Jamya Elliott

2022-01-28

Strange trigonometric simplification?

$\mathrm{cos}}^{-1}(-\frac{1}{3\sqrt{7}})-\frac{1}{2}\pi ={\mathrm{sin}}^{-1}\frac{1}{3\sqrt{7}$

becky4208fj

Beginner2022-01-29Added 10 answers

In the first step

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

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

$\mathrm{arccos}(-\theta )=\pi -\mathrm{arccos}\theta$

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

For the second step note that for$0\le x\le \frac{\pi}{2}$

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

therefore

$\theta =\mathrm{arcsin}x$

we are using that for

which is trivial from the definition of

For the second step note that for

therefore