Prove that: \[ \sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma) \] \[ = 4\sin\frac{\alpha+\beta}{2} \sin\frac{\beta+\gamma}{2} \sin\frac{\gamma+\alpha}{2} \]
Solution
L.H.S.
\[ = \sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma) \]Group first two terms and last two terms.
Use identity:
\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
2\sin\frac{\alpha+\beta}{2}
\cos\frac{\alpha-\beta}{2}
\]
\[
+
\sin\gamma-\sin(\alpha+\beta+\gamma)
\]
Use identity:
\[ \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
=
2\sin\frac{\alpha+\beta}{2}
\cos\frac{\alpha-\beta}{2}
\]
\[
–
2\cos\frac{\alpha+\beta+2\gamma}{2}
\sin\frac{\alpha+\beta}{2}
\]
Take common factor:
\[ = 2\sin\frac{\alpha+\beta}{2} \left[ \cos\frac{\alpha-\beta}{2} – \cos\frac{\alpha+\beta+2\gamma}{2} \right] \]Use identity:
\[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \]
\[
=
2\sin\frac{\alpha+\beta}{2}
\cdot
2\sin\frac{\beta+\gamma}{2}
\sin\frac{\gamma+\alpha}{2}
\]
\[
=
4\sin\frac{\alpha+\beta}{2}
\sin\frac{\beta+\gamma}{2}
\sin\frac{\gamma+\alpha}{2}
\]
Hence Proved.