How to find algebraic expressions?

  • #1
How do you find an equivalent expression free of trig funtions for g(b)=cos(arctan(b))? :confused:
 

Answers and Replies

  • #2
AKG
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Draw a right angled triangle, with one of the non-hypoteneuse sides having length b, and the other non-hypoteneuse having length 1.
 
  • #3
HallsofIvy
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A little more detail on AKG's suggestion: You are looking for [itex]cos(\theta)[/itex] with [itex]\theta= arctan(b)[/itex] or [itex]tan(\theta)= b[/itex].

Since "[itex]tan(\theta)[/itex]" 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 [itex]cos(\theta)[/itex] is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenuse and find [itex]cos(\theta)[/itex].
 
  • #4
arildno
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Alternatively, remember the identity: [itex]\sec^{2}(\theta)=\tan^{2}(\theta)+1[/itex]
 

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