What is the value of (a-b)^2 cos^2 frac{C}{2} | vedclass

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(A) Given expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ $= (a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \fr

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$c^2$

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(A) Given expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ $= (a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2}$ $= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$ Since $\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2} = 1$ and $\cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}$,we have: $= (a^2 + b^2)(1) - 2ab(\cos C)$ Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$: $= a^2 + b^2 - 2ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right)$ $= a^2 + b^2 - (a^2 + b^2 - c^2)$ $= c^2$

What is the value of $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$?

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