Question
In $\triangle \mathrm{ABC}$ prove that $(\mathrm{a}-\mathrm{b})^2 2 \cos ^2 \frac{\mathrm{C}}{2}+(\mathrm{a}+\mathrm{b})^2 \sin ^2 \frac{\mathrm{C}}{2}=c^2$.
✓
Answer
LHS $(a-b)^2 2 \cos ^2 \frac{\mathrm{C}}{2}+(a+b)^2 \sin ^2 \frac{\mathrm{C}}{2}$
$=\left(a^2+b^2-2 a b\right) \cos ^2 \frac{C}{2}+\left(a^2+b^2+2 a b\right) \sin \frac{\mathrm{C}}{2} 2$
$=\left(a^2+b^2\right) \cos ^2 \frac{\mathrm{C}}{2}-2 a b \cos ^2 \frac{\mathrm{C}}{2}+\left(a^2+b^2\right) \sin ^2 \frac{\mathrm{C}}{2}+2 a b \sin ^2 \frac{\mathrm{C}}{2}$
$=\left(a^2+b^2\right)\left(\cos ^2 \frac{\mathrm{C}}{2}+\sin ^2 \frac{\mathrm{C}}{2}\right)-2 a b\left(\cos ^2 \frac{\mathrm{C}}{2}-\sin ^2 \frac{\mathrm{C}}{2}\right)$
$\begin{aligned} & =a^2+b^2-2 a b \cos \mathrm{C} \\ & =c^2=\text { RHS. }\end{aligned}$
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