how to evaluate $\arcsin(x)-\arccos(x)=\arccos(\frac{\sqrt{3}}{2})$

Question

how to evaluate $\arcsin(x)-\arccos(x)=\arccos(\frac{\sqrt{3}}{2})$

Thoughts

Do you cos on both sides to finally do quadratic equation to solve at the end of the problem? so i got $0=4x^2+4x-1+2\sqrt{3}$ on my last step but i dont know if this is the right approach

Note that $\arcsin(x)+\arccos(x) =\pi/2$.

Therefore $\arcsin(x)-\arccos(x) =\arccos(\frac{\sqrt{3}}{2})$ becomes $\arccos(\frac{\sqrt{3}}{2}) =\pi/2-2\arccos(x)$ or $\arccos(x) =(\pi/2-\arccos(\frac{\sqrt{3}}{2}))/2$.

You should be able to take it from here.

You can't take the cosine of both sides, at least not the way you probably did. You must do it as follows:

$$\cos(\arcsin(x)-\arccos(x))=\frac{\sqrt3}2$$

And then use sum of angles identity:

$$\cos(\arcsin(x))\cos(\arccos(x))+\sin(\arcsin(x))\sin(\arccos(x))\\=x\cos(\arcsin(x))+x\sin(\arccos(x))$$

And use pythagorean theorems (let's assume all positive)

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

Can you take it from here?

From the unit circle, $\displaystyle \arccos\frac{\sqrt3}2 = \frac\pi6$.

A known property relating inverse sine and inverse cosine is $\arcsin x + \arccos x = \dfrac\pi2$. Therefore:

\begin{align*} \arcsin x - \arccos x &= \arccos \frac{\sqrt3}2\\[0.3cm] \frac\pi2 - \arccos x - \arccos x &= \frac\pi6\\[0.3cm] -2\arccos x &= -\frac\pi3\\[0.3cm] \arccos x &= \frac\pi6 \end{align*}

Now take the cosine of both sides.

• Same idea as mine, and you went all the way. – marty cohen Jan 20 '17 at 1:53
• @martycohen, I could say likewise (except for the second part) as you posted yours less than 1 minute before mine. And I dunno about all the way, I left a step at the end there XD – tilper Jan 20 '17 at 19:18

I would just solve this using the unit circle. If you're working in degree mode, $\arccos(\frac{\sqrt{3}}{2}) = 30^{\circ}$. So then, can you find a ratio $x$ such that $\arcsin(x) - \arccos(x) = 30^{\circ}$? If you look at a unit circle, you should see an immediate answer...