Solving Trig Identities! HELP!

**Mineko245** · February 5th 2009, 06:50 AM

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.

**red_dog** · February 5th 2009, 06:56 AM

$4(\cos^6x+\sin^6x)=4[(\sin^2x)^3+(\cos^2x)^3]=$

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

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

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

I used $a^3+b^3=(a+b)^3-3ab(a+b)$

**Mineko245** · February 5th 2009, 07:21 AM

Thanks, But I kinda don't understand. Call me dense if you will :P Are you only solving the left side?

**Soroban** · February 5th 2009, 09:01 AM

Hello, Mineko245!

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

$1)\;\;4(\cos^6\!x + \sin^6\!x) \:= \:1 + 3\cos^2\!2x$

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

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

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

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

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

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

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

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

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

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

**Mineko245** · February 5th 2009, 03:24 PM

Thank You!

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