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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 $sin(\pi/2)=1, cos(\pi/2)=0$.2012-09-02
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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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    So I have it down to 0-sinXsin(-1). I am really stumped here. I want to say the sin(x) is 1?2012-09-02
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    Not quite. We know that $\sin(-\pi/2)=-1$--not the same as $\sin(-1)$--so we have $$\cos(x-\pi/2)=0-\sin(x)\sin(-\pi/2)=-(-1)\sin(x)=\sin(x).$$2012-09-02
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    How did you arrive at sin(x)? I had 0-sinXsin(-1) but I am having trouble trying to simplify further.2012-09-02
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    You don't plug -1 into sin at the end of your expression. As Cameron stated, you simply replace $\sin{(-\pi/2)}$ with -1.2012-09-02
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    Oh I see... -sin(x)sin(-pi/2) is the same as -(sin-pi/2)sin(x) which gives us 1sin(x)=sin(x)?2012-09-02
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    That's it, precisely!2012-09-02