Trigonometric Functions Short Questions (Question 7 & 8)


Question 7:
It is given that   sinA= 5 13  and cosB= 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)
tanA= 5 12
(b)
sin( A+B )=sinAcosB+cosAsinB sin( A+B )=( 5 13 )( 4 5 )+( 12 13 )( 3 5 ) cosA= 12 13 sinB= 3 5 sin( A+B )= 4 13 36 65 sin( A+B )= 16 65
(c)
cos( AB )=cosAcosB+sinAsinB cos( AB )=( 12 13 )( 4 5 )+( 5 13 )( 3 5 ) cos( AB )= 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)
tanA= p 1 p 2 tan is negative at second quadrant
(b)
cosA= 1 p 2 cos is negative at second quadrant
(c)
sinA=2sinAcosA sinA=2( p )( 1 p 2 ) sinA=2p 1 p 2