Why does this method of using trigonometric functions to calculate relativistic gamma work?

Question

When I studied special relativity, I noticed that some of the problems and answers on calculating $\gamma$ would be fractions of c that looked like ratios of side lengths in right triangles. For example, when $v = 4/5c$, $\gamma = 5/3.$ By playing around with trigonometric functions, I found an alternative formula to calculate $\gamma$,

$$ \gamma = \frac 1 {\sin(\arccos(\frac{v}{c}))} $$

I am not very familiar with trigonometric identities. Can someone explain why this works?

Answer 1

$$ \sin(\arccos x) = \sqrt{1-x^2} $$ Draw the triangle: the hypotenuse is $1;$ the adjacent side is $x.$

By the Pythagorean theorem, the opposite side is $\sqrt{1-x^2}.$

Answer 2

If you let $\theta=\arccos(\frac{v}{c})$, you have

$$\cos\theta= \frac vc,\>\>\>\>\>\sin\theta = \sqrt{1-\frac {v^2}{c^2}}$$

Then,

$$\sin \left(\arccos\frac{v}{c}\right) = \sin\theta = \sqrt{1-\frac {v^2}{c^2}} =\frac1{\gamma}$$