# Problem Study (Trigonometric Identity)

Given that $\alpha&space;+\beta&space;+\gamma&space;=180^\circ$, Prove that $\cos\alpha&space;+\cos\beta&space;+\cos\gamma&space;=1+4\sin\frac{\alpha&space;}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$.
Solution
$\alpha&space;+\beta&space;+\gamma&space;=180^\circ$
$\alpha&space;+\beta&space;=180^\circ&space;-\gamma$
$\frac{\alpha&space;+\beta}{2}&space;=90^\circ&space;-\frac{\gamma}{2}$
$\frac{\gamma}{2}&space;=90^\circ&space;-\frac{\alpha&space;+\beta}{2}$
$\cos\alpha&space;+\cos\beta&space;+\cos\gamma$
$=(\cos\alpha&space;+\cos\beta)&space;+\cos2\left&space;(&space;\frac{\gamma&space;}{2}&space;\right&space;)$
$=2\cos\frac{\alpha&space;+\beta&space;}{2}\cos\frac{\alpha&space;-\beta&space;}{2}+1-2\sin^2\frac{\gamma&space;}{2}$
$=2\cos(90^\circ-\frac{\gamma&space;}{2}&space;)\cos\frac{\alpha&space;-\beta&space;}{2}+1-2\sin^2(\frac{\gamma&space;}{2})$
$=2\sin(\frac{\gamma&space;}{2})\cos\frac{\alpha&space;-\beta&space;}{2}+1-2\sin^2(\frac{\gamma&space;}{2})$
$=1+2\sin(\frac{\gamma&space;}{2})\cos\frac{\alpha&space;-\beta&space;}{2}-2\sin^2(\frac{\gamma&space;}{2})$
$=1&space;+&space;2\sin&space;\frac{\gamma&space;}{2}\left(&space;{\cos&space;\frac{{\alpha&space;-&space;\beta&space;}}{2}&space;-&space;\sin&space;\frac{\gamma&space;}{2}}&space;\right)$
$=1&space;+&space;2\sin&space;\frac{\gamma&space;}{2}\left(&space;{\cos&space;\frac{{\alpha&space;-&space;\beta&space;}}{2}&space;-&space;\sin&space;\left(&space;{{{90}^&space;\circ&space;}&space;-&space;\frac{{\alpha&space;+&space;\beta&space;}}{2}}&space;\right)}&space;\right)$
$=1&space;+&space;2\sin&space;\frac{\gamma&space;}{2}\left(&space;{\cos&space;\frac{{\alpha&space;-&space;\beta&space;}}{2}&space;-&space;\cos&space;\frac{{\alpha&space;+&space;\beta&space;}}{2}}&space;\right)$
$=1&space;+&space;2\sin&space;\frac{\gamma&space;}{2}\left(&space;{&space;-&space;2\sin&space;\frac{{\frac{{\alpha&space;-&space;\beta&space;}}{2}&space;+&space;\frac{{\alpha&space;+&space;\beta&space;}}{2}}}{2}\sin&space;\frac{{\frac{{\alpha&space;-&space;\beta&space;}}{2}&space;-&space;\frac{{\alpha&space;+&space;\beta&space;}}{2}}}{2}}&space;\right)$
= $&space;1&space;-&space;4\sin&space;\frac{\gamma&space;}{2}\sin&space;\frac{\alpha&space;}{2}\sin&space;\left(&space;{&space;-&space;\frac{\beta&space;}{2}}&space;\right)$
= $&space;1&space;+&space;4\sin&space;\frac{\gamma&space;}{2}\sin&space;\frac{\alpha&space;}{2}\sin&space;\frac{\beta&space;}{2}$  $(\because&space;\sin\left&space;(&space;-\frac{\beta&space;}{2}&space;\right&space;)=-\sin\frac{\beta&space;}{2})$