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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.

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

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  • $\begingroup$ Please write a more specific title. $\endgroup$
    – David G. Stork
    Dec 9, 2021 at 21:02
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    $\begingroup$ Hint: draw a triangle $\endgroup$
    – Zadig
    Dec 9, 2021 at 21:04
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    $\begingroup$ 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... $\endgroup$
    – Eric Towers
    Dec 9, 2021 at 21:06
  • $\begingroup$ 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 $a$. $\endgroup$
    – Enrico M.
    Dec 9, 2021 at 21:10
  • $\begingroup$ Does this answer your question? How to derive compositions of trigonometric and inverse trigonometric functions? $\endgroup$
    – Git Gud
    Dec 9, 2021 at 23:09

3 Answers 3

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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

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

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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.]

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