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I was given the following problem on a quiz:

enter image description here

I put A, C, and D. The answer was A and D. We were taught four relevant equations:

$\sin(x)=-\sin(-x)$

$\cos(x)=\cos(-x)$

$\sin(x)=\cos(x-\frac{\pi}{2})$

$\cos(x)=\sin(x+\frac{\pi}{2})$

Based on my understanding of the unit circle definitions of cosine, and the appearance of the graphs of sine and cosine, I assumed:

$-\cos(x)=\cos(x\pm\pi)$

That's part of how I got C as an answer. I also graphed my answers after the quiz and they all looked the same.

Was my assumption wrong? Is there something I'm missing?

I also checked Wikipedia, which says:

$\cos(\pi-\theta)=-\cos(\theta)$

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

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    $\cos(x+\pi/2)=\cos x \cos \pi/2-\sin x \sin \pi/2$2012-03-15

2 Answers 2

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The correct answer is $A$, $C$ and $D$

First $\sin(-\theta) = -\sin(\theta)$ and second $\cos(\frac{\pi}{2}+\theta) = -\sin(\theta)$.

Also $-3\cos(x-\frac{\pi}{2}) = -3\cos(\frac{\pi}{2}-x) = -3\sin(x)$.

Other answers dont evaluate to a value of $2-3\sin(x)$

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    Alright, just checking because you had changed your answer and I wasn't sure if I was seeing things 0.O2012-03-15
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C is right. (I'm assuming you meant to say that you answered A, C, D).

Your assumption is correct, but you don't need it. From your four formulas, $ \cos(x+\frac\pi2)=\cos(-x-\frac\pi2)=\sin(-x)=-\sin(x). $ Using your second, third, and first equalities, in that order.

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    Oh, I see. Sorry, I was a bit confused. Thanks!2012-03-15