Simplifying composition of trigonometric functions

Question

Sometimes when Integrating I end up with taking:

$$\sin \left(\tan ^{-1}\left(\frac{x}{2}\right)\right)$$, or other similar combinations however thats when I use the computer to extract:

$$\frac{x}{2 \sqrt{\frac{x^2}{4}+1}}$$

However, I want to find out how this is done and if there are any tricks I could use to evaluate similar expressions.

$$\sin\theta=\dfrac{2\tan(\theta/2)}{1+\tan^2(\theta/2)}$$ – Bumblebee Jan 02 '15 at 06:13 — Bumblebee, Jan 02 '15 at 06:13

score 5 · Accepted Answer · answered Jan 02 '15 at 01:57

5

Let $\theta=\tan^{-1}\left(\frac x2\right)$. So, $\tan\theta=\frac x2$. So, draw a right triangle with the opposite side labelled $x$ and the adjacent side labelled $2$. Applying the Pythagorean Theorem, will give that the hypoteneuse is $\sqrt{x^2+4}$. Since $\sin=\frac{\rm{opp}}{\rm{hyp}}$, you should have it.

answered Jan 02 '15 at 01:57

Tim Raczkowski

20,046

In general, the identities are: $\sin=\frac{opp}{hyp}$, $\cos=\frac{adj}{hyp}$, and $\tan=\frac{opp}{adj}$. All are related by:$(hyp)^2=(adj)^2+(opp)^2$. – FundThmCalculus Jan 02 '15 at 01:59
@FundThmCalculus Good point, I was assuming OP was familiar with basic trig. – Tim Raczkowski Jan 02 '15 at 02:00
True, just for completeness for someone else later. :) – FundThmCalculus Jan 02 '15 at 02:24

score 1 · Answer 2 · answered Jan 02 '15 at 02:06

In this particular case, for $-\dfrac{\pi}{2}< \theta < \dfrac{\pi}{2}$, $$ \tan^{-1}(\dfrac{x}{2}){}={}\theta\implies \tan\left(\theta\right){}={}\dfrac{x}{2}\,. $$

And, since,

$$ \tan(\theta){}={}\dfrac{\sin(\theta)}{\cos(\theta)}{}={}\dfrac{\sin(\theta)}{\sqrt{1-\sin^2\left(\theta\right)}}\,, $$ we must have $$ \sin\left(\tan^{-1}(\dfrac{x}{2})\right){}={}\sin\left(\theta\right){}={}\dfrac{x}{2\sqrt{1{}+{}\dfrac{x^2}{4}}}\,. $$

Ming-Tang · Answer 3 · 2015-01-02T02:16:39.810

Here is how I would do it in an exam: Draw out a right triangle, and assign values to the three sides and to the angle.

Here, what's the angle of the right triangle? The inverse trig function will tell you. $\theta = \tan^{-1} (\frac x 2)$ should be the angle.

The inverse trig function $\tan^{-1}$ also tells you something about the triangle: what are the side lengths? We re-express the equation as $\tan \theta = \frac x 2 $. Since $\tan$ is the ratio of opposite/adjacent, we assign the opposite to be x and adjacent to be 2. Now we also see hypotenuse $r = \sqrt { x^2 + 2^2 } $.

Back to the original expression, $\sin \left( \tan^{-1} (\frac x 2) \right)$ is actually $\sin \theta$ in the triangle. And $\sin \theta$ is opposite / hypotenuse, and we know what the values are: $\sin \theta = \frac{x}{\sqrt{x^2 + 2^2}}$.

Factoring so that $2^2$ becomes $1$: $\sin \theta = \frac{x}{2 \sqrt{\frac{x^2}{4} + 1}}$

Here is my potato-quality diagram: https://www.dropbox.com/s/zzv8npbyk3h73tb/2015-01-01%2017.58.17.jpg?dl=0

Simplifying composition of trigonometric functions

3 Answers3