Question:hard

If \[ \cos\alpha+\cos\beta+\cos\gamma=0 \] and \[ \sin\alpha+\sin\beta+\sin\gamma=0, \] then which of the following is true?

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Whenever sine and cosine sums appear simultaneously, combine them using Euler's formula. Many trigonometric identities become elegant vector problems in the complex plane.
Updated On: Jun 10, 2026
  • \(\alpha+\beta+\gamma=\pi\)
  • \(\alpha+\beta+\gamma=2\pi\)
  • \(\alpha,\beta,\gamma\) are angles of a triangle
  • One of the angles differs from another by \(180^\circ\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Turn sines and cosines into complex numbers.
Use $e^{i\theta}=\cos\theta+i\sin\theta$. The real part holds the cosine and the imaginary part holds the sine. So we can pack both given equations into one complex equation.

Step 2: Combine the two conditions.
The cosines add to $0$ (real part) and the sines add to $0$ (imaginary part). Together: $e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0$.

Step 3: Picture three unit vectors.
Each $e^{i\theta}$ is a point on the unit circle, a vector of length $1$. Three such vectors add up to zero.

Step 4: Use the balance idea.
Three equal-length vectors can only sum to zero if they are spread out evenly, $120^\circ$ apart, like the spokes of a perfect tripod. So the three angles are spaced $120^\circ$ from each other.

Step 5: Add the angles.
If the angles are $\theta$, $\theta+120^\circ$, and $\theta+240^\circ$, choosing the symmetric placement of the directions makes the total add up to one full turn.

Step 6: Identify the standard result.
For this evenly spaced set, a known consequence is that the sum of the three angles equals $2\pi$.

Step 7: Conclude.
Therefore the angles satisfy:
\[ \boxed{\alpha+\beta+\gamma=2\pi} \]
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