# Prove that $\frac{\sin3A}{\sin2A-\sin A} = 2\cos A+1$.

Prove $$\frac{\sin3A}{\sin2A-\sin A} = 2\cos A+1$$

I am confused on what to do after changing $\sin 3A$ into $\sin(2A+A)$ and then applying the compound angle rule to make $\sin2A\cos A - \cos2A\sin A$. I also changed the $\sin2A$ in the denominator into $2\sin A\cos A$ using the double angle rule. What do I do now?

We have

$$(\sin(2x)-\sin x)(2\cos x+1)=\left(\frac{e^{2ix}-e^{-2ix}-e^{ix}+e^{-ix}}{2i}\right)\left( e^{ix}+e^{-ix}+1\right)=\frac{e^{3ix}-e^{-3ix}}{2i}=\sin(3x)$$

$$\frac{\sin 3A}{\sin 2A-\sin A}=\frac{\sin 3A}{2\sin A/2\cos 3A/2}\\=\frac{\sin 3A/2}{\sin A/2}=\frac{\sin A\cos A/2+\cos A\sin A/2}{\sin A/2}\\=\sin A\cot A/2+\cos A=2\cos^2A/2+\cos A=\cos A+1+\cos A=2\cos A+1$$

$sin3A = 3sinA - 4sin^3A$, and $sin2A = 2sinAcosA$. So:

$LHS = \dfrac{3 - 4sin^2A}{2cosA - 1} = \dfrac{3 - 4(1 - cos^2A)}{2cosA - 1} = \dfrac{4cos^2A - 1}{2cosA - 1} = 2cosA + 1 = RHS$

Hint:
Prove that: $$\big(2\cos A+1\big)\big(\sin(2A)-\sin A\big)=3\sin a -4\sin^3a\ ,$$ Then use the identity $$\sin (3a)=3\sin a -4\sin^3a$$ to conclude.

Note that his can be done directly as follows: $$\sin 2A - \sin A=\sin A \cdot(2\cos A-1)$$

So you'd like to find a factor $\sin A$ to cancel from the numerator and express the remainder in terms of $\cos A$ - likely possible because the numerator is an odd function.

$$\sin 3A=\sin A \cos 2A+\cos A \sin 2A=\sin A\cdot(2\cos^2A-1)+2\sin A \cos^2 A=$$ now we have the right form, and you should be able to take it from there.