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6.7.3 Trigonometric Functions Short Questions (Question 7 & 8)


Question 7:
It is given that   sin A = 5 13 and cos B = 4 5 , where A is an obtuse angle and B is an acute angle.
Find
(a) tan A
(b) sin (A + B)
(c) cos (A – B) 
 
Solution:
(a)
tan A = − 5 12


(b)
sin ( A + B ) = sin A cos B + cos A sin B sin ( A + B ) = ( 5 13 ) ( 4 5 ) + ( − 12 13 ) ( 3 5 ) ← cos A = − 12 13 sin B = 3 5 sin ( A + B ) = 4 13 − 36 65 sin ( A + B ) = − 16 65


(c)
cos ( A − B ) = cos A cos B + sin A sin B cos ( A − B ) = ( − 12 13 ) ( 4 5 ) + ( 5 13 ) ( 3 5 ) cos ( A − B ) = − 33 65



Question 8:
If sin A = p, and 90° < A < 180°, express in terms of p
(a) tan A
(b) cos A
(c) sin 2A

Solution:
 Using Pythagoras Theorem, Adjacent side = 1 2 − p 2 = 1 − p 2


(a)
tan A = − p 1 − p 2 ← tan is negative at second quadrant


(b)
cos A = − 1 − p 2 ← cos is negative at second quadrant


(c)
sin A = 2 sin A cos A sin A = 2 ( p ) ( − 1 − p 2 ) sin A = − 2 p 1 − p 2

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