simplification of a cos

• Aug 27th 2008, 09:40 PM
ah-bee
simplification of a cos
can anybody show me how to simplify: cos(arctan x), im not too sure on how to evaluate that, thanks!
• Aug 27th 2008, 09:53 PM
Chris L T521
Quote:

Originally Posted by ah-bee
can anybody show me how to simplify: cos(arctan x), im not too sure on how to evaluate that, thanks!

Try to construct a triangle with an angle represented by $\displaystyle \vartheta=\tan^{-1}x$

This implies that our triangle has an opposite of length $\displaystyle x$ and an adjacent of length $\displaystyle 1$. We can now use this to determine the length of the hypotenuse of this triangle.

Using Pythagoras' Theorem, we see that $\displaystyle x^2+1^2=h^2\implies h=\sqrt{x^2+1}$

Thus, $\displaystyle \cos\vartheta=\frac{adj.}{hyp.}$

Since $\displaystyle adj.=1$ and $\displaystyle hyp.=\sqrt{x^2+1}$, we see that $\displaystyle \cos(\tan^{-1}x)=\cos\vartheta=\color{red}\boxed{\frac{1}{\sqr t{x^2+1}}}$

Does this make sense?

--Chris
• Aug 27th 2008, 10:20 PM
ah-bee
omg the way u put it makes it look so easy. very easily understandable!