How do you find the value for sin2theta, cos2theta, and tan2theta and the quadrant in which 2theta lies given tantheta=-15/8 and theta is in quadrant II?

1 Answer
Jan 3, 2017

sin2theta=-240/289, cos2theta=-161/289 and tan2theta=240/161

Explanation:

We can easily use here three identities

sin2theta=(2tantheta)/(1+tan^2theta)=(2xx(-15/8))/(1+(-15/8)^2)

= (-15/4)/(1+225/64)=-15/4xx64/289=-240/289

cos2theta=(1-tan^2theta)/(1-tan^2theta)=(1-(-15/8)^2)/(1+(-15/8)^2)

= (1-225/64)/(1+225/64)=((64-225)/64)/((64+225)/64)=-161/289

and tan2theta=(2tantheta)/(1-tan^2theta)=(2xx(-15/8))/(1-(-15/8)^2)

= (-15/4)/(1-225/64)=-15/4xx-64/161=240/161

Note that as we have used identities, here it does not matter where theta lies, whether in Q2 or Q4.