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amydh
 11 months ago
sin^2 theta  cos^2 theta = 1
amydh
 11 months ago
sin^2 theta  cos^2 theta = 1

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ParthKohli
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.2Just add those two equations to get \(2\sin^2 \theta = 0\).

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

mathslover
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.2why are you typing like satellite does
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