Why is $\sin\left[\arctan\left(\frac{x}{a}\right)\right] = \frac{x}{a\sqrt{\frac{x^2}{a^2} + 1}}$ true? [closed]

My textbook has this expression, but I can't figure (either geometrically or analytically) why is this true and how do I get from one expression to the other.

$$\sin\left[\arctan\left(\frac{x}{a}\right)\right] = \frac{x}{a\sqrt{\frac{x^2}{a^2} + 1}}$$

Solution 1:

I'll show you how to work out anything in trig(inversetrig) form, I'll do $\cos{\tan^{-1}{\frac{x}{a}}}$, and you can figure out your case with much the same method.

Let $\alpha=\tan^{-1}{\frac{x}{a}}$
$\tan{\alpha}=\frac{x}{a}$

Now you know your trig ratios and tan is opposite on adjacent which means your triangle has sides $x$ (opposite), $a$ (adjacent) and $\sqrt{x^2+a^2}$ (hypotenuse, use pythagoras to figure it out)

Therefore $\cos{\alpha}=\frac{a}{\sqrt{x^2+a^2}}=\frac{a}{a\sqrt{\frac{x^2}{a^2}+1}}$ (remember $\cos{\alpha}=\cos{\tan^{-1}{\frac{x}{a}}}$).

Now see if you can apply this method to $\sin{\tan^{-1}{\frac{x}{a}}}$ :D