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I am asked to prove that $\cos(\arcsin(x)) = \sqrt{1-x^2}$

I have used the trig identity to show that $\cos^2(x) = 1 - x^2$

Therefore why isn't the answer denoted with the plus-or-minus sign?

as in $\pm \sqrt{1-x^2}$.

Thank you!

4 Answers 4

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Hope this image helps (pythagorean identity?):

enter image description here

And as to the $\pm$, it simply because $\cos(\arcsin(x))$ is only equal to one value, and this value can be only $+$ or $-$, but not both (so we just choose, and $+$ is just more logical).

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    The triangle is used to show that $\arcsin{x}$ can be seen as $y=\arcsin\displaystyle\frac{x}{1}$ implying $\sin{y}=\displaystyle\frac{x}{1}$. Only the positive value is taken because the sides of the triangle are lengths which are scalar and hence non-negative. (This is to say that the sides are not pointing in any one direction. If the unit circle definition $sin^2{y}+cos^2{y}=1$ was used, then we could take negative solutions depending on the orientation of the triangle in the Cartesian plane.) – 2017-01-15
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    @M.Diggerson not with the usual range restriction for arcsine – 2017-01-15
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Note: this answer does not explain the formula $\sqrt{1-x^2}$, but does address why you don't need $\pm$.

Let $x$ be any number between $-1$ and $1$. Then $\arcsin(x)$ is an angle $\theta$ with $\sin(\theta)=x$. But which angle $\theta$? There are lots of different angles that all have the same sine. By definition, $\arcsin(x)$ is an angle between $-90$° and $90$° (or, if you prefer, between $-\pi/2$ and $\pi/2$ radians).

All right, now we want $\cos(\arcsin(x))$, or $\cos(\theta)$. That means we are taking the cosine of an angle between $-90$° and $90$°. The cosine of such an angle is never negative (we only get a negative cosine from an obtuse angle). So $\cos(\arcsin(x))$ is always non-negative.

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    Actually, $\cos(\arcsin x)$ is always *nonnegative*. – 2017-01-15
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    @N.F.Taussig Very good point -- I will edit. – 2017-01-15
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Let $\arcsin x = \theta$. Then, by definition of the arcsine function, $\sin\theta = x$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and $\cos(\arcsin x) = \cos\theta$. Using the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$, we obtain \begin{align*} \sin^2\theta + \cos^2\theta & = 1\\ \cos^2\theta & = 1 - \sin^2\theta\\ \cos^2\theta & = 1 - x^2\\ |\cos\theta| & = \sqrt{1 - x^2} \end{align*} Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\cos\theta \geq 0$. Thus, $|\cos\theta| = \cos\theta$, whence \begin{align*} \cos\theta & = \sqrt{1 - x^2}\\ \cos(\arcsin x) & = \sqrt{1 - x^2} \end{align*}

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

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    And the question is, why do we take the positive square root? – 2017-01-15