# Proving (Q8B)

• August 20th 2010, 06:40 AM
Punch
Proving (Q8B)
Prove that $\frac{2sin2x-sin4x}{2sin2x+sin4x}=tan^2x$
• August 20th 2010, 06:56 AM
Quote:

Originally Posted by Punch
Prove that $\frac{2sin2x-sin4x}{2sin2x+sin4x}=tan^2x$

sin 2x = 2 sin x cos x

sin 4x = 2 sin 2x cos 2x

Substitute this into the equation, factor whatever is necessary to simplify the expression and see where that leads you.
• August 20th 2010, 09:14 AM
devdel
2 sin(2x) = 4 sin(x)·cos(x)
sin(4x) = 2 sin(2x)·cos(2x) = 4sin(x)·cos(x)·cos(2x)
cos(2x) = cos(x)-sin(x)

Ok, with that solved we can now begin.

(2 sin(2x)-sin(4x))/(2sin(2x)+sin(4x)) =
(4sin(x)·cos(x)-4sin(x)·cos(x)·cos(2x))/(4(sin(x)·cos(x)+4sin(x)·cos(x)·cos(2x) =
(4sin(x)·cos(x)(1-cos(2x)))/(4sin(x)·cos(x)(1+cos(2x)) =
(1-cos(2x))/(1+cos(2x)) =
(1-(cos(x)-sin(x))/(1+cos(x)-sin(x)) =
(sin(x)+cos(x)-cos(x)+sin(x))/(sin(x)+cos(x)+cos(x)-sin(x) =
2sin(x)/2cos(x) = tan(x)