Trig Identities...

[Qualinoshei]

hi i am in need of help in proving this trig identity, any insight is greatly appreciated thx.

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

ive tried factoring and other methods but i mustve been doing something wrong...

 

[arthur ohlsten]

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

but:
cos^2x=1-sin^2x

cos^4x=[1-sin^2x]^2
cos^4x=1-2sin^2x+sin^4x

substitute
sin^2x-sin^4x=1-sin^2x-[1-2sin^2x+sin^4x]
sin^2x-sin^4x= sin^2x-sin^4x

Arthur

 

[Mrspi]

hi i am in need of help in proving this trig identity, any insight is greatly appreciated thx.

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

ive tried factoring and other methods but i mustve been doing something wrong...

I'll work with the left-hand side:

sin[sup:29vm435g]2[/sup:29vm435g] x - sin[sup:29vm435g]4[/sup:29vm435g] x

Remove a common factor of sin[sup:29vm435g]2[/sup:29vm435g] x:

sin[sup:29vm435g]2[/sup:29vm435g] x (1 - sin[sup:29vm435g]2[/sup:29vm435g] x)

Now, looking at the right side, we see that we REALLY would like cos x.....
What is sin[sup:29vm435g]2[/sup:29vm435g] x? Well, it is (1 - cos[sup:29vm435g]2[/sup:29vm435g] x). And what is (1 - sin[sup:29vm435g]2[/sup:29vm435g] x)? Ah...that's cos[sup:29vm435g]2[/sup:29vm435g] x!

Substitute (1 - cos[sup:29vm435g]2[/sup:29vm435g] x) for sin[sup:29vm435g]2[/sup:29vm435g] x. Substitute cos[sup:29vm435g]2[/sup:29vm435g] x for (1 - sin[sup:29vm435g]2[/sup:29vm435g] x):

(1 - cos[sup:29vm435g]2[/sup:29vm435g] x)*cos[sup:29vm435g]2[/sup:29vm435g] x

Expand, and compare to the right side!

 

[Qualinoshei]

oh alright i get it thank you!