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


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

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

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

  • $\begingroup$ $x = \frac{\sqrt 3}{ 2} $ works $\endgroup$ – Will Jagy Mar 1 '17 at 22:21

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 $x<0$

  • $\begingroup$ Thanks for helping! Why is the length of the opposite leg equal to √(1-x^2)? What identity, or logic is used to reach that conclusion? Sorry if this is a newbie question. Will accept and 1+ if you can answer this :) $\endgroup$ – GiantSpruce Mar 1 '17 at 22:41
  • 1
    $\begingroup$ Pythagorean Theorem. $x^2 + (\sqrt{1-x^2})^2 = 1$ Or perhaps even more to the point.... $\cos^2 u + \sin^2 u = 1$ $\endgroup$ – Doug M Mar 1 '17 at 22:44


if $u=\arccos (x)$ than

$\cos u=x$


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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.