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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:\[\cos^2 \theta = 1  \sin^2 \theta\]\[\sin^2 \theta = 1 \cos^2 \theta\]
 one month 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
 one month ago

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

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

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

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