If alpha, beta, gamma are any three angles,th | vedclass

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(B) We use the sum-to-product formulas: $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We use the sum-to-product formulas: $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.Given expression: $E = (\cos \alpha + \cos \beta) - (\cos \gamma + \cos (\alpha + \beta + \gamma))$.Applying the formulas:$E = 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} - 2 \cos \frac{\alpha+\beta+2\gamma}{2} \cos \frac{\alpha+\beta}{2}$.Factor out $2 \cos \frac{\alpha+\beta}{2}$:$E = 2 \cos \frac{\alpha+\beta}{2} [\cos \frac{\alpha-\beta}{2} - \cos \frac{\alpha+\beta+2\gamma}{2}]$.Using $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$:$E = 2 \cos \frac{\alpha+\beta}{2} [2 \sin \frac{\alpha+\gamma}{2} \sin \frac{\beta+\gamma}{2}]$.Thus,$E = 4 \cos \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2}$.

If $\alpha, \beta, \gamma$ are any three angles,then $\cos \alpha + \cos \beta - \cos \gamma - \cos (\alpha + \beta + \gamma) =$

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