MHB What is the ratio of sin 5x to sin x in this Trigonometric Challenge?

[anemone]

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

 

[kaliprasad]

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

Spoiler

$\frac{\sin\, 3x}{\sin\, x}=\frac{6}{5}$ given
subtract 1 from both sides
$\frac{\sin\, 3x - \sin\, x}{\sin\, x}=\frac{1}{5}$
or $\frac{2 \sin\, x \cos\, 2x}{\sin\, x}=\frac{1}{5}$
or $\cos \, 2x = \frac{1}{10}\cdots(1)$
now
$\frac{\sin\, 5x + \sin \,x }{\sin x} = \frac{2 \sin\,3x \cos \,2x }{\sin x}$
$=2 \cos \,2x \frac{\sin\,3x}{\sin x}$
$=2 * \frac{1}{10} * \frac{6}{5} = \frac{6}{25}$
hence $\frac{\sin\, 5x}{\sin x} = \frac{6}{25} - 1 = - \frac{19}{25}$

 

[anemone]

Spoiler

$\frac{\sin\, 3x}{\sin\, x}=\frac{6}{5}$ given
subtract 1 from both sides
$\frac{\sin\, 3x - \sin\, x}{\sin\, x}=\frac{1}{5}$
or $\frac{2 \sin\, x \cos\, 2x}{\sin\, x}=\frac{1}{5}$
or $\cos \, 2x = \frac{1}{10}\cdots(1)$
now
$\frac{\sin\, 5x + \sin \,x }{\sin x} = \frac{2 \sin\,3x \cos \,2x }{\sin x}$
$=2 \cos \,2x \frac{\sin\,3x}{\sin x}$
$=2 * \frac{1}{10} * \frac{6}{5} = \frac{6}{25}$
hence $\frac{\sin\, 5x}{\sin x} = \frac{6}{25} - 1 = - \frac{19}{25}$

Good job kaliprasad!(Cool)