Prove the identity

**Yuroichi** · August 3rd 2009, 10:10 AM

$4sin2xcos2x=2sin4x$

$6cot^22x-csc^22x=6cos^32x$

**e^(i*pi)** · August 3rd 2009, 10:46 AM

Originally Posted by Yuroichi

$4sin2xcos2x=2sin4x$

$6cot^22x-csc^22x=6cos^32x$

1. Use the double angle identity $sin(2u) = 2sin(u)cos(u)$ .
In this case $u=2x$

2. This is not an identity - try $x = 1.25\pi$

$\frac{6cos^2(2x)-1}{sin^2(2x)}$

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