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$\displaystyle 4sin2xcos2x=2sin4x$ $\displaystyle 6cot^22x-csc^22x=6cos^32x$
Quote: Originally Posted by Yuroichi $\displaystyle 4sin2xcos2x=2sin4x$ $\displaystyle 6cot^22x-csc^22x=6cos^32x$ 1. Use the double angle identity $\displaystyle sin(2u) = 2sin(u)cos(u)$. In this case $\displaystyle u=2x$ 2. This is not an identity - try $\displaystyle x = 1.25\pi$ $\displaystyle \frac{6cos^2(2x)-1}{sin^2(2x)} $