 Turnseeuw

2022-01-23

$\mathrm{sin}\left(\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)\right)=\frac{\sqrt{2}}{\sqrt{3}}$ proof this equation vasselefa

Let $\theta =\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)$, so $\mathrm{cos}\theta =\frac{1}{\sqrt{3}}$
You want to find $\mathrm{sin}\theta$
Use ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$ Roman Stevens

${\mathrm{sin}}^{2}\left(\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)\right)+{\mathrm{cos}}^{2}\left(\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)\right)={\mathrm{sin}}^{2}\left(\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)\right)+\frac{13}{=}1$
or
Solution sketch: Draw a right triangle, with 1 as the hypotenuse and $\frac{1}{\sqrt{3}}$ as one of the legs (or $\sqrt{3}$ as the hypotenuse and 1 as one of the legs, same thing). Then see what $\mathrm{arccos}\left(\frac{1}{\sqrt{3}}\right)$ corresponds to in that triangle, then see what sin of that represents. Once you know what you are actually after, you do the calculations.

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