sin4x-sin2x/sin2x=cos3x/cosx
i got left side equals right side..which is 1-4sin^2x
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?
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.
Prove: .$\displaystyle \frac{\sin4x-\sin2x}{\sin2x} \;=\;\frac{\cos3x}{\cos x}$
The right side is: . $\displaystyle \frac{4\cos^3\!x - 3\cos x}{\cos x} \;\;=\;\;4\cos^2\!x - 3 \;\;=\;\;4\left(\frac{1+\cos2x}{2}\right)-3$
. . . . . $\displaystyle = \;\;2 + 2\cos2x - 3 \;\;=\;\;2\cos2x - 1$
Multiply by $\displaystyle \frac{\sin2x}{\sin2x}$
. . $\displaystyle \frac{\sin2x}{\sin2x}\cdot(2\cos2x - 1) \;\;=\;\;\frac{2\sin2x\cos2x - \sin2x}{\sin2x} \;\;=\;\;\frac{\sin4x - \sin2x}{\sin2x}$