# `cos^2(beta) + cos^2(pi/2 - beta) = 1` Verfiy the identity.

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### 1 Answer

`cos(pi/2-b)=sin(b)` for all b.

So the initial identity is equivalent to

`cos^2(b)+sin^2(b)=1,`

which is true.

*print*Print*list*Cite

`cos(pi/2-b)=sin(b)` for all b.

So the initial identity is equivalent to

`cos^2(b)+sin^2(b)=1,`

which is true.

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