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cos(2θ)=cos^2(θ)−sin^2(θ)?
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Yup. Remember the identity \(\cos(a+b) = \cos a\cos b - \sin a \sin b\)? Just let \(a = b = \theta\)
http://www.biology.arizona.edu/biomath/tutorials/trigonometric/graphics/Trigidenteqn6.gif
\[\cos2\theta=\cos(\theta+\theta)\] \[=( cos \theta*\cos \theta)-(\sin \theta*\sin \theta)\] \[=\cos ^{2}\theta-\sin ^{2}\]
lol its sin ^2 theta
then why does r^2 [cos 2theta]=1
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at r=1 and \[\theta=0\] r^2 [cos 2theta] would be 1
hey, if @ParthKohli says its "yup", then its yup !
lol -_-
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