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anonymous
 5 years ago
(1cos²θ)(1+cos²θ)=2sin²θsin⁴θ. Need to show this is an identity.
anonymous
 5 years ago
(1cos²θ)(1+cos²θ)=2sin²θsin⁴θ. Need to show this is an identity.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin^2\theta+\cos^2\theta=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use this wherever you see 1 then simplify it till you get your right hand side equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0best bet may be to write first as \[1cos^4(\theta)\] then replace \[cos^4(\theta)\] by \[(1sin^2(\theta))^2\] then multiply out and i believe this gives it to you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0will work it out if you like. it is algebra from there on in.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh my, could you work it out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure i will multiply out \[(1sin^2(\theta))^2\] but it is just the same as \[(1x^2)^2\] with x replaced by sine. \[(1x^2)^2=12x^2+x^4\] so \[(1sin^2(\theta))^2=12sin^2(\theta)+sin^4(\theta)\]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't forget that originally you have \[1cos^4(\theta)\] so now you have \[1(12sin^2(\theta)+sin^4(\theta))=2sin^2(\theta)sin^4(\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that enough detail? if not i let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i don't know why, but i just can't see it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(\sin^2\theta+\cos^2\theta\cos^2\theta)(\sin^2\theta+\cos^2\theta+\cos^2\theta)=(\sin^2\theta)(sina^2\theta+2\cos^2\theta)=\sin^4\theta+2\sin^2\theta(\cos^2\theta)=\sin^4\theta+2(\sin^2\theta)(1\sin^2\theta)=sina^4\theta+2\sin^\theta2\sin^4\theta=2\sin^2\theta\sin^4\theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets to slowly. first of all, the left hand side of the equation is \[(1cos^2(\theta))(1cos^2(\theta))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so our first job is to multiply this out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it clear that this is the same as multiplying out \[(1x)(1+x)\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, when you write it like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[or rather (xx^2)(1+x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so lets multiply out \[(1cos^2(\theta))(1+cos^2(\theta))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(1x^2)(1+x^2)=1x^4\] so\[((1cos^2(\theta))(1+cos^2(\theta))=1cos^4(\theta)\] so far so good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok now you recall that \[sin^2(\theta)+cos^2(\theta)=1\] so \[cos^2(\theta)=1sin^2(\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and of course \[cos^4(\theta)=(cos^2(\theta))^2\] so replace \[cos^2(\theta)\]by \[(1sin^2(\theta))\] in this expression to get \[1(1sin^2(\theta))^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now multiply out and you get your answer exactly. this is like \[1(1x^2)^2=1(12x^2+x^4)=2x^2x^4\] with x replaced by sine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is when you replace the sin/cos with the 'x'. easier to see

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes of course. the only little bit of trig here was replacing \[cos^2(\theta)\]by \[(1sin^2(\theta)\] every other step was algebra. multiply, collect like terms etc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wanna do the next one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not sure where to start
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