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

Question

parthkohli · Answer

Let $\arctan(x)$ be $\rm y$.$$\rm \cos (y)$$Can you find an expression for $\rm cos(y)$ in terms of $\rm sine$?

anonymous · Answer

cosy=sin(y+pi/2)

parthkohli · Answer

I'd do this instead...$$\rm \sqrt{1 - \sin^2(y)}$$

anonymous · Answer

OK

parthkohli · Answer

I'm stuck on this too. :p

anonymous · Answer

$$1-s^2arctanx=1/(x^2+1)$$ No, it seems too messy

turingtest · Answer

|dw:1350743050996:dw|you must be working with this triangle

anonymous · Answer

Not knowingly, but I suppose so.

anonymous · Answer

Genius solution! Thanks.

turingtest · Answer

$$\tan\theta=\frac x1=x\implies\theta=\tan^{-1}x$$$$\cos\theta=\frac1{\sqrt{x^2+1}}\implies\cos(\tan^{-1}x)=\frac1{\sqrt{x^2+1}}$$not sure where this triangle is coming from
welcome!

anonymous · Answer

How often is it very useful (i.e. hard not to solve while not) thinking geometrically with trig identities? That is, is it worth doing often or only when I see sqrt(1+x^2) lying about?

turingtest · Answer

I guess you did an integration and substituted x=tan(u) or something. Whenever I do trig substitutions I *always* draw the triangle to reduce things into simpler terms of x at the end, but I don't memorize anything except the trig definitions ant the Pythagorean theorem for that.

turingtest · Answer

and*

anonymous · Answer

OK- it was in fact Taylor seriesing for arctan. I'm not sure where the cosine popped up, though