How do you find the exact value of #cos(arcsin(5/13))#?

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Answer 1 · 2016-12-26T11:04:24

The answer is #=12/13#

We need

#cos^2theta+sin^2theta=1#

Let #y=arcsin(5/13)#

So,

#siny=5/13#

#cos^2y=1-sin^2y=1-25/169=144/169#

#cosy=12/13#

Therefore,

#cos(arcsin(5/13))=12/13#