Understanding the Given Trigonometric Identities

We are given two conditions involving trigonometric functions of angles $\alpha$, $\beta$, and $\gamma$:

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

We need to find the value of $\sum \sin (\beta + \gamma)$, which is equivalent to $\sin (\beta + \gamma) + \sin (\gamma + \alpha) + \sin (\alpha + \beta)$. This involves sums of sines of angle combinations, and can be elegantly solved using properties related to trigonometric identities and complex numbers.

Solving using Complex Numbers and Trigonometric Identities

A powerful technique to handle sums of sines and cosines is using complex numbers. We can define three complex numbers based on the angles:

• Let $x = \cos \alpha + i \sin \alpha = e^{i\alpha}$
• Let $y = \cos \beta + i \sin \beta = e^{i\beta}$
• Let $z = \cos \gamma + i \sin \gamma = e^{i\gamma}$

The given conditions $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ can be combined in terms of these complex numbers:

$(\cos \alpha + \cos \beta + \cos \gamma) + i (\sin \alpha + \sin \beta + \sin \gamma) = 0 + i \cdot 0$

This means $x + y + z = 0$.

Since $x, y, z$ are of the form $e^{i\theta}$, their magnitudes are $|x|=|y|=|z|=1$. A key property relating complex numbers and trigonometric identities is that if $x+y+z=0$ and $|x|=|y|=|z|=1$, then $xy + yz + zx = 0$. Let's see why.

If $x+y+z=0$, then $1/x + 1/y + 1/z = \bar{x} + \bar{y} + \bar{z}$ because for a complex number $w$ with $|w|=1$, $1/w = \bar{w}$.

$\bar{x} + \bar{y} + \bar{z} = (\cos \alpha - i \sin \alpha) + (\cos \beta - i \sin \beta) + (\cos \gamma - i \sin \gamma)$

$= (\cos \alpha + \cos \beta + \cos \gamma) - i (\sin \alpha + \sin \beta + \sin \gamma)$

Using the given conditions, this is $0 - i \cdot 0 = 0$.

So, $1/x + 1/y + 1/z = 0$. Multiplying by $xyz$ (which is non-zero as $|xyz|=1$), we get $yz + xz + xy = 0$. This is a direct result derived using trigonometric identities and complex number properties.

Evaluating the Trigonometric Sum

Now, let's look at the expression $xy + yz + zx$ in terms of the original angles:

• $xy = e^{i\alpha} e^{i\beta} = e^{i(\alpha+\beta)} = \cos(\alpha+\beta) + i \sin(\alpha+\beta)$
• $yz = e^{i\beta} e^{i\gamma} = e^{i(\beta+\gamma)} = \cos(\beta+\gamma) + i \sin(\beta+\gamma)$
• $zx = e^{i\gamma} e^{i\alpha} = e^{i(\gamma+\alpha)} = \cos(\gamma+\alpha) + i \sin(\gamma+\alpha)$

Adding these three terms:

$xy + yz + zx = (\cos(\alpha+\beta) + i \sin(\alpha+\beta)) + (\cos(\beta+\gamma) + i \sin(\beta+\gamma)) + (\cos(\gamma+\alpha) + i \sin(\gamma+\alpha))$

$= (\cos(\alpha+\beta) + \cos(\beta+\gamma) + \cos(\gamma+\alpha)) + i (\sin(\alpha+\beta) + \sin(\beta+\gamma) + \sin(\gamma+\alpha))$

$= \sum \cos(\alpha+\beta) + i \sum \sin(\beta+\gamma)$.

We established earlier using complex numbers and trigonometric identities that $xy + yz + zx = 0$. Substituting this into the equation above:

$\sum \cos(\alpha+\beta) + i \sum \sin(\beta+\gamma) = 0$.

For a complex number to be equal to zero, both its real part and its imaginary part must be zero.

• Real part: $\sum \cos(\alpha+\beta) = \cos(\alpha+\beta) + \cos(\beta+\gamma) + \cos(\gamma+\alpha) = 0$.
• Imaginary part: $\sum \sin(\beta+\gamma) = \sin(\beta+\gamma) + \sin(\gamma+\alpha) + \sin(\alpha+\beta) = 0$.

The value we need to find is the imaginary part of this sum, which is $\sum \sin (\beta + \gamma)$. From the above analysis, this value is 0. This result demonstrates the power of combining complex numbers with trigonometric identities to solve such problems involving trigonometric sum conditions.

Thus, $\sum \sin (\beta + \gamma) = 0$.