 1tan1.2

−

= 39.81° or 210.81° since cosine is positive in the 1st and

4th quadrants (check CAST)

© 2014, John Bird

711

Hence, in the range 180 180x− ° ≤ ≤ ° , x = 39.81° or – 140.19°

9. Solve for x in the range 180 180x− ° ≤ ≤ ° : cosec x = –2

Since cosec x = –2 then sin x = 1 0.52= −

−

Hence, x = ( )1sin 0.5− − = 210° or 330° since sine is negative in the 3rd and

4th quadrants (check CAST)

Hence, in the range 180 180x− ° ≤ ≤ ° , x = –30° or –150°

10. Solve for θ in the range 0 360θ° ≤ ≤ ° : 3sin 2cosθ θ=

Since 3sin 2cosθ θ= then sin 2cos 3

θθ= i.e. tan θ = 2

3

Hence, θ = 1 2tan3

−

= 33.69° or 213.69° since tangent is positive in the 1st and

3rd quadrants (check CAST)

11. Solve for θ in the range 0 360θ° ≤ ≤ ° : 5 cos θ = – sin θ

Since 5 cos θ = – sin θ then sin 5cos

θθ= − i.e. tan θ = –5

Hence, θ = 1tan ( 5)− − = 101.31° or 281.31° since tangent is negative in the 2nd and

4th quadrants (check CAST)

© 2014, John Bird

712

EXERCISE 177 Page 479

1. Solve for angles between 0° and 360°: 5 sin2 y = 3

Since 25sin 3y = then 2 3sin 0.605

y = = and sin 0.60 0.774596...y = = ±

and y = 1sin (0.774596...) 50.77− = ° Since sine y is both positive and negative, a value for y occurs in each of the four quadrants, as shown in the diagram below

Hence the values of y between 0° and 360° are: 50.77°, 180° – 50.77° = 129.23°, 180° + 50.77° = 230.77° and 360° – 50.77° = 309.23° 2. Solve for angles between 0° and 360°: 2cos θ = 0.25

Since 2cos θ = 0.25 then cos 0.25 0.5θ = = ±

and θ = 1cos (0.5) 60− = ° Since cos θ is both positive and negative, a value for θ occurs in each of the four quadrants Hence, θ = 60° or 120° or 240° or 300° 3. Solve for angles between 0° and 360°: 2tan x = 3

Since 2tan x = 3 then tan x = 3 1.7321= ± and x = ( )1tan 1.7321− ± = 60° or 120° or 240° or 300° Since tan x is both positive and negative, a value for x occurs in each of the four quadrants

© 2014, John Bird

713

4. Solve for angles between 0° and 360°: 5 + 3 cosec2 D = 8

Since 25 3cosec 8D+ = then 23cosec 8 5 3D = − = i.e. 2cosec 1D =

Hence, 2

1 1sin D

= and 2sin 1D = from which, sin D = 1 1= ±

and D = ( )1sin 1− ±

There are two values of D between 0° and 360° which satisfy this equation, as shown in the

sinusoidal waveform below

Hence, D = 90° and 270° 5. Solve for angles between 0° and 360°: 2 cot2 θ = 5

Since 2 cot2 θ = 5 then 2 5cot 2.52

θ = =

Hence, 2

1 5tan 2θ

= and 2 2tan 0.45

θ = =

from which, tan θ = 0.4 0.63246= ±

and θ = ( )1tan 0.63246− ± = 32.31° or 147.69° or 212.31° or 327.69° Since tan θ is both positive and negative, a value for θ occurs in each of the four quadrants

© 2014, John Bird

714

EXERCISE 178 Page 480

1. Solve for angles between 0° and 360°: 15 sin2 A + sin A – 2 = 0

Since 15 sin2 A + sin A – 2 = 0 then ( )( )5sin 2 3sin 1 0θ θ+ − =

i.e. 5 sin A + 2 = 0 from which, sin A = 2 0.45

− = − and 1sin ( 0.4)A −= − = –23.58°

and 3 sin θ – 1 = 0 from which, sin θ = 1 0.3333...3= and 1sin (0.33333...)A −= =

