How do I find the exact value of cos(2sin^-1(12/13))?

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Reviewing for a trig exam and saw this. I don't remember learning anything like it. Thanks in advance!

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Answer 1 · 2017-10-17T05:40:57

#-119/169.#

Write, #theta=sin^-1(12/13)," such that, "sintheta=12/13.....(1).#

Hence, #"the required value="cos(2sin^-1(12/13))=cos2theta,#

#=1-2sin^2theta,#

#=1-2(12/13)^2...........[because, (1)],#

#=1-288/169.#

#:. "the required value="-119/169.#

I hope that helps!