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ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\[\left(\cos^2 \theta  \sin^2 \theta\right) = 1\]\[\implies \cos(2\theta) = 1\]\[\implies \cos(2\theta) = 1\]Or use either of the two identities:\[\cos^2 \theta = 1  \sin^2 \theta\]\[\sin^2 \theta = 1 \cos^2 \theta\]
 2 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
or ... ANOTHER WAY! \[\sin^2(\theta)\cos^2(\theta)=1 \] Recall \[\sin^2(\theta)+\cos^2(\theta)=1\] So add \[\cos^2(\theta)\] on both sides to use the identity that is being recalled \[\sin^2(\theta)+\cos^2(\theta)\cos^2(\theta)=1+\cos^2(\theta) \] Now use the identity \[1\cos^2(\theta)=1+\cos^2(\theta)\] \[2=2\cos^2(\theta)\] \[1=\cos^2(\theta) \] which results into two equations
 2 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Just add those two equations to get \(2\sin^2 \theta = 0\).
 2 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
oh yeah that is pretty too
 2 months ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
This problem can be done in several ways. I was going to write some of them here, but I guess, they will all be actually similar to each other. But @myininaya and @ParthKohli have done well to find 2 of them.
 2 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
3 of them actually
 2 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
parth did one way at first then he found another way off what i was doing
 2 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
why are you typing like satellite does
 2 months ago
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