What is $\cos[2\tan ^{-1}(x^2)]$

Question

What is $\cos[2\tan ^{-1}(x^2)]$?

I tried $$y= \tan x =\frac{\sin x}{\cos x}=\frac{\sqrt{1-\cos^2{x}}}{\cos x}$$ then $$x=\tan^{-1} y = \tan^{-1} \frac{\sqrt{1-\cos^2{x}}}{\cos x}$$ However I don't know how to deal with the $2$ in the problem to make it simpler, any help? Thanks!

Use the cosine double angle formula and draw a right angled triangle with an angle of arctan(x^2). Then use Pythagoras' theorem to find the other side and then the sines and cosines of the angle can be found. — Vishal Gupta, Oct 06 '16 at 16:28

score 4 · Accepted Answer · answered Oct 06 '16 at 16:27

4

Let $\tan^{-1}x^2=u\implies\tan u=x^2$

$\cos\{2\tan^{-1}x^2\}=\cos2u=\dfrac{1-\tan^2u}{1+\tan^2u}=?$

answered Oct 06 '16 at 16:27

lab bhattacharjee

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score 0 · Answer 2 · answered Oct 06 '16 at 16:33

Basic trigonometry tells us that $\cos(2\theta) = \cos^2\theta - \sin^2\theta = (\cos\theta)^2 - (\sin\theta)^2$. That's how you deal with the $2$.

Then you have $\cos(\arctan a) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac 1 {\sqrt{1+a^2}}.$ This is in a triangle in which the "opposite" side is $a$ and the "adjacent" is $1$, so that the angle is $\arctan a = \arctan \dfrac a 1.$ The Pythagorean theorem gives us the hypotenuse.

And you can deal similarly with $\sin(\arctan a)$, but with "opposite" rather than "adjacent".

Finally, put $x^2$ in place of $a$.

What is $\cos[2\tan ^{-1}(x^2)]$

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