Answer: see proof below
Step-by-step explanation:
Use the Triple Angle Identity: sin 3A = 3sinA - 4sin³A
[tex]\text{Given:}\quad \sin\bigg(\dfrac{\theta}{3}\bigg)=\dfrac{1}{2}\bigg(m+\dfrac{1}{m}\bigg)[/tex]
Proof LHS → RHS
LHS: sin Ф
Let A = Ф/3 sin 3A
Triple Angle Identity: 3sinA - 4sin³A
Substitute Ф/3 = A: 3sin(Ф/3) - 4sin³(Ф/3)
Substitute sin(Ф/3): [tex]3\bigg(\dfrac{1}{2}\bigg(m+\dfrac{1}{m}\bigg)\bigg)-4\bigg(\dfrac{1}{2}\bigg(m+\dfrac{1}{m}\bigg)\bigg)^3[/tex]
[tex]\text{Simplify:}\qquad \qquad \dfrac{3m}{2}+\dfrac{3}{2m}-4\bigg(\dfrac{1}{8}\bigg(m^3+3m+\dfrac{3}{m}+\dfrac{1}{m^3}\bigg)\bigg)\\\\.\qquad \qquad \qquad =\dfrac{3m}{2}+\dfrac{3}{2m}-\dfrac{m^3}{2}-\dfrac{3m}{2}-\dfrac{3}{2m}-\dfrac{1}{2m^3}\\\\.\qquad \qquad \qquad =-\dfrac{m^3}{2}-\dfrac{1}{2m^3}\\\\\text{Factor:}\qquad \qquad -\dfrac{1}{2}\bigg(m^3+\dfrac{1}{m^3}\bigg)[/tex]
[tex]\text{LHS = RHS:}\quad -\dfrac{1}{2}\bigg(m^3+\dfrac{1}{m^3}\bigg)=-\dfrac{1}{2}\bigg(m^3+\dfrac{1}{m^3}\bigg)\quad \checkmark\\[/tex]
