If sintheta=sqrt403/22 and pi/2<theta<pi, how do you find tan2theta?

1 Answer
Nov 20, 2016

tan2theta=(9sqrt403)/161

Explanation:

If theta in [pi/2,pi], costheta<0, so:
costheta=-sqrt{1-sin^2theta}
costheta=-sqrt{1-403/484}=-sqrt{81/484}=-9/22

tantheta=sintheta/costheta=(sqrt{403}/22)/(-9/22)=-sqrt403/9
tan2theta=(2tantheta)/(1-tan^2theta)=(-(2sqrt403)/9)/(1-403/81)=(-(2sqrt403)/9)/(-322/81)=(9sqrt403)/161