<s>[INST] Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
 [/INST] By C-S we obtain:
$$v=$$
$$=\sqrt{a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a^2\sin^2x+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}=$$
$$=\sqrt{a^2+b^2+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}\geq$$
$$\geq\sqrt{a^2+b^2+2|ab|(\cos^2x+\sin^2x)}=|a|+|b|.$$
$$v\leq\sqrt{2(a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a^2\sin^2x)}=\sqrt{2(a^2+b^2)}.$$
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