If csc Ø = 13/12, the sinØ = 12/13
so we are dealing with the 12-5-13 right-angled triangle in II
then cosØ = -5/13
sin 2Ø = 2sinØcosØ
= 2(12/13)(-5/13)
= -120/169
cos 2Ø = cos^2 Ø - sin^2 Ø
= 25/169 - 144/169
= -119/169
Find sin2theta, cos2theta if csctheta=13/12 and lies in quadrant II.
Thank you!
1 answer