Find $\cos2B+\cos2C$ in a triangle

Question

In triangle $ABC$ we have $2a^2=b^2+c^2$. The measure of angle $A$ is $30^{\circ}$. Find $\cos2B+\cos2C$

Answer 1

By the law of cosines we obtain: $$a^2=b^2+c^2-2bc\cdot \frac{\sqrt3}{2},$$ which gives $$2(b^2+c^2-\sqrt3bc)=b^2+c^2$$ or $$b^2+c^2=2\sqrt3bc.$$ Also, we have $$a^2=\sqrt3bc.$$ Thus, $$\cos2\beta+\cos2\gamma=2\left(\frac{a^2+c^2-b^2}{2ac}\right)^2+2\left(\frac{a^2+b^2-c^2}{2ab}\right)^2-2=$$ $$=2\left(\frac{\frac{b^2+c^2}{2}+c^2-b^2}{2ac}\right)^2+2\left(\frac{\frac{b^2+c^2}{2}+b^2-c^2}{2ab}\right)^2-2=$$ $$=\frac{(3c^2-b^2)^2}{4c^2(b^2+c^2)}+\frac{(3b^2-c^2)^2}{4b^2(b^2+c^2)}-2=$$ $$=\frac{b^6+c^6+3b^4c^2+3c^4b^2}{4b^2c^2(b^2+c^2)}-2=\frac{(b^2+c^2)^2}{4b^2c^2}-2=3-2=1.$$

Answer 2

Let $O$ be the center of circumcircle with radius $r$. Then $\angle BOC = 60^{\circ}$ so $BOC$ is equilateral, so $a=r$. Now use the cosinus theorem in triangles $ABO$ and $ACO$ and we get:

$$\cos 2B +\cos 2C = {2r^2-b^2\over 2r^2}+{2r^2-c^2\over 2r^2}={4r^2-(b^2+c^2)\over 2r^2} = 1$$