use an addition formula to derive the identity…

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

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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that's right. then use $sin(\pi/2)=1, cos(\pi/2)=0$. –  i. m. soloveichik Sep 2 '12 at 2:22
You are nearly there. Do you know that $\cos(\pm\pi/2) = 0$ and $\sin(\pi/2) = 1$? –  Dilip Sarwate Sep 2 '12 at 2:22

Not quite. We know that $\sin(-\pi/2)=-1$--not the same as $\sin(-1)$--so we have $$\cos(x-\pi/2)=0-\sin(x)\sin(-\pi/2)=-(-1)\sin(x)=\sin(x).$$ –  Cameron Buie Sep 2 '12 at 2:57
You don't plug -1 into sin at the end of your expression. As Cameron stated, you simply replace $\sin{(-\pi/2)}$ with -1. –  Nicolás Kim Sep 2 '12 at 3:11