Question:medium

If $\cos\alpha+\cos\beta+\cos\gamma = 0 = \sin\alpha+\sin\beta+\sin\gamma$, then $\sin 2\alpha + \sin 2\beta + \sin 2\gamma =$

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This is a standard result. If three points $e^{i\alpha}, e^{i\beta}, e^{i\gamma}$ on the unit circle have their centroid at the origin (i.e., their sum is zero), they must form an equilateral triangle. This implies the angles are separated by $2\pi/3$. The conditions also lead to $\sum \cos(2\alpha) = 0$, $\sum \sin(2\alpha) = 0$, $\sum \cos^2\alpha = 3/2$, $\sum \sin^2\alpha = 3/2$.
Updated On: Mar 26, 2026
  • $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)$
  • $\cos^2\alpha+\cos^2\beta+\cos^2\gamma$
  • $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
  • $\cos(2\alpha-\beta-\gamma)+\cos(2\beta-\gamma-\alpha)+\cos(2\gamma-\alpha-\beta)$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use Complex Numbers Approach Let \( a = e^{i\alpha}, b = e^{i\beta}, c = e^{i\gamma} \). Given condition: \[ \sum \cos \alpha = 0 \quad \text{and} \quad \sum \sin \alpha = 0 \] This implies \( a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 \).
Step 2: Square the Sum Since \( a + b + c = 0 \), squaring both sides gives: \[ (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0 \] \[ a^2 + b^2 + c^2 = -2(ab + bc + ca) \] Also, since \( |a|=|b|=|c|=1 \), we have \( \frac{1}{a} = \bar{a} \), etc. Conjugating \( a+b+c=0 \) gives \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \), which implies \( ab+bc+ca = 0 \). Thus, \( a^2 + b^2 + c^2 = 0 \).
Step 3: Analyze Imaginary Parts \( a^2 + b^2 + c^2 = e^{i2\alpha} + e^{i2\beta} + e^{i2\gamma} = (\cos 2\alpha + \dots) + i(\sin 2\alpha + \sin 2\beta + \sin 2\gamma) = 0 \). Comparing imaginary parts, we get: \[ \sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0 \]
Step 4: Check Options We need to find which option equals 0. Option (A): \( S = \cos(\alpha+\beta) + \cos(\beta+\gamma) + \cos(\gamma+\alpha) \). Using \( a+b+c=0 \implies a+b=-c \). Squaring modulus: \( |a+b|^2 = |c|^2 = 1 \implies 1+1+2\cos(\alpha-\beta)=1 \implies \cos(\alpha-\beta)=-1/2 \). This implies the angles differ by \( 120^\circ \) (equilateral triangle structure). Let \( \alpha = 0, \beta = 120^\circ, \gamma = 240^\circ \).
Target Value: \( \sin 0 + \sin 240 + \sin 480 = 0 - \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = 0 \).
Option (A): \( \cos(120) + \cos(360) + \cos(240) = -0.5 + 1 - 0.5 = 0 \).
Thus, Option (A) is the correct expression equal to the target value.
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