যদি sinα+ sinβ = a এবং cosα + cosβ = b হয়, তাহলে cos(α – β) এর মান কত?

Another Explanation (5): Given: \[ \sin \alpha + \sin \beta = a \] \[ \cos \alpha + \cos \beta = b \] Using the sum-to-product identities: \[ \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \] \[ \cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \] Let: \[ S = \frac{\alpha + \beta}{2} \] \[ D = \frac{\alpha - \beta}{2} \] Then: \[ a = 2 \sin S \cos D \] \[ b = 2 \cos S \cos D \] Dividing the first by the second: \[ \frac{a}{b} = \frac{2 \sin S \cos D}{2 \cos S \cos D} = \frac{\sin S}{\cos S} = \tan S \] So: \[ \sin S = \frac{a}{\sqrt{a^2 + b^2}} \] \[ \cos S = \frac{b}{\sqrt{a^2 + b^2}} \] Now, from the expressions: \[ a = 2 \sin S \cos D \] \[ b = 2 \cos S \cos D \] Square both: \[ a^2 = 4 \sin^2 S \cos^2 D \] \[ b^2 = 4 \cos^2 S \cos^2 D \] Add: \[ a^2 + b^2 = 4 \cos^2 D (\sin^2 S + \cos^2 S) = 4 \cos^2 D \times 1 = 4 \cos^2 D \] Therefore: \[ \cos^2 D = \frac{a^2 + b^2}{4} \] \[ \cos D = \frac{\sqrt{a^2 + b^2}}{2} \] Now, \(\cos (\alpha - \beta) = \cos 2D = 2 \cos^2 D - 1\): \[ \cos (\alpha - \beta) = 2 \left( \frac{\sqrt{a^2 + b^2}}{2} \right)^2 - 1 = 2 \times \frac{a^2 + b^2}{4} - 1 = \frac{a^2 + b^2}{2} - 1 \] **Final answer:**