Prove that sin x + 60 ° + cos x + 30 ° = 3 cos x .

To prove: sin x + 60 ° + cos x + 30 ° LHS : sin x + 60 ° + cos x + 30 ° = sin x cos 60 ° + cos x sin 60 ° + cos x cos 30 ° - sin x sin 30 ° = 1 2 sin x + 3 2 cos x + 3 2 cos x - 1 2 sin x = 3 cos x = RHS Hence, showed.

Prove that sin x + 60 ° + cos x + 30 ° ≡ 3 cos x . Hence solve sin x + 60 ° + cos x + 30 ° = 3 2 for 0 ° < x < 360 ° .

To prove: sin x + 60 ° + cos x + 30 ° ≡ 3 cos x LHS : sin x + 60 ° + cos x + 30 ° ≡ sin x cos 60 ° + cos x sin 60 ° + cos x cos 30 ° - sin x sin 30 ° ≡ 1 2 sin x + 3 2 cos x + 3 2 cos x - 1 2 sin x ≡ 3 cos x = RHS Hence, showed. Now, 3 cos x = 3 2 cos x = 3 2 3 = 3 2 ⇒ x = 30 ° or x = 360 ° - 30 ° = 330 ° .

Prove that sin 60 ° - x + cos 30 ° - x ≡ 3 cos x

To prove: sin 60 ° - x + cos 30 ° - x ≡ 3 cos x LHS : sin 60 ° - x + cos 30 ° - x ≡ sin 60 ° cos x - cos 60 ° sin x + cos 30 ° cos x + sin 30 ° sin x ≡ 3 2 cos x - 1 2 sin x + 3 2 cos x + 1 2 sin x ≡ 3 cos x = RHS Hence, showed.

Prove that sin 60 ° - x + cos 30 ° - x = 3 cos x . Hence solve the equation sin 60 ° - x + cos 30 ° - x = 2 5 sec x for 0 ° < x < 360 ° .

To prove: sin 60 ° - x + cos 30 ° - x ≡ 3 cos x LHS : sin 60 ° - x + cos 30 ° - x ≡ sin 60 ° cos x - cos 60 ° sin x + cos 30 ° cos x + sin 30 ° sin x ≡ 3 2 cos x - 1 2 sin x + 3 2 cos x + 1 2 sin x ≡ 3 cos x = RHS Hence, showed. Now, 2 5 sec x = 3 cos x ⇒ 2 5 cos x = 3 cos x ⇒ cos 2 x = 2 5 3 ⇒ cos x = 2 5 3 or cos x = - 2 5 3 ⇒ x = 61 . 3 ° or x = 360 ° - 61 . 3 ° = 298 . 7 ° or ⇒ x = 118 . 7 ° or x = 360 ° - 118 . 7 ° = 241 . 3 °

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 2

EASY

AS and A Level

IMPORTANT

Earn 100

By expressing the equation $cosec θ = 3 \sin θ + \cot θ$ in terms of $\cos θ$ only, solve the equation for $0 ° < θ < 180 ° .$ If $a$ and $b$ are the values of $θ$ in degrees. Find $a + b$ .(Answer without units)

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Important Questions on Trigonometry

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 3

Given that

\cos A = \frac{1}{4},

where

270 ° < A < 360 °,

exact value of

\sin 2 A

is

- \frac{\sqrt{a}}{8}

. Find

a

.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 4

Solve the equation

2 \tan^{2} x + \sec x = 1

for

0 ° ⩽ x ⩽ 360 ° .

iF the values of

x

are

a, b, c, d

then find

a + b + c + d

in degrees.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 5

Given

2 \cot^{2} x + 5 cosec x = 10

, find the value of

x

.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 6

Prove that $\sin (x + 60 °) + \cos (x + 30 °) = \sqrt{3} \cos x .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 6

Prove that $\sin (x + 60 °) + \cos (x + 30 °) \equiv \sqrt{3} \cos x .$ Hence solve $\sin (x + 60 °) + \cos (x + 30 °) = \frac{3}{2}$ for $0 ° < x < 360 °$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 7

Prove that $\sin (60 ° - x) + \cos (30 ° - x) \equiv \sqrt{3} \cos x$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 7

Prove that $\sin (60 ° - x) + \cos (30 ° - x) = \sqrt{3} \cos x$ . Hence solve the equation $\sin (60 ° - x) + \cos (30 ° - x) = \frac{2}{5} \sec x$ for $0 ° < x < 360 ° .$