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OpenStudy (anonymous):
=1-2sin^2(Π/16)cos^2(Π/16)+1-2sin^2(3Π/16)cos^2(3Π/16) =2-(sinΠ/4)^2/2-(sin3Π/4)^2/2 Can anyone explain how we got the second expression from the first one ?
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OpenStudy (anonymous):
@Peter14
OpenStudy (anonymous):
try factoring stuff and see what happens?
OpenStudy (anonymous):
I used 2sinAcosA = sin2A I got 2 - 1/2*sinΠ/8 - 1/2*sin3Π/8 What do I do next ?
OpenStudy (yttrium):
What do you need to do with this eq'n? Are you going to prove it or what?
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