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

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

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Answer 1 · 2016-11-20T21:18:15

#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#

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

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Explanation:

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