# How to find algebraic expressions?

1. Feb 20, 2007

### princiebebe57

How do you find an equivalent expression free of trig funtions for g(b)=cos(arctan(b))?

2. Feb 20, 2007

### AKG

Draw a right angled triangle, with one of the non-hypoteneuse sides having length b, and the other non-hypoteneuse having length 1.

3. Feb 21, 2007

### HallsofIvy

Staff Emeritus
A little more detail on AKG's suggestion: You are looking for $cos(\theta)$ with $\theta= arctan(b)$ or $tan(\theta)= b$.

Since "$tan(\theta)$" is "opposite side over near side" you can think of b as b/1. Draw a right triangle have "opposite side" b and "near side" 1. Now $cos(\theta)$ is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenuse and find $cos(\theta)$.

4. Feb 21, 2007

### arildno

Alternatively, remember the identity: $\sec^{2}(\theta)=\tan^{2}(\theta)+1$