# Determining compositions of trig functions by knowing Euler's identity etc

How does one determine: $$\cos^2(\arctan(x))?$$

I know what it is equal to, since its in the tables. But without working with many trigonometric identities, its not clear how to find such things.

How would you see this with the minimal number of trig identities?

$\cos^2(\arctan(x))=\cos(\arctan(x))\cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+x^2}}=\frac1{1+x^2}.$

The one identity I used here, I didn't know. It seems in similar situations, on an exam, I would have massive trouble without these identities. Can all of these sorts of things be solved by knowing something about Eulers identity and such?

Define the point $P(\theta)=(1,x)$ so that $\tan(\theta) = x$. Note that we need to have $\theta$ in the first or fourth quadrant in order for $\theta = \arctan x$ to be true.

From the picture we see that $\cos^2(\arctan x) = \cos^2 \theta = \dfrac{1}{1+x^2}$.

I find that the easiest way to solve these is to let $\theta$ be $\arctan{x}$, so that $\tan{\theta}=x$. Then we have \begin{align*} \frac{\sin{\theta}}{\cos{\theta}}&=x\\ \cos{\theta}&=\frac{\sin{\theta}}{x}\\ \cos{\theta}&=\frac{\sqrt{1-\cos^2{\theta}}}{x}\\ x\cos{\theta}&=\sqrt{1-\cos^2{\theta}}\\ x^2\cos^2{\theta}&=1-\cos^2{\theta}\\ (x^2+1)\cos^2{\theta}&=1\\ \cos^2{\theta}&=\frac{1}{x^2+1}\\ \end{align*} This may look tougher, but after the third step, we only need to use algebraic manipulations, and don't need to worry about trigonometry anymore. The two nicest ways to make a trig problem easier to solve are getting rid of inverses like I have in this problem, and writing everything in terms of sine and cosine.

• Is $\sin\theta$ always non negative like $$\sqrt{1-\cos^2\theta}$$ Nov 13 '17 at 4:08
• Best way to fix that problem is to already square after the second step. Nov 13 '17 at 6:55
• I think that the OP main concern is not to solve the problem posted. That is just an example. He seems to look for a general way of solving trig functions with some sort of simplifying tool like for example Euler’s identity. Nov 13 '17 at 7:35

From the well-known trig identity $$\sec^2y=1+\tan^2y$$ and $$\sec y=\frac{1}{\cos y}$$ one can easily find $$\cos^2y=\frac{1}{1+\tan^2y}$$ Plugging $y=\arctan x$ implies $$\cos^2(\arctan x)=\frac{1}{1+x^2}$$

Firstly, lets check the range of commonly defined $\arctan (x)$ because our result will depend on this. It is generally taken to be $(-\frac{\pi}{2}, \frac{\pi}{2})$. So $\cos$ will be positive over this domain.

Here this analysis is not of much use as we have squared the $\cos$ term, which will always be positive anyway.

Now to get the value of $\cos(\arctan(x))$ for acute angle $\arctan(x)$, let this angle be $\phi$.

\begin{align} \phi &= \arctan (x) \tag{1}\\ \tan(\phi) &= x\\ \frac{1}{1+\tan^2 \phi} &= \frac{1}{1+x^2}\\ \cos^2 \phi &= \frac{1}{1+x^2}\\ \implies \cos^2 (\arctan x) &=\frac{1}{1+x^2} \end{align}

There are (as is often the case with trigonometric identities) several ways to solve this, and the already-posted answers are just fine. Here's another, which I like because it uses a very general-purpose tool.

The tool in question: the famous half-angle formulae. These are extremely useful and reduce many trigonometric identities to routine algebraic manipulation. In case you don't already know them, here they are. Let $t=\tan\frac12\theta$; then $\cos\theta=\frac{1-t^2}{1+t^2}$, and $\sin\theta=\frac{2t}{1+t^2}$, and $\tan\theta=\frac{2t}{1-t^2}$.

Why are these useful here? Because $\cos^2\phi$ is very closely related to $\cos2\phi$, which means we have cosine of twice an arctangent, which if you think about it for a moment you'll see is exactly what the half-angle formula for cos tells us how to handle.

So. We're taking $\tan^{-1}x$ and we want (in effect) the cosine of twice this, so write $x=\tan\frac12\theta$. The formula above says that $\cos\theta=\frac{1-x^2}{1+x^2}$; call this $c$. The thing we were actually asked for is $\cos^2\frac12\theta$ and we have $c=2\cos^2\frac12\theta-1$ so $\cos^2\frac12\theta=\frac{c+1}2=\frac1{1+t^2}$.