The Derivative of $\cos(x-2)$

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I would think that the solution would be $-\sin(x-2)$, but when i use WolframAlpha it says that the answer is $\sin(2-x)$. Are these $2$ answers equivalent or I am missing some fact here? Thanks in advance.

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2 Answers

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Since $\sin(x)$ is an odd function, $\sin(-x)=-\sin(x)$.

In particular, $\frac{d}{dx}[\cos(x-2)]=-\sin(x-2)=\sin(2-x)$.

Likewise, $\cos(x)$ is an even function, so $\cos(-x)=\cos(x)$.

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By the chain rule,

$$ \frac{d}{dx} \left( \cos(x - 2) \right) = -\sin(x - 2) \cdot \frac{d}{dx}\left(x - 2\right) = -\sin(x - 2) $$

Since $\sin(x)$ is an odd function, $\sin(-x) = -\sin(x)$ so $-\sin(x - 2) = \sin(2 - x)$

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