# Trigonometric identities

• Apr 29th 2010, 05:16 PM
Caturdayz
Trigonometric identities
$\displaystyle Cos^4(theta)+Cos^2(theta)*sin^2(theta)+sin^2(theta )=1$
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This is being introduced at the end of a high-school Algebra II course, and is by far the most difficult, at least in my case, thing we've done all year.

I've been having luck with some of the more basic of these proofs despite an arduous start, but as we progress further into this section, I'm getting pretty lost.

Now, obviously there is factoring to be done here, but I really am unsure of where to start.

Any help would be greatly appreciated.
• Apr 29th 2010, 05:24 PM
Quote:

Originally Posted by Caturdayz
$\displaystyle Cos^4(theta)+Cos^2(theta)*sin^2(theta)+sin^2(theta )=1$
___

This is being introduced at the end of a high-school Algebra II course, and is by far the most difficult, at least in my case, thing we've done all year.

I've been having luck with some of the more basic of these proofs despite an arduous start, but as we progress further into this section, I'm getting pretty lost.

Now, obviously there is factoring to be done here, but I really am unsure of where to start.

Any help would be greatly appreciated.

I'll do a small part of it. Hopefully you'll see how to finish...

$\displaystyle \cos^4(\theta) + \cos^2(\theta) \sin^2(\theta) = \cos^2(\theta)(\cos^2(\theta) + \sin^2(\theta)) = \cos^2(\theta)$

Since $\displaystyle \cos^2(\theta) + \sin^2(\theta) = 1$
• May 1st 2010, 03:26 AM
mrmohamed
[quote=Caturdayz;503961]$\displaystyle Cos^4(theta)+Cos^2(theta)*sin^2(theta)+sin^2(theta )=1$

HI all
another way
cos^4 a = (1- sin^2a)^2
=(1- sin^2a)^2 + sin^2a(1-sin^2a)+sin^2a
=1 - 2 sin^2a +sin^4a+sin^2a - sin^4a + sin^2a
= 1- 2 sin^2a+2sin^2a - sin^4a + sin^4a = 1
mrmohamed