How do you find $sin ({tan}^{- 1} (\frac{a}{3}))$ $sin(tan^-1(a/3))$ ?

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How do you find $sin ({tan}^{- 1} (\frac{a}{3}))$ $sin(tan^-1(a/3))$ ?

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Answer 1 · 2016-09-27T23:24:10

$sin ({tan}^{- 1} (\frac{a}{3})) = \frac{a}{\sqrt{a^{2} + 9}}$ $sin(tan^(-1)(a/3)) = a/(sqrt(a^2+9)$

Explanation:

Note that the range of ${tan}^{- 1} (y)$ $tan^(-1)(y)$ is $(- \frac{π}{2}, \frac{π}{2})$ $(-pi/2, pi/2)$ .

If $θ \in (- \frac{π}{2}, \frac{π}{2})$ $theta in (-pi/2, pi/2)$ then $cos θ > 0$ $cos theta > 0$ .

If $\frac{a}{3} > 0$ $a/3 > 0$ then consider a right angled triangle with sides $\frac{a}{3}$ $a/3$ , $1$ $1$ and $\sqrt{\frac{a^{2}}{9} + 1}$ $sqrt(a^2/9+1)$

We find:

$sin ({tan}^{- 1} (\frac{a}{3})) = \frac{\frac{a}{3}}{\sqrt{\frac{a^{2}}{9} + 1}} = \frac{a}{\sqrt{a^{2} + 9}}$ $sin(tan^(-1)(a/3)) = (a/3)/sqrt(a^2/9+1) = a/(sqrt(a^2+9)$

This identity continues to hold for $\frac{a}{3} < 0$ $a/3 < 0$ since $sin θ$ $sin theta$ and $tan θ$ $tan theta$ are odd functions.

It also holds for $a = 9$ $a=9$ since $sin ({tan}^{- 1} (0)) = sin (0) = 0$ $sin(tan^(-1)(0)) = sin(0) = 0$

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