how to evaluate $\arcsin(x)-\arccos(x)=\arccos(\frac{\sqrt{3}}{2})$
By Sarah Smith
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
$\endgroup$4 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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?
$\endgroup$ $\begingroup$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.
$\endgroup$ 2 $\begingroup$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...
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