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.

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

• Please write a more specific title. Commented Dec 9, 2021 at 21:02
• Hint: draw a triangle Commented Dec 9, 2021 at 21:04
• Specifically, a right triangle in standard position with legs $x$ and $a$. Note that you should probably draw two triangles, to capture the possibilities for the signs... Commented Dec 9, 2021 at 21:06
• Note that if $a\in\mathbb{R}-\{0\}$ you can simplify it into $$\pm \frac{x}{\sqrt{x^2 + a^2}}$$ according to the sign of $a$. Commented Dec 9, 2021 at 21:10
• Does this answer your question? How to derive compositions of trigonometric and inverse trigonometric functions? Commented Dec 9, 2021 at 23:09

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

Draw a right triangle and label one of the acute angles $$\theta$$. Now, label the side opposite $$\theta$$ as 'x' and the side adjacent to $$\theta$$ as 'a'. From the pythagorean theorem, the hypotenuse is then $$\sqrt{x^2+a^2}$$. We then see directly that $$sin(\theta) = \frac{x}{\sqrt{x^2+a^2}}$$.

Now, let's express $$\theta$$ using the arctangent function. Since $$\tan(\theta) = x/a$$, it follows that $$\theta = \arctan(\frac{x}{a})$$. If we substitute $$\theta$$ into the sine function, we have the result you're looking for:

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

All that's left is to factor an $$a^2$$ from the argument of the square root and you're there. $$\sin\left(\arctan\left(\frac{x}{a}\right)\right) = \frac{x}{a\sqrt{\frac{x^2}{a^2}+1}}.$$

If we assume that $$\dfrac xa=\tan\theta$$, the identity $$\sin\left(\arctan\left(\frac xa\right)\right)=\frac{\dfrac xa}{\sqrt{\left(\dfrac xa\right)^2+1}}$$

becomes

$$\sin(\theta)=\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}=\frac{\sin\theta}{\sqrt{\sin^2\theta+\cos^2\theta}}.$$

[Note that the signs are correct because in the range of the arc tangent, the cosine is positive.]