## Tuesday, June 21, 2016

### Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

My solution:

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

[MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}---(1)[/MATH]

[MATH]\frac{\sin 3x-\sin x}{\sin x}=\frac{6-5}{5}[/MATH]

[MATH]\frac{2\cos \left(\frac{3x+x}{2}\right)\sin \left(\frac{3x-x}{2}\right)}{\sin x}=\frac{1}{5}[/MATH]

[MATH]\frac{2\cos 2x\sin x}{\sin x}=\frac{1}{5}[/MATH]

[MATH]2\cos 2x=\frac{1}{5}[/MATH]

[MATH]\cos 2x=\frac{1}{10}[/MATH]

Now we let [MATH]\frac{\sin 5x}{\sin x}=k---(2)[/MATH].

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

[MATH]\frac{\sin 5x-\sin 3x}{\sin x}=k-\frac{6}{5}[/MATH]

[MATH]\frac{2\cos \left(\frac{5x+3x}{2}\right) \sin \left(\frac{5x-3x}{2}\right)}{\sin x}=k-\frac{6}{5}[/MATH]

[MATH]\frac{2\cos 4x \sin x}{\sin x}=k-\frac{6}{5}[/MATH]

[MATH]2\cos 4x =k-\frac{6}{5}[/MATH]

[MATH]2(2\cos^2 2x-1) =k-\frac{6}{5}[/MATH]

[MATH]2\left(2\left(\frac{1}{10}\right)^2 -1\right) =k-\frac{6}{5}[/MATH]

[MATH]k=2\left(2\left(\frac{1}{10}\right)^2 -1\right) +\frac{6}{5}=-\frac{19}{25}[/MATH], therefore

[MATH]\frac{\sin 5x}{\sin x}=k=-\frac{19}{25}[/MATH]