$\arcsin x- \arccos x= \pi/6$, difficult conclusion

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I do not understand a conclusion in the following equation. To understand my problem, please read through the example steps and my questions below.

Solve:

$\arcsin x - \arccos x = \frac{\pi}{6}$

$\arcsin x = \arccos x + \frac{\pi}{6}$

Substitute $u$ for $\arccos x$

$\arcsin x = u + \frac{\pi}{6}$

$\sin(u + \frac{\pi}{6})=x$

Use sum identity for $\sin (A + B)$

$\sin(u + \frac{\pi}{6}) = \sin u \cos\frac{\pi}{6}+\cos u\sin\frac{\pi}{6}$

$\sin u \cos \frac{\pi}{6} + \cos u\sin\frac{\pi}{6} =x$

I understand the problem up to this point. The following conclusion I do not understand:

$\sin u=\pm\sqrt{(1 - x^2)}$

How does this example come to this conclusion? What identities may have been used? I have pondered this equation for a while and I'm still flummoxed. All advice is greatly appreciated.

Note: This was a textbook example problem

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2 Answers

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The easy way to do this is to start with:

$\arccos x = \frac {\pi}{2} - \arcsin x$

Then your problem becomes

$2\arcsin x - \frac {\pi}{2} = \frac {\pi}{6}\\ x = \sin \frac {\pi}{3} = \frac {\sqrt3}2$

Now what have you done?

$\sin (u + \frac \pi6) = x\\ \frac 12 \cos u + \frac {\sqrt{3}}{2}\sin u = x\\ \frac 12 x + \frac {\sqrt{3}}{2}\sqrt{1-x^2} = x$

Why does $\sin (\arccos x) =\sqrt {1-x^2}?$

Draw a right triangle with base $= x$ and hypotenuse $= 1.$
For that angle $u, \cos u = x.$ What is the length of the opposite leg? $\sqrt {1-x^2}$

The range of $\arccos x$ is $[0, \pi]$ so $\sin (\arccos x) \ge 0$

$\sqrt{3}\sqrt{1-x^2} = x\\ \sqrt{3}(1-x^2) = x^2\\ 4x^2 = 3\\ x = \sqrt{\frac 34} $

We can reject the negative root as $x$ must be greater than $0$ as $\arcsin x - \arccos x < 0$ when $x<0$

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Hint:

if $u=\arccos (x)$ than

$\cos u=x$

and

$\sin u=\pm\sqrt{1-\cos^2u}=\pm \sqrt{1-x^2}$

substitute and square your equation and you have

$ \frac{3}{4}(1-x^2)=\frac{1}{4}x^2 $

that can be solved, (with care to improper solutions).

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