Jamya Elliott

2022-01-28

Strange trigonometric simplification?
${\mathrm{cos}}^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac{1}{2}\pi ={\mathrm{sin}}^{-1}\frac{1}{3\sqrt{7}}$

becky4208fj

In the first step
${\mathrm{cos}}^{-1}\left(-\frac{1}{3\sqrt{7}}\right)-\frac{1}{2}\pi =\left(\pi -{\mathrm{cos}}^{-1}\left(\frac{1}{3\sqrt{7}}\right)\right)-\frac{1}{2}\pi$
we are using that for $0\le x\le \pi$
$\mathrm{arccos}\left(-\theta \right)=\pi -\mathrm{arccos}\theta$
which is trivial from the definition of $\mathrm{arccos}\theta$ dependin for the fact that $\mathrm{cos}\theta =\mathrm{cos}\left(-\theta \right)$
For the second step note that for $0\le x\le \frac{\pi }{2}$
$\theta =\frac{\pi }{2}-\mathrm{arccos}x⇒\mathrm{sin}\theta =\mathrm{sin}\left(\frac{\pi }{2}-\mathrm{arccos}x\right)=\mathrm{cos}\left(\mathrm{arccos}x\right)=x$
therefore
$\theta =\mathrm{arcsin}x$

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