## camilasanchez Group Title verify identity in terms of sine and cosines one year ago one year ago

$\frac{ 1+ \cos^2 3 \theta }{ \sin^2 3 \theta } = 2 \csc^2 3 \theta$
Because 3θ is everywhere, it plays no special part. No need to break down into smaller angles. Just rewrite 1+cos²3θ as 1+1-sin²3θ=2-sin²3θ, so the left hand side becomes:$\frac{ 2-\sin^23θ }{ \sin^2 3θ }=\frac{ 2 }{ \sin^23θ }-1$ Hmm... I see a problem here: right hand side is equal to:$2\csc^23θ=2 \cdot \frac{ 1 }{ \sin^23θ }=\frac{ 2 }{ \sin^23θ }$So this "identity" seems to be false! We're left with the -1...