# How to find algebraic expressions?

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

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

HallsofIvy
Homework Helper
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 $cos(\theta)$.

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