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Prove that:

Question:

Prove that:

(i) $\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)=4 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\beta+\gamma}{2}\right) \sin \left(\frac{\gamma+\alpha}{2}\right)$

(ii) cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos B cos C

Solution:

(i) Consider LHS :

$\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{\gamma+\alpha+\beta+\gamma}{2}\right) \sin \left(\frac{\gamma-\alpha-\beta-\gamma}{2}\right)$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{2 \gamma+\alpha+\beta}{2}\right) \sin \left(\frac{-\alpha-\beta}{2}\right)$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{2 \gamma+\alpha+\beta}{2}\right) \sin \left[-\left(\frac{\alpha+\beta}{2}\right)\right]$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right)\left[\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{2 \gamma+\alpha+\beta}{2}\right)\right]$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right)\left[-2 \sin \left(\frac{\alpha-\beta+2 \gamma+\alpha+\beta}{4}\right) \sin \left(\frac{\alpha-\beta-2 \gamma-\alpha-\beta}{4}\right)\right]$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right)\left[-2 \sin \left(\frac{\alpha+\gamma}{2}\right) \sin \left(\frac{-\beta-\gamma}{2}\right)\right]$

$=2 \sin \left(\frac{\alpha+\beta}{2}\right)\left[2 \sin \left(\frac{\alpha+\gamma}{2}\right) s \operatorname{in}\left(\frac{\beta+\gamma}{2}\right)\right]$

$=4 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha+\gamma}{2}\right) \sin \left(\frac{\beta+\gamma}{2}\right)$

= RHS

Hence, LHS = RHS

(ii) Consider LHS :

$\cos (A+B+C)+\cos (A-B+C)+\cos (A+B-C)+\cos (-A+B+C)$

$=2 \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}-\mathrm{C}}{2}\right)+2 \cos \left(\frac{\mathrm{A}+\mathrm{B}-\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}+\mathrm{B}-\mathrm{C}+\mathrm{A}-\mathrm{B}-\mathrm{C}}{2}\right)$

$=2 \cos (A+\mathrm{C}) \cos B+2 \cos B \cos (A-C)$

$=2 \cos B[\cos (A+\mathrm{C})+\cos (A-C)]$

$=2 \cos B\left[2 \cos \left(\frac{A+C+A-C}{2}\right) \cos \left(\frac{A+C-A+C}{2}\right)\right]$

$=2 \cos B[2 \cos A \cos C]$

$=4 \cos A \cos B \cos C$

= RHS

Hence, LHS = RHS

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