Question

how do you solve tan(sin^-1 x/2)? (show steps)

Answer 1

First: Equations are solved. $\tan\left(\sin^{-1}\dfrac x2\right)$ is an expression, and expressions can be simplified.

Now suppose $y=\sin^{-1}\dfrac x2$. Imagine a right triangle with one angle (not the one opposite the hypotenuse, of course) labeled $y$. This relation suggests that

$\sin y=\dfrac x2$

and under the hood, we're assuming that $-\dfrac\pi2<y<\dfrac\pi2$, because otherwise we wouldn't be able to properly alternate from $x$ to $y$ so seamlessly.

Now,

$\tan\left(\sin^{-1}\dfrac x2\right)=\tan y$

so the given expression is the same as the tangent of that same angle $y$.

If $\sin y=\dfrac x2$, then in the triangle, the leg opposite the angle $y$ occurs with the length of the hypotenuse in a ratio of $x$ to 2. By the Pythagorean theorem, the remaining side - call it $z$ - satisifies

$z^2+x^2=2^2\implies z=\sqrt{4-x^2}$

and we have

$\tan y=\dfrac xz=\dfrac x{\sqrt{4-x^2}}$