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?

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Answer 1 · 2017-01-03T04:10:03

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

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.