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प्रश्न
If (cos α + cos β)2 + (sin α + sin β)2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\], write the value of λ.
उत्तर
(cos α + cos β)2 + (sin α + sin β)2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Consider LHS:
(cos α + cos β)2 + (sin α + sin β)2
\[= \left[ 2\cos \left( \frac{\alpha + \beta}{2} \right) cos \left( \frac{\alpha - \beta}{2} \right) \right]^2 + \left[ 2\sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \right]^2 \]
\[ = 4 \cos^2 \left( \frac{\alpha + \beta}{2} \right) \cos^2 \left( \frac{\alpha - \beta}{2} \right) + 4 \sin^2 \left( \frac{\alpha + \beta}{2} \right) \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
\[ = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\left[ \cos^2 \left( \frac{\alpha + \beta}{2} \right) + \sin^2 \left( \frac{\alpha + \beta}{2} \right) \right]\]
\[ = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
= RHS
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