Sinα+sinβ+sinγ-sin(α+β+γ) = 4 sin α+β/2 sin β+γ/2 sin γ +α/2

Sinα+sinβ+sinγ-sin(α+β+γ) = 4 sin α+β/2 sin β+γ/2 sin γ +α/2 || $Sin\alpha + sin\beta + sin\gamma – sin(\alpha + \beta + \gamma )$ = $4sin\frac{\alpha + \beta }{2}sin\frac{\beta + \gamma }{2}sin\frac{\gamma + \alpha }{2}$

Sinα+sinβ+sinγ-sin(α+β+γ) = 4 sin α+β/2 sin β+γ/2 sin γ +α/2

LHS= Sinα+sinβ+sinγ-sin(α+β+γ)

= 2 sin $\frac{\alpha+\beta}{2}$ cos $\frac{\alpha-\beta}{2}$ + 2 cos $\frac{\gamma  + \alpha  + \beta  + \gamma }{2}$ sin $\frac{\gamma  – \alpha  – \beta  – \gamma }{2}$

= 2 sin $\frac{\alpha+\beta}{2}$ cos $\frac{\alpha-\beta}{2}$ – 2 cos $\frac{\alpha  + \beta  + 2\gamma }{2}$ sin $\frac{\alpha+\beta}{2}$

=2 sin $\frac{\alpha+\beta}{2}$ {cos $\frac{\alpha-\beta}{2}$ – cos $\frac{\alpha  + \beta  + 2\gamma }{2}$}  

= 2 sin $\frac{\alpha+\beta}{2}$ ( 2 sin $\frac{\frac{\alpha  – \beta }{2} + \frac{\alpha  + \beta  + 2\gamma }{2}}{2}$ sin $\frac{\frac{\alpha  + \beta  + 2\gamma }{2} – \frac{\alpha  – \beta }{2}}{2}$)

= 4 sin $\frac{\alpha+\beta}{2}$ sin $\frac{\alpha  + \beta  + 2\gamma  + \alpha  – \beta }{4}$ sin $\frac{\alpha  + \beta  + 2\gamma  – \alpha  + \beta }{4}$

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

=RHS [Proved]

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