# Equivalent sine / cosine functions?

I was given the following problem on a quiz:

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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$\cos(x+\pi/2)=\cos x \cos \pi/2-\sin x \sin \pi/2$ –  pedja Mar 15 '12 at 6:54

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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You initially said D wasn't correct right? –  mowwwalker Mar 15 '12 at 20:28
Yes that was by mistake, the problem I had was with my latex editor, so I had to go back and forth, and also when I looked at the first answer I assumed he was saying only $C$ was correct. –  Kirthi Raman Mar 15 '12 at 20:48
Alright, just checking because you had changed your answer and I wasn't sure if I was seeing things 0.O –  mowwwalker Mar 15 '12 at 20:54
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.
Do you mean the fourth equality? In the third equality, $\sin(x)=\cos(x-\frac{\pi}{2})$, so $\sin(-x)=\cos(-x+\frac{\pi}{2})$ and not $\cos(-x-\frac\pi2)$ –  mowwwalker Mar 15 '12 at 6:52