$\displaystyle 4sin2xcos2x=2sin4x$
$\displaystyle 6cot^22x-csc^22x=6cos^32x$
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$\displaystyle 4sin2xcos2x=2sin4x$
$\displaystyle 6cot^22x-csc^22x=6cos^32x$
1. Use the double angle identity $\displaystyle sin(2u) = 2sin(u)cos(u)$.Quote:
Originally Posted by Yuroichi
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)} $
