Trig Proof

[HopeWelch]

So this is probably another simple proof I'm missing, maybe someone can help. I'm only allowed to manipulate the left side of the equation.

sin^4x - cos^4x = 2sin^2x - 1

So I tried factoring into

(sin^2x - cos^2x)(sin^2x + cos^2x)

Then substituting in 1 - sin^2x for both the positive and negative cos^2x

For the first I'm getting -2sin^2 - 1
and the second -1

Maybe my algebra is off and I'm messing up or something. I can't see how turning (sin^2x + cos^2x) into 1 would do anything...

If someone could give me a hint or find my error I would appreciate it.

 

[srmichael]

You factored correctly, but now you have the mother of all trig identities: \(\displaystyle \sin^2x+\cos^2x = 1\)

Use this and the \(\displaystyle \cos^2x=1-\sin^2x\) identity you mention (which, by the way, comes from the "mother" identity) and you should be right as rain.

 

[HopeWelch]

Gosh I feel silly haha. So I tried one and I tried the other but not both at the same time...gah

Thanks so much for the reply!!

 

[HopeWelch]

I went back and retried this using what you said in your post and now I realize why I couldn't get that to work. I was simply screwing up the distribution of the negatives.

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

When I tried it before I kept doing sin^2x - sin^2x -1 leaving me with -1.
Silly me...