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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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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
up vote 1 down vote accepted

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

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So I have it down to 0-sinXsin(-1). I am really stumped here. I want to say the sin(x) is 1? – Kyle H Sep 2 '12 at 2:51
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
How did you arrive at sin(x)? I had 0-sinXsin(-1) but I am having trouble trying to simplify further. – Kyle H Sep 2 '12 at 3:02
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
Oh I see... -sin(x)sin(-pi/2) is the same as -(sin-pi/2)sin(x) which gives us 1sin(x)=sin(x)? – Kyle H Sep 2 '12 at 3:21

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