If sin 'theta'/3 = 1/2 (m + 1/m), prove that: sin 'theta' = -1/2(m³ + 1/m³). | Trigonometry

Solution:

Given,

$sin \frac{\theta}{3} = \frac{1}{2} \left ( m + \dfrac{1}{m} \right )$

Now,

$sin \theta$

$= 3sin \frac{\theta}{3} - 4sin^3 \frac{\theta}{3}$

$= sin \frac{\theta}{3} \left ( 3 - 4 sin^2 \frac{\theta}{3} \right )$

$= sin \frac{\theta}{3} \left ( 3 - 4 \left [ \dfrac{1}{2} \left ( m + \frac{1}{m} \right ) \right ]^2 \right )$

$= sin \frac{\theta}{3} \left ( 3 - 4 × \dfrac{1}{4} \left ( m^2 + 2 + \frac{1}{m^2} \right ) \right )$

$= sin \frac{\theta}{3} \left (3 - m^2 - 2 - \frac{1}{m^2} \right )$

$= \dfrac{1}{2} \left (m + \dfrac{1}{m} \right ) × - \left ( \m^2 -1 + \dfrac{1}{m^2} \right )$

[Using (a + b)(a²-ab+b²) = a³+b³]

$= - \dfrac{1}{2} \left ( m^3+ \dfrac{1}{m^3} \right )$

RHS

#proved