6
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Hint: rewrite your function as $\sqrt u + \sqrt v$, where $u+v=a^2+b^2$, and find derivative (using that $u'+v'=0$) $\endgroup$ – Michael Galuza Jul 17 '15 at 6:36
4
$\begingroup$

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 $$u^2_{min}=(\left |a\right |+\left |b\right |)^2$$ $$u^2_{max}=2(a^2+b^2)$$

$\endgroup$
  • $\begingroup$ Same as mine almost! ;-)) Watch the signs on the $a$ and $b$ for the minimum. $\endgroup$ – Mark Viola Jul 17 '15 at 6:55
  • $\begingroup$ @Dr.MV thanks for mentioning that.I would correct that. $\endgroup$ – user2838619 Jul 17 '15 at 7:03
  • $\begingroup$ You're welcome! My pleasure. And isn't it good to see that the different approaches led to the same result? $\endgroup$ – Mark Viola Jul 17 '15 at 7:05
  • $\begingroup$ @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. $\endgroup$ – user2838619 Jul 17 '15 at 7:12
  • 1
    $\begingroup$ 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 $(|a|+|b|)^2$. $\endgroup$ – David K Jul 17 '15 at 12:20
5
$\begingroup$

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 $\sin 2x =0$,

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

$\endgroup$
  • $\begingroup$ Very Nice Solution. Dr. MV $\endgroup$ – juantheron Jul 19 '15 at 4:54
  • $\begingroup$ @juantheron Thank you!! Very much appreciative. $\endgroup$ – Mark Viola Jul 19 '15 at 5:02
3
$\begingroup$

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 $a^2b^2 \leq \dfrac{(a^2+b^2)^2}{4}$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.