Trigonometry - How do I simplify this expression? We have the expression $$ 13 \sin [ \tan ^{-1} (\dfrac{12}{5}) ] $$
Apparently the answer is 12, and I have to simplify it, and I'm assuming it means I have to show it's 12, without using a calculator.
Normally I show my own work in the questions, but in this case I have absolutely no clue how to. The only thing I know that might help is that $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$.
 A: Hint: There is a right triangle with side lengths $5$, $12$ and $13$. Draw this triangle, and choose one of the two non-right angles $t$ for which
$$\tan{t} = \frac{12}{5}$$
Recall that the tangent is the opposite side over the adjacent side. 
A: Here is an approach. Let
$$ y=13 \sin \left( \tan ^{-1} (\dfrac{12}{5}) \right) \implies \sin \left( \tan ^{-1} (\dfrac{12}{5}) \right)=\frac{y}{13} \longrightarrow (1)$$
$$ \implies \cos \left( \tan ^{-1} (\dfrac{12}{5}) \right)=\sqrt{1-\left(\frac{y}{13}\right)^2} \longrightarrow (2). $$
Now, dividing $(1)$ by $(2)$ and simplifying gives
$$ 12^213^3-13^2y^2=0\implies y=\pm 12 \implies y=12. $$
Note: We used the identity

$$ \tan(\tan^{-1}(x))=x. $$

A: Have the Pytagoric triangle of sides $5$, $12$ and $13$, and use the definitions of the trigonometric functions in a rectangle triangle.
A: Let $\displaystyle\tan^{-1}\frac{12}5=\theta$
$\displaystyle\implies(i)\tan\theta=\frac{12}5$
and $\displaystyle(ii)-\frac\pi2\le\theta\le\frac\pi2 $  (using the definition of principal value)
$\displaystyle\implies\cos\theta\ge0$
$\displaystyle\implies\cos\theta=\frac1{\sec\theta}=+\frac1{\sqrt{1+\tan^2\theta}}=\cdots$
$\displaystyle\implies \sin\theta=\tan\theta\cdot\cos\theta=\cdots$
A: $\tan^{-1} \left({\frac{12}{5}}\right)\in (0,\frac{\pi}{2})$ since $\frac{12}{5}>0$ now we see $\tan^{-1}\left({\frac{12}{5}}\right)=\sin^{-1}\left({+\frac{12}{13}}\right)$ .So the given expression is $13.\frac{12}{13}=12$ and proved.
