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The proofs of the derivatives of the trig functions typically use the identity

$$\cos^2(x) + \sin^2(x) = 1$$

but I usually see something like this

$$\sin^2(x) = 1 - \cos^2(x)$$ $$\implies \sin(x) = \sqrt{1-\cos^2(x)}$$

to prove, say, the derivative of $\cos^{-1}(x)$.

Why do we take the positive square root of $\sin(x)$? I'm sure I'm missing something really obvious.

(For one thing, the RHS must be non negative ... )

Thanks.

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  • $\begingroup$ This is not how the argument typically goes, at least not how I learned it. One can find the derivative of $\arccos x$ by noting $\cos(\arccos x)=x$ for $x\in[-1,1]$. Differentiating both sides using the chain rule gives $-\sin(\arccos x)(\arccos x)'=1$. Now use the fact that $\sin(\arccos x)=\sqrt{1-x^2}$. Whatever way you do if, you can see we take the positive square root simply because $\arccos x\in[0, \pi]$, and on that interval $\sin$ is positive. $\endgroup$ – symplectomorphic Nov 10 '17 at 12:25
  • $\begingroup$ Yes, I did it that way @symplectomorphic, using the chain rule exactly as you had said. I just mainly had a question of the domain the range, answered below. Thanks, $\endgroup$ – D.Hutchinson Nov 10 '17 at 12:27
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It only depends on the definition. For example, the definition of $\arccos$ states that $\arccos x \in [0,\pi]$. Hence the corresponding values of $\sin$ are positive, hence the positive square root.

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  • $\begingroup$ ah, good point. Silly question: $\cos(x)$ is one-to-one on $[\pi, 2\pi]$, then shouldn't we consider its range, too? $\endgroup$ – D.Hutchinson Nov 10 '17 at 12:22
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    $\begingroup$ You have to choose. The standard definition is $\arccos \colon [-1,1] \to [0,\pi]$. $\endgroup$ – Siminore Nov 10 '17 at 12:24
  • $\begingroup$ Ok, got it -- thanks so much! :) $\endgroup$ – D.Hutchinson Nov 10 '17 at 12:25

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