Trigonometric Functions Short Questions (Question 9 & 10)


Question 9:
Given that sin θ = 3 5 , where θ is an acute angle, without using tables or a calculator, find the values of
(a) sin (180º + θ),
(b) cos (180º – θ),
(c) tan (360º + θ).

Solution:
(a)


sin θ = 3 5 cos θ = 4 5 tan θ = 3 4

sin (180º + θ)
= sin 180º cos θ + cos 180º sin θ
= (0) cos θ + (– 1) sin θ
= – sin θ
= 3 5


(b)
cos (180º – θ)
= cos 180º cos θ + sin 180º sin θ
= (– 1) cos θ + (0) sin θ
= – cos θ
4 5


(c)
tan ( 360 + θ ) = tan 360 + tan θ 1 tan 360 tan θ = 0 + tan θ 1 ( 0 ) ( tan θ ) = tan θ = 3 4


Question 10:
Prove each of the following trigonometric identities.
(a) cot2 x – cot2 x cos2x = cos2 x
(b) sec x sec x cos x = cos e c 2 x

Solution:
(a)
LHS: cot 2 x cot 2 x cos 2 x = cot 2 x ( 1 cos 2 x ) = cot 2 x ( s i n 2 x ) = cos 2 x s i n 2 x ( s i n 2 x ) = cos 2 x (RHS)


(b)
LHS: sec x sec x cos x = 1 cos x 1 cos x cos x = 1 cos x 1 cos x cos 2 x cos x = 1 cos x 1 cos 2 x cos x = 1 cos x × cos x 1 cos 2 x = 1 1 cos 2 x = 1 s i n 2 x = cos e c 2 x (RHS)

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