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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Prove that:
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$
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Answer

We have,
 $\text{LHS}=\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)$
$=\ (\sin\alpha+\sin\beta)+(\sin\gamma-\sin(\alpha+\beta+\gamma))$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{\gamma-(\alpha+\beta+\gamma)}{2}\Big)\cos\Big(\frac{\gamma+\alpha+\beta+\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{-\alpha-\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)-2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[\cos\Big(\frac{\alpha-\beta}{2}\Big)-\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)\Big]$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Bigg[-2\sin\frac{\Big[\frac{\alpha-\beta}{2}+\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\sin\frac{\Big[\frac{\alpha-\beta}{2}-\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\Bigg]$
$$$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[-2\sin\Big[\frac{2\alpha+2\gamma}{2\times2}\Big]\sin\Big[\frac{-2\beta-2\gamma}{2\times2}\Big]\Big]$
$=\ -4\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[\sin\Big(\frac{\alpha+\gamma}{2}\Big)\sin\Big[\frac{-(\beta+\gamma)}{2}\Big]\Big]$
$=\ 4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\alpha+\gamma}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)$
$=\ \text{RHS}$
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$ Hence proved.

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