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How does that bit work?

How is

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

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Make a right triangle with sides $x$, $\sqrt{1-x^2}$ and hypotenuse $1$. – anon Mar 28 '13 at 20:00
do you want proof? or just a way to remember? my high school teacher always told us to draw a right-angled (is this english?) triangle (lengths x, 1 and $\sqrt{1-x^2}$) to remember the formula... – long tom Mar 28 '13 at 20:02
up vote 14 down vote accepted

You know that "$\textrm{cosine} = \frac{\textrm{adjacent}}{\textrm{hypotenuse}}$" (so the cosine of an angle is the adjacent side over the hypotenuse), so now you have to imagine that your angle is $y = \arcsin x$. Since "$\textrm{sine} = \frac{\textrm{opposite}}{\textrm{hypotenuse}}$", and we have $\sin y = x/1$, draw a right triangle with a hypotenuse of length $1$ and adjacent side to your angle $y$ with length $x$. Then, use the pythagorean theorem to see that the other side must have length $\sqrt{1 - x^2}$. Once you have that, use the picture to deduce $$ \cos y = \cos\left(\arcsin x\right) = \sqrt{1 -x^2}. $$ Here's a picture of the scenario:


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This is a nice proof. I'm trying to prove it from first principles. I have questions. Why to draw a right angled triangle in the first place? Why not an other kind of triangle and use the law of cosines afterwards? Also, why did you choose the hypotenuse to be 1? I know that you did so arbitrarily, but who gave us the right to make arbitrary assumptions in mathematics? If e.g. i choose the hypotenuse to be 2, i will get $\sqrt{4-x^2}$. Will i be wrong if i choose to do so? Thanks in advance. – RestlessC0bra Sep 30 '15 at 19:33
@RestlessC0bra We draw a right triangle because that's kind of how sine and cosine are defined in the first place (or at least, that's the definition we're using), as a ratio of lengths of right triangles based on a certain angle in the triangle. This also has the benefit of not needing to prove the law of cosines or appeal to it, and it is more complicated in an arbitrary triangle than the simple right triangle case. If you didn't do it with a right triangle first, you might see that your law of cosines proof simplifies drastically when the triangle is a right triangle. – Stahl Sep 30 '15 at 21:11
@RestlessC0bra As for why we can make arbitrary choices, we're appealing to the well-defined-ness of sine and cosine here. The claim that we're implicitly assuming is that sine and cosine are well defined, or that if you have two right triangles with the same angle, then the ratio of specific sides is the same, no matter the lengths of the sides. If you choose the hypotenuse to be 2, you need the $\sin(y)$ to be $2x$ instead (remember, sine is opp/hyp = $2x/2$), and you get $\sqrt{2^2 - (2x)^2}/2 = \sqrt{1 - x^2}$. You could do this with any side length $\ell$ and it works the same way. – Stahl Sep 30 '15 at 21:14
You are right. I also made a fundamental mistake about not knowing exactly how the sine and cosine are defined. Thank you VERY much for clearing this up for me. – RestlessC0bra Sep 30 '15 at 21:31

If $\arcsin(x) = a \in [-\pi/2,\pi/2]$, we then have $\sin(a) = x \in [-1,1]$. Since $\cos^2(a) + \sin^2(a) = 1$, we have $$\cos(a) = \pm \sqrt{1-\sin^2(a)} = \pm \sqrt{1-x^2}$$ But since $a \in [-\pi/2,\pi/2]$, we have $\cos(a) \geq 0$. Hence, we get that $$\cos(a) = \sqrt{1-x^2}$$

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$\cos^2 (\arcsin x)+\sin^2 (\arcsin x)=1$, and the same for the other one.

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$$\left(\cos\arcsin x\right)'\stackrel{\text{chain rule}}=\frac{-\sin\arcsin x}{\sqrt{1-x^2}}=-\frac{x}{\sqrt{1-x^2}}=\left(\sqrt{1-x^2}\right)'\iff$$

$$\iff \cos\arcsin x=\sqrt{1-x^2}+K\;,\;\;K=\text{ a constant}\,$$

Now just choose a nice value of $\,x\,$ to find out $\,K\,$ ...

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Note that $$\arcsin(x)=y\Longrightarrow \sin(y)=x=\frac{x}{1}.$$ Using the fact that $$\text{sine}=\frac{\text{opposite}}{\text{hypotenuse}}$$we can form a right triangle where the hypotenuse has length $1$ and the leg opposite the angle has length $x$. Using the Pythagorean theorem, we see that the adjacent leg has length $\sqrt{1-x^2}$. As above, use the identity $$\text{cosine}=\frac{\text{adjacent}}{\text{hypotenuse}}$$and we get $$\cos(\arcsin(x))=\sqrt{1-x^2}.$$

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$$\begin{align*} \sin(\arcsin(x))&=x\\ \cos(\arcsin(x))^2&=1-\sin(\arcsin(x))^2\\ \cos(\arcsin(x))&=\sqrt{1-\sin(\arcsin(x))^2}\\ \cos(\arcsin(x))&=\sqrt{1-x^2} \end{align*}$$

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