# Solving Trig Identities! HELP!

• Feb 5th 2009, 06:50 AM
Mineko245
Solving Trig Identities! HELP!
I have these questions in my Trig Identities homework but I don't understand how to do them.

1) 4(Cos^6X + Sin^6X) = 1 + 3 Cos^(2)2X

I don't know what to do when the Cos or Sin has an exponent more than 2.
• Feb 5th 2009, 06:56 AM
red_dog
$\displaystyle 4(\cos^6x+\sin^6x)=4[(\sin^2x)^3+(\cos^2x)^3]=$

$\displaystyle =4[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)]=$

$\displaystyle =4(1-3\sin^2x\cos^2x)=4-12\sin^2x\cos^2x=4-3\sin^22x=$

$\displaystyle =4-3(1-\cos^22x)=1+3\sin^22x$

I used $\displaystyle a^3+b^3=(a+b)^3-3ab(a+b)$
• Feb 5th 2009, 07:21 AM
Mineko245
Thanks, But I kinda don't understand. Call me dense if you will :P Are you only solving the left side?
• Feb 5th 2009, 09:01 AM
Soroban
Hello, Mineko245!

A slight variation on red-dog's solution . . . done in baby-steps.

Quote:

$\displaystyle 1)\;\;4(\cos^6\!x + \sin^6\!x) \:= \:1 + 3\cos^2\!2x$
We have: .$\displaystyle 4\bigg[(\cos^2\!x)^3 + (\sin^2\!x)^3\bigg]$ . . . sum of cubes

. . $\displaystyle = \;4\bigg[\underbrace{\left(\cos^2\!x + \sin^2\!x\right)}_{\text{This is 1}}\left(\cos^4\!x - \cos^2\!x\sin^2\!x + \sin^4\!x\right)\bigg]$

. . $\displaystyle = \;4\bigg[\cos^4\!~x - \cos^2\!x\sin^2\!x + \sin^4\!x\bigg]$

Add and subtract $\displaystyle 3\cos^2\!x\sin^2\!x$

. . $\displaystyle 4\bigg[\cos^4\!x - \cos^2\!x\sin^2\!x \:{\color{blue}+\: 3\cos^2\!x\sin^2\!x} + 1 \:{\color{blue}- \:3\cos^2\!x\sin^2\!x}\bigg]$

. . $\displaystyle = \;\;4\bigg[\left(\cos^4\!x + 2\cos^2\!x\sin^2\!x\right) + \sin^4\!x - 3\cos\!x\sin^2\!x\bigg]$

. . $\displaystyle =\;\;4\bigg[\underbrace{(\cos^2\!x + \sin^2\!x)^2}_{\text{This is 1}} - 3\cos^2\!x\sin^2\!x\bigg] \;\;=\; \;4\bigg[1 - 3\sin^2\!x\cos^2\!x\bigg]$

. . $\displaystyle =\;\;4 - 12\sin^2\!x\cos^2\!x \;\;=\;\;4-12\left(\frac{1-\cos2x}{2}\right)\left(\frac{1+\cos2x}{2}\right)$

. . $\displaystyle = \;\;4 - 12\left(\frac{1-\cos^2\!2x}{4}\right) \;\;=\;\;4 - 3(1-\cos^2\!2x)$

. . $\displaystyle =\;\;4-3+3\cos^2\!2x \;\;=\;\;1 + 3\cos^2\!2x \quad\hdots There!$

• Feb 5th 2009, 03:24 PM
Mineko245
Thank You!