# i need help proving this identity

• Aug 23rd 2008, 07:21 PM
nithu
i need help proving this identity
sin4x-sin2x/sin2x=cos3x/cosx

i got left side equals right side..which is 1-4sin^2x
• Aug 23rd 2008, 07:38 PM
TKHunny
1) Please add parentheses to clarify meaning.

Is it as you have written it or did you mean (sin(4x)-sin(2x))/sin(2x)?

2) Please show your work. It's nice that you managed some result, but how did you do that?

3) When you get left side = right side, you are done.

Where does that leave us?
• Aug 24th 2008, 06:37 AM
Soroban
Hello, nithu!

Working from one side only (the recommended method) is difficult.
I worked on both sides and made them "meet in the middle" (as you did)
. . then devised a way to connect them.

Quote:

Prove: . $\frac{\sin4x-\sin2x}{\sin2x} \;=\;\frac{\cos3x}{\cos x}$

The right side is: . $\frac{4\cos^3\!x - 3\cos x}{\cos x} \;\;=\;\;4\cos^2\!x - 3 \;\;=\;\;4\left(\frac{1+\cos2x}{2}\right)-3$

. . . . . $= \;\;2 + 2\cos2x - 3 \;\;=\;\;2\cos2x - 1$

Multiply by $\frac{\sin2x}{\sin2x}$

. . $\frac{\sin2x}{\sin2x}\cdot(2\cos2x - 1) \;\;=\;\;\frac{2\sin2x\cos2x - \sin2x}{\sin2x} \;\;=\;\;\frac{\sin4x - \sin2x}{\sin2x}$

• Aug 24th 2008, 10:12 AM
nithu
can u show me the left side and right side work...thx
• Aug 24th 2008, 01:11 PM
Soroban
Hello, nithu!

For the left-side work, read up from the bottom of my solution.

You should be able to see how they "meet in the middle."