19.47°

From the diagram below, the four values of θ between 0° and 360° are: 19.47°, 180° – 19.47° = 160.53°, 180° + 23.58° = 203.58° and 360° – 23.58° = 336.42°

2. Solve for angles between 0° and 360°: 8 tan2 θ + 2 tan θ = 15

Since 28 tan 2 tan 15θ θ+ = then 28 tan 2 tan 15 0θ θ+ − = i.e. ( )( )4 tan 5 2 tan 3 0θ θ− + =

i.e. 4 tan θ – 5 = 0 from which, tan θ = 5 1.254= and 1tan 1.25θ −= = 51.34°

and 2 tan θ + 3 = 0 from which, tan θ = 3 1.52

− = − and 1tan 1.5θ −= − = –56.31°

From the diagram below, the four values of θ between 0° and 360° are: 51.34°, 180° – 56.31° = 123.69°, 180° + 51.34° = 231.34° and 360° – 56.31° = 303.69°

© 2014, John Bird

715

3. Solve for angles between 0° and 360°: 2 cosec2 t – 5 cosec t = 12

Since 22cosec 5cosec 12t t− = then 22cosec 5cosec 12 0t t− − = and (2 cosec t + 3)(cosec t – 4) = 0

i.e. 2 cosec t + 3 = 0 from which, cosec t = 32

− and sin t = 23

− from which, t = –41.81°

and cosec t – 4 = 0 from which, cosec t = 4 and sin t = 14

from which, t = 14.48°

From the diagram below, the four values of θ between 0° and 360° are: 14.48°, 180° – 14.48° = 165.52°, 180° + 41.81° = 221.81° and 360° – 41.81° = 318.19°

4. Solve for angles between 0° and 360°: 2 cos2 θ + 9 cos θ – 5 = 0

Factorizing 2 cos2 θ + 9 cos θ – 5 = 0 gives: (2 cos θ – 1)(cos θ + 5) = 0 from which, either (2 cos θ – 1) = 0 or (cos θ + 5) = 0

i.e. cos θ = 12

or cos θ = –5 (which has no solution)

Hence, θ = 1 1cos2

−

= 60° or 300° since cosine is positive in the 1st and

4th quadrants

© 2014, John Bird

716

EXERCISE 179 Page 481

1. Solve for angles between 0° and 360°: 22cos sin 1θ θ+ = Since 2 2cos sin 1θ θ+ = then 2 2cos 1 sinθ θ= − Hence 22cos sin 1θ θ+ = becomes 22(1 sin ) sin 1θ θ− + = i.e. 22 2sin sin 1θ θ− + = i.e. 22sin sin 1 0θ θ− − = Factorizing gives: (2 sin θ + 1)(sin θ – 1) = 0 from which, either (2 sin θ + 1) = 0 or (sin θ – 1) = 0

i.e. sin θ = 12

− or sin θ = 1

Hence, θ = 1 1sin2

− −

= 210° or 330° (since sine is negative in the 3rd and 4th

quadrants or θ = 1sin 1− = 90° 2. Solve for angles between 0° and 360°: 24cos 5sin 3t t+ = Since 2 2cos sin 1t t+ = then 2 2cos 1 sint t= − Hence 24cos 5sin 3t t+ = becomes 24(1 sin ) 5sin 3t t− + = i.e. 24 4sin 5sin 3t t− + = i.e. 24sin 5sin 1 0t t− − =

Using the quadratic formula: 25 ( 5) 4(4)( 1) 5 41sin

2(4) 8t

− − ± − − − ±= =

= 1.4254… (which has no solution) or –0.17539… Hence, t = ( )1sin 0.17539...− − = 191.1° or 349.9° (since sine is negative in the 3rd and 4th

quadrants

© 2014, John Bird

717

3. Solve for angles between 0° and 360°: 22cos 4sin 0θ θ− =

Since 2 2cos sin 1θ θ+ = then 2 2sin 1 cosθ θ= − Hence, 22cos 4sin 0θ θ− = becomes 22cos 4(1 cos ) 0θ θ− − = i.e. 22cos 4 4cos 0θ θ− + = i.e. 24cos 2cos 4 0θ θ+ − = or 22cos cos 2 0θ θ+ − =

