Question

Please someone help me..!!!!​

Answer 1

Answer: see proof below

Explanation:

Use the Triple Angle Identity: sin 3A = 3sinA - 4sin³A

`\text{Given:}\quad \sin\bigg((\theta)/(3)\bigg)=(1)/(2)\bigg(m+(1)/(m)\bigg)`

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):
`3\bigg((1)/(2)\bigg(m+(1)/(m)\bigg)\bigg)-4\bigg((1)/(2)\bigg(m+(1)/(m)\bigg)\bigg)^3`

`\text{Simplify:}\qquad \qquad (3m)/(2)+(3)/(2m)-4\bigg((1)/(8)\bigg(m^3+3m+(3)/(m)+(1)/(m^3)\bigg)\bigg)\\\\.\qquad \qquad \qquad =(3m)/(2)+(3)/(2m)-(m^3)/(2)-(3m)/(2)-(3)/(2m)-(1)/(2m^3)\\\\.\qquad \qquad \qquad =-(m^3)/(2)-(1)/(2m^3)\\\\\text{Factor:}\qquad \qquad -(1)/(2)\bigg(m^3+(1)/(m^3)\bigg)`

`\text{LHS = RHS:}\quad -(1)/(2)\bigg(m^3+(1)/(m^3)\bigg)=-(1)/(2)\bigg(m^3+(1)/(m^3)\bigg)\quad \checkmark\\`