# Solutions of triangles problem

Prove that $$\left(a-b\right)^2\cos^2 \left(\frac{C}{2}\right) + \left(a+b\right)^2\sin^2\left(\frac{C}{2}\right)=c^2$$

My solution:-

I don't know what to do next..please give some hint. Or point out if I am wrong somewhere..

• :you forgot to write coeffeciients of Cos$C$ . – P.Styles Aug 18 '16 at 4:48

$$\text{L.H.S}=(a^2 +b^2)\left(\cos^2 \frac{C}{2} +\sin^2\frac{C}{2}\right)-2ab\left(\cos^2 \frac{C}{2} -\sin^2\frac{C}{2}\right)$$ $$=a^2 +b^2 - 2ab\cos C =c^2$$
Using the Cosine, and "double angle" formulas we have:$$(a-b)^2\cos^2(\frac{C}{2})+(a+b)^2\sin^2(\frac{C}{2})=(a-b)^2\left(\dfrac{1+\cos C}{2}\right)+(a+b)^2\left(\dfrac{1-\cos C}{2}\right)=\dfrac{(a-b)^2+(a+b)^2}{2}+\dfrac{(a-b)^2-(a+b)^2}{2}\cdot \cos C=a^2+b^2-2ab\cos C = c^2$$
We have $$\left( \frac{a-b}{c} \right)^2 \cos^2 \frac{C}{2}+ \left( \frac{a+b}{c} \right)^2 \sin^2 \frac{C}{2}=1$$