given: sinØ = 3/4, and Ø is in quadrant II
then by Pythagoras, y= 3, x = ? , r = 4
x^2 + 9 = 16
x = ±√5 , but in II x = -√7
so cosØ = -√7/4
sin 2Ø = 2sinØcosØ
= 2(3/4)(-√7/4) = -6√7/16
= -3√7/8
cos 2Ø = cos^2 Ø - sin^2 Ø = 7/16 - 9/16
= -2/16
= - 1/8
tan 2Ø = sin 2Ø / cos 2Ø
= (-3√7/8) / (-1/8)
= 3√7
Find the exact values of sin 2 theta, cos 2 theta and tan 2thea using the double angle formula
sin theta = 3/4 , pi/2 < theta < pi
1 answer