# The maximum and minimum values of the expression

Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$

My try:I have just normally squared the expression and got

$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$

$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$

I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?

• Hint: rewrite your function as $\sqrt u + \sqrt v$, where $u+v=a^2+b^2$, and find derivative (using that $u'+v'=0$) – Michael Galuza Jul 17 '15 at 6:36

Expanding $u^2$ more:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^4+b^4)+a^2b^2(\sin^4x+\cos^4x)}$$

Using trigonometric identity $\sin^2x+\cos^2x=1$ we can derive that:$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$$

Rewrite $u^2$ again:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^2-b^2)^2+a^2b^2}$$

The minimum value of $\sin^2x\cos^2x$ is $0$ and its maximum value is (using AM-GM) $$\frac{\sin^2x+\cos^2x}{2}=\frac{1}{2}\ge\sin x\cos x$$ $$\frac{1}{4}\ge\sin^2x\cos^2x$$ Also you can find it this way using trigonometric identities $$\sin^2x\cos^2x = \frac{\sin^2(2x)}{4}\Rightarrow \max(\sin^2x\cos^2x)=\max \left(\frac{\sin^2(2x)}{4}\right)=\frac{1}{4}$$ So $$u^2_{min}=(\left |a\right |+\left |b\right |)^2$$ $$u^2_{max}=2(a^2+b^2)$$

• Same as mine almost! ;-)) Watch the signs on the $a$ and $b$ for the minimum. – Mark Viola Jul 17 '15 at 6:55
• @Dr.MV thanks for mentioning that.I would correct that. – user2838619 Jul 17 '15 at 7:03
• You're welcome! My pleasure. And isn't it good to see that the different approaches led to the same result? – Mark Viola Jul 17 '15 at 7:05
• @Dr.MV sure. Using trigonometry identities at first was a very good idea for having a nice and shorter solution. I wanted to continue OP's work so he would understand he can find the answer in that way too. – user2838619 Jul 17 '15 at 7:12
• If $a=2$ and $b=1$ then $$u=\sqrt{4\cos^2x+\sin^2x} + \sqrt{4\sin^2x+\cos^2x},$$ but if $a=2$ and $b=\color{red}{-1}$ then $$u=\sqrt{4\cos^2x+\sin^2x} + \sqrt{4\sin^2x+\cos^2x}.$$ You get the same value of $u$ for any $x$, and therefore the same minimum and maximum values of $u^2$, regardless of the sign of $b$. You can just as easily show that the sign of $a$ is irrelevant. So there must be no "depending on the signs of $a$ and $b$" in the answer. Since $2\sqrt{a^2b^2}=2|a|\cdot|b|$, the minimum is $(|a|+|b|)^2$. – David K Jul 17 '15 at 12:20

Write

$$\cos^2x=\frac{1+\cos 2x}{2}$$

and

$$\sin^2x=\frac{1-\cos 2x}{2}$$

Then, we have

$$u=\sqrt{A+B\cos 2x}+\sqrt{A-B\cos 2x}\tag 1$$

where

$$A=\frac{a^2+b^2}{2}$$

$$B=\frac{a^2-b^2}{2}$$

Taking the derivative of u in $(1)$ and setting the derivative equal to zero reveals

$$\frac{-B\sin 2x}{\sqrt{A+B\cos 2x}}+\frac{B\sin 2x}{\sqrt{A-B\cos 2x}}=0$$

whereupon solving reveals that either $\sin 2x=0$ or $\cos 2x=0$. When $\cos 2x=0$,

$$\bbox[5px,border:2px solid #C0A000]{u=\sqrt{2(a^2+b^2)} \,\,\text{is the maximum}}$$

and when $\sin 2x =0$,

$$\bbox[5px,border:2px solid #C0A000]{u=|a|+|b|\,\,\,\text{is the minimum}}$$

• Very Nice Solution. Dr. MV – juantheron Jul 19 '15 at 4:54
• @juantheron Thank you!! Very much appreciative. – Mark Viola Jul 19 '15 at 5:02

Assume WLOG $a > b > 0$, $A = \sqrt{a^2\cos^2x+b^2\sin^2x}, B = \sqrt{a^2\sin^2x+b^2\cos^2x}\Rightarrow A^2+B^2 = a^2+b^2\Rightarrow u^2 = (1\cdot A+1\cdot B)^2\leq (1^2+1^2)(A^2+B^2)=2(a^2+b^2)\Rightarrow u^2_{max} = 2(a^2+b^2)$. To find $u^2_{min}$, you need to find the min of $(a^2\cos^2x+b^2\sin^2x)(a^2\sin^2x+b^2\cos^2x)=f(\cos^2 x)=(a^2-(a^2-b^2)t)(b^2+(a^2-b^2)t), t = \cos^2x, 0 \leq t \leq 1=f(p) = (a^2-p)(b^2+p), p = (a^2-b^2)t, 0 \leq p \leq a^2-b^2\to f(p) = a^2b^2 + (a^2-b^2)p - p^2\Rightarrow f'(p) = a^2-b^2 - 2p=0 \iff p = \dfrac{a^2-b^2}{2}\Rightarrow f\left(\dfrac{a^2-b^2}{2}\right)=\dfrac{(a^2+b^2)^2}{4}$. At end points $p = 0, a^2-b^2, f(0) = a^2b^2, f(a^2-b^2) = a^2b^2$. Thus $u^2_{min} = a^2+b^2 + 2\sqrt{a^2b^2}=(a+b)^2$, since $a^2b^2 \leq \dfrac{(a^2+b^2)^2}{4}$.

For $\min$

Using $\triangle$ Inequality::

Let $z_{1} = a\cos x+i b\sin x$ and $z_{2} = b\cos x+i a\sin x$

So $$|z_{1}|+|z_{2}|\geq |z_{1}+z_{2}|$$

So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\geq \sqrt{(a+b)\cos^2 x+(a+b)^2\sin^2 x}=|a+b|$$

For $\max$ Same as Deepsea

$$\left[\left(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}\right)^2+\left(\sqrt{a^2\sin^2 x+b^2 \cos^2 x}\right)^2\right]\cdot \left[1^2+1^2\right]\geq \bigg(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\bigg)^2$$

So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$

So $$|a+b|\leq \sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$