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Verify that (1-cos^2x)(1+cos^2x)=2sin^2x-sin^4x is a trig identity

Mathematics
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By a certain Pythagorean identity we can say 1-cos^2(x)=?
Sin^2x?
yep! :)

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Other answers:

so this means we now have \[\sin^2(x)(1+\cos^2(x))=2\sin^2(x)-\sin^4(x)\] now this is an identity we are trying to prove now we have one side in terms of sin and another side in terms of mixture of sin and cos
We could use that same identity to rewrite cos^2(x)
\[1-\cos^2(x)=\sin^2(x) => \cos^2(x)=\]
that is a blank for you to fill in
I'm lost
Ok I want to rewrite cos^2(x) because the other side is just in terms of sin I know an identity that has cos^2(x) and sin^2(x) in it.
\[\cos^2(x)+\sin^2(x)=1 \]
So cos^2(x)=?
Sin^x+ 1
1-sin^2(x)
Sin^2x+1?
I was close
1-cos^2(x)=sin^2(x) 1-sin^2(x)=cos^2(x) cos^2(x)+sin^2(x)=1
So we can back to what we are trying to prove and replace the cos^2(x) with 1-sin^2(x) \[\sin^2(x)(1+(1-\sin^2(x))=2\sin^2(x)-\sin^4(x)\]
I replaced mr.cos^2(x) with mrs. (1-sin^2(x))
now we don't need those parenthesis since there is a + in front of that parenthesis lets drop that extra stuff giving us: \[\sin^2(x)(1+1-\sin^2(x))=2\sin^2(x)-\sin^4(x) \]
now 1+1=2 which I know you know (:p) so now you distribute on that left hand side
I gtg I can figure out what's left. Friend me on Facebook @ Christopher McElhannon. I will need your help again :)
I will be here most likely. I'm not much of a facebook user. I'm on facebook free diet right now. :p

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