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Advocard [28]
3 years ago
9

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

Mathematics
1 answer:
andrew11 [14]3 years ago
4 0

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, 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}}

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