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Mathematics 6 Online
OpenStudy (anonymous):

cos(2θ)=cos^2(θ)−sin^2(θ)?

Parth (parthkohli):

Yup. Remember the identity \(\cos(a+b) = \cos a\cos b - \sin a \sin b\)? Just let \(a = b = \theta\)

OpenStudy (anonymous):

\[\cos2\theta=\cos(\theta+\theta)\] \[=( cos \theta*\cos \theta)-(\sin \theta*\sin \theta)\] \[=\cos ^{2}\theta-\sin ^{2}\]

OpenStudy (anonymous):

lol its sin ^2 theta

OpenStudy (anonymous):

then why does r^2 [cos 2theta]=1

OpenStudy (anonymous):

at r=1 and \[\theta=0\] r^2 [cos 2theta] would be 1

OpenStudy (anonymous):

hey, if @ParthKohli says its "yup", then its yup !

Parth (parthkohli):

lol -_-

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