Why is $\cos(\arctan(3x)) = \frac{1}{\sqrt{1 + 9x^2}}?$

Question

Why is $\cos(\arctan(3x)) = \frac{1}{\sqrt{1 + 9x^2}}$? I've tried messing around with identities but still I'm not sure what's going on here.

Thanks.

Answer 1

$$\tan^2 u + 1 = \sec^2 u$$

Apply that to $u = \arctan(3x)$ to get $$ (3x)^2 + 1 = \frac{1}{\cos^2 (\arctan(3x))} $$ Now invert: $$ \frac{1}{(3x)^2 + 1} = {\cos^2 (\arctan(3x))} $$

and take a square root.

Answer 2

Either go for a trigonometic identity connecting $\tan$ with $\cos$; or:

Hint: in a right triangle, $\tan$ of an angle is equal to the ratio of the lengths of the opposite side to the adjacent side. For an angle $\theta = \arctan(3x)$, you have $\tan \theta = 3x$ so imagine such a triangle with opposite side $3x$ and adjacent side $1$.

What is the length of the hypotenuse (Pythagoras!)?

Then what is $\cos \theta$?

Answer 3

If we use the well-known formula $$\forall X\in \mathbb R$$ $$\cos^2(X)=\frac{1}{1+\tan^2(X)}$$

with $$X=\arctan(3x),$$ we get

$$\cos^2(\arctan(3x))=\frac{1}{1+\tan^2(\arctan(3x))}.$$

$$=\frac{1}{1+9x^2}$$

in the end, notice that

$$-\frac{\pi}{2}<\arctan(X)<\frac{\pi}{2}$$ and $$\cos(\arctan(X))>0.$$ thus

$$\cos(\arctan(3x))=\frac{1}{\sqrt{1+9x^2}}.$$