Solving Trig Identities! HELP!

Hello, Mineko245!

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

Quote:

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

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 $3\cos^2\!x\sin^2\!x$

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

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

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

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

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

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