Question

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

• A. $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)$
• B. $\cos^2\alpha+\cos^2\beta+\cos^2\gamma$
• C. $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
• D. $\cos(2\alpha-\beta-\gamma)+\cos(2\beta-\gamma-\alpha)+\cos(2\gamma-\alpha-\beta)$

Hint:

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 $\sum \cos(2\alpha) = 0$, $\sum \sin(2\alpha) = 0$, $\sum \cos^2\alpha = 3/2$, $\sum \sin^2\alpha = 3/2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This problem can be elegantly solved using complex numbers. The given conditions can be interpreted as the sum of three complex numbers on the unit circle being zero.
Step 2: Key Formula or Approach
Let $x = \cos\alpha + i\sin\alpha = e^{i\alpha}$, $y = \cos\beta + i\sin\beta = e^{i\beta}$, and $z = \cos\gamma + i\sin\gamma = e^{i\gamma}$. The given conditions translate to $x+y+z=0$. We can then use algebraic identities to find the required expression, which is the imaginary part of $x^2+y^2+z^2$.
Step 3: Detailed Explanation
From the given information:
\[ (\cos\alpha+\cos\beta+\cos\gamma) + i(\sin\alpha+\sin\beta+\sin\gamma) = 0 + i(0) = 0 \] This is equivalent to:
\[ (\cos\alpha+i\sin\alpha) + (\cos\beta+i\sin\beta) + (\cos\gamma+i\sin\gamma) = 0 \] Let $x=e^{i\alpha}, y=e^{i\beta}, z=e^{i\gamma}$. The condition is $x+y+z=0$.
Since $x, y, z$ are on the unit circle, their reciprocals are their conjugates:
$1/x = e^{-i\alpha} = \cos\alpha - i\sin\alpha$
$1/y = e^{-i\beta} = \cos\beta - i\sin\beta$
$1/z = e^{-i\gamma} = \cos\gamma - i\sin\gamma$
Let's find the sum of the reciprocals:
\[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = (\cos\alpha+\cos\beta+\cos\gamma) - i(\sin\alpha+\sin\beta+\sin\gamma) = 0 - i(0) = 0 \]
Taking a common denominator, we get $\frac{yz+xz+xy}{xyz} = 0$, which implies $xy+yz+zx = 0$.
Now, consider the identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$.
Since $x+y+z=0$ and $xy+yz+zx=0$, we have:
\[ 0^2 = x^2+y^2+z^2 + 2(0) \implies x^2+y^2+z^2 = 0 \] Let's expand $x^2+y^2+z^2$ back into trigonometric form:
\[ (e^{i\alpha})^2 + (e^{i\beta})^2 + (e^{i\gamma})^2 = 0 \] \[ e^{i2\alpha} + e^{i2\beta} + e^{i2\gamma} = 0 \] \[ (\cos 2\alpha + i\sin 2\alpha) + (\cos 2\beta + i\sin 2\beta) + (\cos 2\gamma + i\sin 2\gamma) = 0 \] \[ (\cos 2\alpha + \cos 2\beta + \cos 2\gamma) + i(\sin 2\alpha + \sin 2\beta + \sin 2\gamma) = 0 \]
For this complex number to be zero, both its real and imaginary parts must be zero.
Therefore, $\sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0$.
Now we must check which of the options is also equal to 0.
Let's evaluate Option (A): $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)$.
This expression is the real part of $e^{i(\alpha+\beta)} + e^{i(\beta+\gamma)} + e^{i(\gamma+\alpha)}$.
$e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta} = xy$
$e^{i(\beta+\gamma)} = e^{i\beta}e^{i\gamma} = yz$
$e^{i(\gamma+\alpha)} = e^{i\gamma}e^{i\alpha} = zx$
So, the expression is $\text{Re}(xy+yz+zx)$.
We already proved that $xy+yz+zx=0$.
Therefore, $\text{Re}(0) = 0$.
So, $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha) = 0$.
Step 4: Final Answer
The value of the expression in the question, $\sin 2\alpha + \sin 2\beta + \sin 2\gamma$, is 0. The value of the expression in Option (A), $\cos(\alpha+\beta)+\cos(\beta+\gamma)+\cos(\gamma+\alpha)$, is also 0. Therefore, they are equal.