Respuesta :

Answer:

It is proved .

Step-by-step explanation:

[tex]\alpha +\beta +\gamma =\pi.....(1)\\\\sin\alpha + sin\beta +sin\gamma=4 cos\frac{\alpha}{2}cos\frac{\beta}{2}cos\frac{\gamma}{2}[/tex]

Take LHS

[tex]=sin\alpha + sin\beta +sin\gamma\\\\=2 sin \frac{\alpha +\beta }{2} cos\frac{\alpha -\beta }{2} + sin\gamma\\\\=2 sin\frac{180-\gamma}{2}cos \frac{\alpha - \beta}{2}+2 sin\frac{\gamma}{2}cos\frac{\gamma}{2}\\\\=2 cos\frac{\gamma}{2}cos \frac{\alpha - \beta}{2}+2 sin\frac{\gamma}{2}cos\frac{\gamma}{2}\\\\=2 cos\frac{\gamma}{2}\left [ cos \frac{\alpha - \beta}{2}+sin\frac{180-\alpha-\beta}{2}\\ \right ]\\\\=2 cos\frac{\gamma}{2}\times 2 cos\frac{\alpha}{2}cos\frac{\beta}{2}\\[/tex]

[tex]=4 cos\frac{\alpha}{2}cos\frac{\beta}{2}cos\frac{\gamma}{2}[/tex]

Hence proved.

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