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ParthKohli
 8 months ago
Best 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:\[\cos^2 \theta = 1  \sin^2 \theta\]\[\sin^2 \theta = 1 \cos^2 \theta\]

myininaya
 8 months ago
Best ResponseYou've already chosen the best response.2or ... 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

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

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

mathslover
 8 months ago
Best ResponseYou've already chosen the best response.0This 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.

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

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