In any triangle ABC,(a + b)^2 sin^2 frac{C}{2 | vedclass

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Question

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

Vedclass · Accepted Answer

$c^2$

Vedclass · Answer

(A) Expanding the expression: $(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2}$ $= (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2} + (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2}$ $= a^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + b^2(\sin^2 \frac{C}{2} + \cos^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$ $= a^2 + b^2 - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ $= a^2 + b^2 - 2ab \cos C$ Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$. Therefore,the expression equals $c^2$.

In any triangle $ABC$,$(a + b)^2 \sin^2 \frac{C}{2} + (a - b)^2 \cos^2 \frac{C}{2} =$

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