Question

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

Answer 1

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

We know that $\sin 2\theta =2\sin \theta \cos \theta$

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

$\Rightarrow a{\sin }^2\theta +\frac{b}{2}\sin 2\theta +c{\cos }^2\theta +c{\sin }^2\theta -c{\sin }^2\theta$ [add and substract $c{\sin }^2\theta$]

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

We know ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

$\Rightarrow (a-c)\frac{(1-\cos 2\theta )}{2}+\frac{b}{2}\sin 2\theta +c$

$\Rightarrow \underset{\underset{\frac{a-c}{2}+c}{\downarrow }}{\frac{(a-c)}{2}}-\frac{(a-c)}{2}\cos 2\theta +\frac{b}{2}\sin 2\theta +c$

$\Rightarrow \underset{constant}{\frac{a+c}{2}}+\underset{maxandminvalue}{\frac{b}{2}\sin 2\theta -\left(\frac{a-c}{2}\right)\cos 2\theta }$

We know $a\sin x-b\cos x$

$max=\sqrt{a^2+b^2}$

$min=-\sqrt{a+b^2}$

$max\Rightarrow \frac{a+c}{2}+\sqrt{(\frac{b}{2}{)}^2+(\frac{(a-c)}{2}{)}^2}$

$min\Rightarrow \frac{a+c}{2}-\sqrt{(\frac{b}{2}{)}^2+(\frac{(a-c)}{2}{)}^2}$

max value of expansion $=\frac{a+c}{2}+\frac{1}{2}\sqrt{b^2+(a-c{)}^2}$

min value of expansion $=\frac{a+c}{2}-\frac{1}{2}\sqrt{b^2+(a-c{)}^2}$

Hence proved.