# Trig Identity and Pythagorean

## Homework Statement

3(sin(x)^4+cos(x)^4)-2(sin(x)^6+cos(x)^6)=1

(these are sinx raised to the 4 and 6 powers, not x^4or6)

## Homework Equations

Pythagorean Identities

## The Attempt at a Solution

I've tried using pythagorean identities to change everything to terms of sine or cosine. I've been hoping to manipulate it enough to get enough cos(x)^2+sin(x)^2 to try and turn all trig functions into a numerical value. Maybe this is right and I'm missing something on the way or not going far enough. I have figured out that the 3 and the 2 are necessary to equal 1, and other values such as 2 and 1 respectively will not equate to 1, therefore (sin(x)^4+cos(x)^4)-(sin(x)^6+cos(x)^6) =/= 1-1 (although I am not sure if this is relevant.)

If anyone can give me a hint at how to correctly approach the problem that would be nice, I don't want anyone to actually work the problem out for me. Thank you

## Answers and Replies

Try setting $y=(\sin x)^2$, and writing everything in terms of $y$.

Apphysicist
I rewrote it as 3((sin2)2+cos4)-2(sin2*sin4+cos6)

And then using sin2=1-cos2, and some FOILing, the messy algebra worked out nicely.