Using the quadratic formula: 21 1 4(2)( 2) 1 17cos2(2) 4

θ− ± − − − ±

= =

= 0.7807764… or –1.280776… (which has no solution) Hence, θ = 1cos 0.7807764...− = 38.67° or 321.33° since cosine is positive in the 1st and 4th

quadrants

4. Solve for angles between 0° and 360°: 23cos 2sin 3θ θ+ =

Since 2 2cos sin 1θ θ+ = then 2 2sin 1 cosθ θ= − Hence, 23cos 2sin 3θ θ+ = becomes 23cos 2(1 cos ) 3θ θ+ − = i.e. 23cos 2 2cos 3θ θ+ − = i.e. 22cos 3cos 1 0θ θ− + = Factorizing gives: (2 cos θ – 1)(cos θ – 1) = 0 from which, either (2 cos θ – 1) = 0 or (cos θ – 1) = 0

i.e. cos θ = 12

or cos θ = 1

Hence, θ = 1 1cos2

−

= 60° or 300° (since cosine is positive in the 1st and 4th quadrants

and θ = 1cos 1− = 0° or 360° Hence, θ = 0° or 60° or 300° or 360°

© 2014, John Bird

718

5. Solve for angles between 0° and 360°: 12 sin2 θ – 6 = cos θ

Since 212sin 6 cosθ θ− = then ( )212 1 cos 6 cosθ θ− − =

i.e. 212 12cos 6 cosθ θ− − =

i.e. 212cos cos 6 0θ θ+ − =

Factorizing gives: (4 cos θ + 3)(3 cos θ – 2) = 0

i.e. 4 cos θ + 3 = 0 from which, cos θ = 3 0.754

− = − and θ = 1cos ( 0.75)− − = –41.41°

and 3 cos θ – 2 = 0 from which, cos θ = 23

and θ = 1 2cos3

−

= 48.19°

From the diagram below, the four values of θ between 0° and 360° are: 48.19°, 180° – 41.41° = 138.59°, 180° + 41.41° = 221.41° and 360° – 48.19° = 311.81°

6. Solve for angles between 0° and 360°: 16 sec x – 2 = 14 tan2 x

1 + tan2 x = sec2 x from which, tan2 x = sec2 x – 1 Substituting for tan2 x in 16 sec x – 2 = 14 tan2 x gives: 16 sec x – 2 = 14(sec2 x – 1) i.e. 16 sec x – 2 = 14 sec2 x – 14 14 sec2 x – 16 sec x – 12 = 0

Using the quadratic formula: 216 ( 16) 4(14)( 12) 16 928sec

2(14) 28x

− − ± − − − ±= =

= 1.659396 or –0.516539

If sec x = 1.659396, then cos x = 1 0.6026301.659396

=

© 2014, John Bird

719

and x = ( )1cos 0.602630−

= 52.94º or 307.06º, since cosine is positive in the 1st

and 4th quadrants

or, if sec x = –0.516539 then cos x = 1 1.935960.516539

= −−

which is not possible

Hence, x = 52.94° or 307.06°

7. Solve for angles between 0° and 360°: 4 cot2 A – 6 cosec A + 6 = 0

Since 24cot 6cosec 6 0A A− + = then ( )24 cosec 1 6cosec 6 0A A− − + =

and 24cosec 6cosec 2 0A A− + =

Factorizing gives: (2 cosec A – 1) (2 cosec A – 2) = 0

i.e. 2 cosec A – 1 = 0 from which, cosec A = 12

and sin A = 2 which has no solutions

and 2 cosec A – 2 = 0 from which, cosec A = 1 and sin A = 1, which has only one solution

between 0° and 360°, i.e. A = 90°

8. Solve for angles between 0° and 360°: 5 sec t + 2 tan2 t = 3

1 + tan2 x = sec2 x from which, tan2 x = sec2 x – 1 Substituting for tan2 x in 5 sec t + 2 tan2 t = 3

gives: 5 sec t + 2(sec2 x – 1) = 3 i.e. 5 sec t + 2sec2 x – 2 = 3 2 sec2 x + 5 sec x – 5 = 0

