Identity on the three angles of a triangle

Problem

Prove that $\sin\alpha + \sin\beta + \sin\gamma = 4\cos\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}\cos\tfrac{\gamma}{2}$ when $\alpha+\beta+\gamma=\pi$ (for example if $\alpha,\beta,\gamma$ are the angles of a triangle). Hint: apply the sum-to-product formulae twice.

Solution

We group the first two terms with the sum-to-product formula $\sin\alpha+\sin\beta = 2\sin\!\tfrac{\alpha+\beta}{2}\cos\!\tfrac{\alpha-\beta}{2}$. Since $\alpha+\beta = \pi-\gamma$, we have $\tfrac{\alpha+\beta}{2} = \tfrac{\pi}{2}-\tfrac{\gamma}{2}$, so $\sin\!\tfrac{\alpha+\beta}{2} = \cos\tfrac{\gamma}{2}$: $\sin\alpha+\sin\beta = 2\cos\tfrac{\gamma}{2}\cos\tfrac{\alpha-\beta}{2}.$ For the third term we use the double-angle formula $\sin\gamma = 2\sin\tfrac{\gamma}{2}\cos\tfrac{\gamma}{2}$; moreover $\tfrac{\gamma}{2} = \tfrac{\pi}{2}-\tfrac{\alpha+\beta}{2}$, hence $\sin\tfrac{\gamma}{2} = \cos\tfrac{\alpha+\beta}{2}$: $\sin\gamma = 2\cos\tfrac{\gamma}{2}\cos\tfrac{\alpha+\beta}{2}.$ Adding and factoring out $2\cos\tfrac{\gamma}{2}$: $\sin\alpha+\sin\beta+\sin\gamma = 2\cos\tfrac{\gamma}{2}\left[\cos\tfrac{\alpha-\beta}{2}+\cos\tfrac{\alpha+\beta}{2}\right].$ The bracket is again a sum of cosines: with the sum-to-product formula $\cos p+\cos q = 2\cos\tfrac{p+q}{2}\cos\tfrac{p-q}{2}$, taking $p=\tfrac{\alpha-\beta}{2}$ and $q=\tfrac{\alpha+\beta}{2}$, one obtains $2\cos\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}$ (the cosine is even). Therefore $\sin\alpha+\sin\beta+\sin\gamma = 4\cos\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}\cos\tfrac{\gamma}{2}. \qquad\blacksquare$

Links

Topics: Prostaferesi werner
Concepts: Formule di prostaferesi
Functions: Coseno · Seno
Methods: Formule prostaferesi
Skills: Dimostrare · Usare formule
Exercise type: Dimostrazione