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myininaya
 one year ago
Best ResponseYou've already chosen the best response.1By a certain Pythagorean identity we can say 1cos^2(x)=?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1so 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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1We could use that same identity to rewrite cos^2(x)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1\[1\cos^2(x)=\sin^2(x) => \cos^2(x)=\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1that is a blank for you to fill in

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos^2(x)+\sin^2(x)=1 \]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.11cos^2(x)=sin^2(x) 1sin^2(x)=cos^2(x) cos^2(x)+sin^2(x)=1

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1So we can back to what we are trying to prove and replace the cos^2(x) with 1sin^2(x) \[\sin^2(x)(1+(1\sin^2(x))=2\sin^2(x)\sin^4(x)\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I replaced mr.cos^2(x) with mrs. (1sin^2(x))

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1now we don't need those parenthesis since there is a + in front of that parenthesis lets drop that extra stuff giving us: \[\sin^2(x)(1+1\sin^2(x))=2\sin^2(x)\sin^4(x) \]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1now 1+1=2 which I know you know (:p) so now you distribute on that left hand side

ChristopherW28
 one year ago
Best ResponseYou've already chosen the best response.0I gtg I can figure out what's left. Friend me on Facebook @ Christopher McElhannon. I will need your help again :)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I 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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