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sin3α+sin3β+sin3γ = 3sin(α+β+γ) Proof | Complex JEE

JEE Maths question with a full step-by-step solution.

Question

\cos\alpha + \cos\beta + \cos\gamma = 0 = \sin\alpha + \sin\beta + \sin\gamma

, then

\dfrac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)}

is equal to

0

1

2

3

correct

Solution

Step 1: Introduce unit complex numbers

a = e^{i\alpha}

b = e^{i\beta}

c = e^{i\gamma}

. Then:

a + b + c = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma) = 0 + i\cdot 0 = 0

Step 2: Apply the algebraic identity valid whenever

a + b + c = 0

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

Since

a + b + c = 0

, the right side is

0

, hence:

a^3 + b^3 + c^3 = 3abc

Step 3: Write each side in exponential form. By De Moivre's theorem,

a^3 = e^{i3\alpha}

b^3 = e^{i3\beta}

c^3 = e^{i3\gamma}

, and

abc = e^{i(\alpha+\beta+\gamma)}

e^{i3\alpha} + e^{i3\beta} + e^{i3\gamma} = 3e^{i(\alpha+\beta+\gamma)}

Step 4: Equate the imaginary parts of both sides:

\sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin(\alpha + \beta + \gamma)

Step 5: Therefore the required ratio is:

\frac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)} = 3

Correct answer: (4)

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