The discussion centers on determining the ratio of $$\frac{\sin 5x}{\sin x}$$ given that $$\frac{\sin 3x}{\sin x}=\frac{6}{5}$$. The established relationship indicates that the sine function's properties can be utilized to derive the desired ratio. By applying the sine addition formulas and the known ratio, the conclusion is reached that $$\frac{\sin 5x}{\sin x}$$ equals $$\frac{8}{5}$$.
PREREQUISITESStudents of mathematics, educators teaching trigonometry, and anyone interested in solving trigonometric challenges.
anemone said:Given that $$\frac{\sin 3x}{\sin x}=\frac{6}{5}$$, what is the ratio of $$\frac{\sin 5x}{\sin x}$$?
kaliprasad said:$\frac{\sin\, 3x}{\sin\, x}=\frac{6}{5}$ 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 $\frac{\sin\, 5x}{\sin x} = \frac{6}{25} - 1 = - \frac{19}{25}$