Strange trigonometric simplification? $\cos^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac12\pi = \sin^{-1}\frac{1}{3\sqrt{7}}$

Question

I don't understand this simplification:

$$\begin{align} \cos^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac12\pi &= \left(\pi - \cos^{-1}\left(\frac{1}{3\sqrt{7}}\right)\right) - \frac12\pi \\[4pt] &= \frac12 \pi - \cos^{-1}\left(\frac{1}{3\sqrt{7}}\right) \\[4pt] &= \sin^{-1}\frac{1}{3\sqrt{7}} \end{align}$$

Am I not noticing any simple trigonometric identity?

@dxiv You are right. https://math.stackexchange.com/questions/1224415/how-do-i-prove-that-arccosx-arccos-x-pi-when-x-in-1-1 — Ramana, Sep 03 '18 at 07:29
@Ramana Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE — user, Oct 23 '18 at 20:53

score 1 · Accepted Answer · answered Sep 03 '18 at 07:03

In the first step

$$\cos^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac12\pi = \left(\pi - \cos^{-1}\left(\frac{1}{3\sqrt{7}}\right)\right) - \frac12\pi $$

we are using that for $0\le x\le \pi$

$$\arccos (-\theta)=\pi-\arccos \theta$$

which is trivial from the definition of $\arccos \theta$ dependin for the fact that $\cos \theta = \cos (-\theta)$.

For the second step note that for $0\le x\le \pi/2$

$$\theta = \frac{\pi}2-\arccos x\implies \sin \theta=\sin\left(\frac{\pi}2-\arccos x\right)=\cos (\arccos x)=x$$

therefore

$$\theta = \arcsin x$$

Strange trigonometric simplification? $\cos^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac12\pi = \sin^{-1}\frac{1}{3\sqrt{7}}$

1 Answers1