# One Trigonometric Formula and Its Consequences

The formula below that holds for any angles $\alpha,$ $\beta,$ $\gamma$ has a two step proof and great many consequences:

(*)

$\displaystyle\sin\alpha +\sin \beta +\sin \gamma -\sin (\alpha + \beta + \gamma )=4\sin\frac{\alpha +\beta}{2}\sin\frac{\beta +\gamma}{2}\sin\frac{\gamma +\alpha}{2}.$

Proof

From the Addition formulas for sine and cosine one can easily get important formulas for the sum and difference of two sines or of two cosines:

(1)

$\displaystyle\sin\alpha +\sin \beta = 2\sin\frac{\alpha +\beta}{2}\cos\frac{\alpha -\beta}{2},$

(2)

$\displaystyle\sin\alpha -\sin \beta = 2\sin\frac{\alpha -\beta}{2}\cos\frac{\alpha +\beta}{2},$

(3)

$\displaystyle\cos\alpha +\cos \beta = 2\cos\frac{\alpha +\beta}{2}\cos\frac{\alpha -\beta}{2},$

(4)

$\displaystyle\cos\alpha -\cos \beta = 2\sin\frac{\alpha +\beta}{2}\sin\frac{\beta-\alpha}{2},$

For example, since

$\sin(a \pm b) =\sin (a) \cos (b) \pm \cos (a)\sin (b)$

then

$\sin(a + b) +\sin (a - b) = 2\sin (a) \cos (b).$

Now (1) follows by setting $\alpha = a + b$ and $\beta = a - b$ and solving the system for $a$ and $b:$

\displaystyle\begin{align} a &= (\alpha + \beta ) / 2,\\ b &= (\alpha - \beta ) / 2. \end{align}

To prove (*),

(*)

$\displaystyle\sin\alpha +\sin \beta +\sin \gamma -\sin (\alpha + \beta + \gamma )=4\sin\frac{\alpha +\beta}{2}\sin\frac{\beta +\gamma}{2}\sin\frac{\gamma +\alpha}{2}.$

we apply (1) and (2) to the left side and subsequently (4) to the result:

\displaystyle\begin{align} \sin\alpha &+\sin \beta +\sin \gamma -\sin (\alpha + \beta + \gamma )\\ &=2\sin\frac{\alpha + \beta}{2} \cos\frac{\alpha - \beta}{2} - 2\sin\frac{\alpha + \beta}{2} \cos\frac{\gamma + (\alpha + \beta )}{2}\\ &=2\sin\frac{\alpha + \beta}{2} \left[\cos\frac{\alpha - \beta}{2} - \cos\frac{\gamma + (\alpha + \beta )}{2}\right]\\ &=4\sin\frac{\alpha + \beta}{2}\sin\frac{\beta + \gamma}{2}\sin\frac{\gamma + \alpha}{2}, \end{align}

as required.

When $\alpha,$ $\beta,$ $\gamma$ are angles of a triangle, i.e., when, in particular, $\alpha + \beta + \gamma = \pi,$ we may observe that, for example,

$\displaystyle\frac{\alpha + \beta}{2} = \frac{\pi}{2} - \frac{\gamma}{2},$

which allows to reduces (*) to

(**)

$\displaystyle\sin\alpha +\sin \beta +\sin \gamma = 4 \cos\frac{\alpha}{2} \cos\frac{\beta}{2} \cos\frac{\gamma}{2},$

because $\displaystyle\sin (\frac{\pi}{2} - x) = \cos (x).$

The latter identity (**) has further consequences. Right now, I'll give a second, more direct proof of (**) [Stanford Problem Book, #47.4]. It's not easier than the above and I do not know if it admits a simpler proof than the more general (*).

Let $\alpha = 2u,$ $\beta = 2v,$ $\gamma = \pi - 2u - 2v.$ Then, using the formula for the double argument $\sin(2x) = 2\sin (x)\cos (x),$ (**) is transformed into

\begin{align} \sin(2u) +\sin (2v) &= 2[2\cos (u)\cos (v) - \cos (u + v)]\sin(u + v)\\ &= 2[\cos (u)\cos (v) +\sin (u)\sin(v)]\sin(u + v)\\ &= 2\cos (u - v)\sin (u + v) \end{align}

which is actually (1) and hence true.

### Reference

1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009
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