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$\cos(x-\pi/2)=\sin x$

I understand I have to use $\cos(A+B)=\cos A\cos B-\sin A\sin B$. Am I wrong to let $A=x$ and $B= -\pi/2$? Can someone please explain to me in detail how to solve this? Thank you.

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    You are nearly there. Do you know that $\cos(\pm\pi/2) = 0$ and $\sin(\pi/2) = 1$?2012-09-02

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You're on exactly the right track. Good work!

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    That's it, precisely!2012-09-02