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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
If cos⁡α+cos⁡β+cos⁡γ=0=sin⁡α+sin⁡β+sin⁡γ\cos\alpha + \cos\beta + \cos\gamma = 0 = \sin\alpha + \sin\beta + \sin\gamma, then sin⁡3α+sin⁡3β+sin⁡3γsin⁡(α+β+γ)\dfrac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)} is equal to
A00
B11
C22
D33correct
Solution
Step 1: Introduce unit complex numbers a=eiαa = e^{i\alpha}, b=eiβb = e^{i\beta}, c=eiγc = e^{i\gamma}. Then:
a+b+c=(cos⁡α+cos⁡β+cos⁡γ)+i(sin⁡α+sin⁡β+sin⁡γ)=0+i⋅0=0a + 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=0a + b + c = 0:
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
 Since a+b+c=0a + b + c = 0, the right side is 00, hence:
a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc
 Step 3: Write each side in exponential form. By De Moivre's theorem, a3=ei3αa^3 = e^{i3\alpha}, b3=ei3βb^3 = e^{i3\beta}, c3=ei3γc^3 = e^{i3\gamma}, and abc=ei(α+β+γ)abc = e^{i(\alpha+\beta+\gamma)}:
ei3α+ei3β+ei3γ=3ei(α+β+γ)e^{i3\alpha} + e^{i3\beta} + e^{i3\gamma} = 3e^{i(\alpha+\beta+\gamma)}
 Step 4: Equate the imaginary parts of both sides:
sin⁡3α+sin⁡3β+sin⁡3γ=3sin⁡(α+β+γ)\sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3\sin(\alpha + \beta + \gamma)
 Step 5: Therefore the required ratio is:
sin⁡3α+sin⁡3β+sin⁡3γsin⁡(α+β+γ)=3\frac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)} = 3
 Correct answer: (4)
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