Using the quadratic formula: 25 (5) 4(2)( 5) 5 65sec

2(2) 4t

− ± − − − ±= =

= 0.765564 or –3.2655644

If sec t = 0.765564, then cos t = 1 1.30622650.765564

= which is not possible

and sec t = –3.2655644, then cos t = 1 0.306225783.2655644

= −−

© 2014, John Bird

720

and t = ( )1cos 0.30622578− − = 107.83° or 252.17°, since cosine is negative in the 2nd

and 3rd quadrants

Hence, t = 107.83° or 252.17°

9. Solve for angles between 0° and 360°: 2.9 cos2 a – 7 sin a + 1 = 0

Since 22.9cos 7sin 1 0a a− + = then 22.9(1 sin ) 7sin 1 0a a− − + =

i.e. 22.9 2.9sin 7sin 1 0a A− − + =

and 22.9sin 7sin 3.9 0a a+ − =

Hence, sin a = 27 7 4(2.9)( 3.9) 7 94.242(2.9) 5.8

− ± − − − ± = = 0.46685 or –2.88064, which has

no solution

Thus, a = 1sin (0.46685)− = 27.83°

and, from the diagram below, 180° – 27°50′ = 152.17°

10. Solve for angles between 0° and 360°: 3 cosec2 ß = 8 – 7 cot ß

2 2cot 1 cosecβ β+ = and substituting for 2cosec β in 23cosec 8 7cotβ β= − gives:

( )23 cot 1 8 7 cotβ β+ = − i.e. 23cot 3 8 7cotβ β+ = − and 23cot 7 cot 5 0β β+ − = Using the quadratic formula:

27 7 4(3)( 5) 7 109cot 0.573384 or 2.9067182(3) 6

β − ± − − − ± = = = −

© 2014, John Bird

721

from which, 1 1 1cot 0.573384 tan0.573384

tβ − − = =

= 60.17° and 240.17°

(i.e. 1st and 3rd quadrants)

and 1 1 1cot ( 2.906718) tan2.906718

β − − = − = − = 161.02° and 341.02°

(i.e. 2nd and 4th quadrants) 11. Solve for angles between 0° and 360°: cot sinθ θ=

Since 1 coscottan sin

θθθ θ

= = then cos sinsin

θ θθ= i.e. 2cos sinθ θ=

2 2cos sin 1θ θ+ = from which, 2 2sin 1 cosθ θ= −

Hence, 2cos sinθ θ= becomes: cos θ = 21 cos θ−

i.e. 2cos cos 1 0θ θ+ − =

cos θ = 21 1 4(1)( 1) 1 52(1) 2

− ± − − − ± = = 0.61803398… or – 1.61803398…, which has

no solution

Hence, θ = ( )1cos 0.61803398..− = 51.83° and 308.17°

12. Solve for angles between 0° and 360°: tan 3cot 5secθ θ θ+ =

tan 3cot 5secθ θ θ+ = is the same as: sin cos 13 5cos sin cos

θ θθ θ θ

+ =

Multiplying each term by sin θ cos θ gives:

( ) ( ) ( )sin cos 1sin cos 3 sin cos 5 sin coscos sin cos

θ θθ θ θ θ θ θθ θ θ

+ =

Cancelling gives: 2 2sin 3cos 5sinθ θ θ+ = i.e. 2 2sin 3(1 sin ) 5sinθ θ θ+ − = i.e. 2 2sin 3 3sin 5sinθ θ θ+ − = and 22sin 5sin 3 0θ θ+ − =

© 2014, John Bird

722

Factorizing gives: (2 sin θ – 1)(sin θ + 3) = 0 from which, (2 sin θ – 1) = 0 or (sin θ + 3) = 0

i.e. sin θ = 12

or sin θ = –3 (which has no solution)

Hence, θ = 1 1sin2

− = 30° and 150°