Question

Show that for all real values of $\theta$, the expression a ${\sin }^2\theta +b\sin \theta \cos \theta +c{\cos }^2\theta$ lies between $\frac{1}{2}(a+c)-\frac{1}{2}\sqrt{b^2+{(a-c)}^2}$ $\frac{1}{2}(a+c)+\frac{1}{2}\sqrt{b^2+{(a-c)}^2}$

Answer 1

$a{\sin }^2\theta +b\sin \theta \cos \theta +c{\cos }^2\theta$

$=\frac{a}{2}{\sin }^2\theta +\frac{c}{2}{\cos }^2\theta +\frac{a}{2}{\sin }^2\theta +\frac{c}{2}{\cos }^2\theta +b\sin \theta \cos \theta$

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

$=(\frac{a}{2}-\frac{c}{2}){\sin }^2\theta +\frac{c}{2}+(\frac{c}{2}-\frac{a}{2}){\cos }^2\theta +\frac{a}{2}+\frac{b}{2}\sin 2\theta$

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

$=\frac{a+c}{2}+\left(\cos 2\theta \right)\frac{(c-a)}{2}+\frac{b}{2}\sin 2\theta$

$\sqrt{A^2+B^2}\le A\cos \alpha +B\sin \alpha \le +\sqrt{A^2+B^2}$

$\therefore -\sqrt{{\left(\frac{c-a}{2}\right)}^2+{\left(\frac{b}{c}\right)}^2}\le \left(\cos 2\theta \right)\frac{(c-a)}{2}+\frac{b}{2}\sin 2\theta \le \sqrt{{\left(\frac{c-a}{2}\right)}^2+{\left(dfracb2\right)}^2}$

$\therefore \frac{-1}{2}\sqrt{b^2(a-c{)}^2}\le \left(\frac{c-a}{2}\right)\cos 2\theta +\frac{b}{2}\sin 2\theta \le \frac{1}{2}\sqrt{b^2+(a-c{)}^2}$

$\therefore \frac{a+c}{2}-\frac{\sqrt{b^2+(a-c{)}^2}}{2}\le \underset{a{\sin }^2\theta +b\sin \theta \cos \theta +c{\cos }^2\theta }{\underset{}{\frac{a+c}{2}\left(2\cos \theta \right)\frac{(c-a)}{2}+b2\sin 2\theta }}\le \frac{a+c}{2}+\frac{1}{2}\sqrt{b^2+(a-c{)}^2}$