Given that $\displaystyle \frac{\sin 3x}{\sin x}=\frac{6}{5}$, what is the ratio of $\displaystyle \frac{\sin 5x}{\sin x}$?

Given that $\displaystyle \frac{\sin 3x}{\sin x}=\frac{6}{5}$, what is the ratio of $\displaystyle \frac{\sin 5x}{\sin x}$?

My solution:

From the "Componendo And Dividendo Rule", we have:

$\displaystyle \frac{\sin 3x}{\sin x}=\frac{6}{5}---(1)$

$\displaystyle \frac{\sin 3x-\sin x}{\sin x}=\frac{6-5}{5}$

$\displaystyle \frac{2\cos \left(\frac{3x+x}{2}\right)\sin \left(\frac{3x-x}{2}\right)}{\sin x}=\frac{1}{5}$

$\displaystyle \frac{2\cos 2x\sin x}{\sin x}=\frac{1}{5}$

$\displaystyle 2\cos 2x=\frac{1}{5}$

$\displaystyle \cos 2x=\frac{1}{10}$

Now we let $\displaystyle \frac{\sin 5x}{\sin x}=k---(2)$.

Subtracting the equations (1) from (2) we get:

$\displaystyle \frac{\sin 5x-\sin 3x}{\sin x}=k-\frac{6}{5}$

$\displaystyle \frac{2\cos \left(\frac{5x+3x}{2}\right) \sin \left(\frac{5x-3x}{2}\right)}{\sin x}=k-\frac{6}{5}$

$\displaystyle \frac{2\cos 4x \sin x}{\sin x}=k-\frac{6}{5}$

$\displaystyle 2\cos 4x =k-\frac{6}{5}$

$\displaystyle 2(2\cos^2 2x-1) =k-\frac{6}{5}$

$\displaystyle 2\left(2\left(\frac{1}{10}\right)^2 -1\right) =k-\frac{6}{5}$

$\displaystyle k=2\left(2\left(\frac{1}{10}\right)^2 -1\right) +\frac{6}{5}=-\frac{19}{25}$, therefore

$\displaystyle \frac{\sin 5x}{\sin x}=k=-\frac{19}{25}$