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)
sinθ= 3 5           cosθ= 4 5          tanθ= 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 +tanθ 1tan 360 tanθ = 0+tanθ 1( 0 )( tanθ ) =tanθ = 3 4


Question 10:
Prove each of the following trigonometric identities.
(a) cot2 x – cot2 x cos2 x = cos2 x
(b)  secx secxcosx =cose 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( si n 2 x ) = cos 2 x si n 2 x ( si n 2 x ) = cos 2 x (RHS)

(b)
LHS:  secx secxcosx = 1 cosx 1 cosx cosx = 1 cosx 1 cosx cos 2 x cosx = 1 cosx 1 cos 2 x cosx = 1 cosx × cosx 1 cos 2 x = 1 1 cos 2 x = 1 si n 2 x =cose c 2 x (RHS)