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Show that, whatever the value of $\theta$, the expression

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

Lies between

$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$


My try:

The given expression can be reduced as sum of sine functions as:

$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$

Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.

Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,

$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$

For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!

So what's the problem here? Can it be solved?

Thanks :)


Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.

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  • $\begingroup$ The fraction tends to $c$ as $t \to \pm \infty$. Why do you think its range is $(-\infty, \infty)$? $\endgroup$ – Kabo Murphy Jun 3 at 8:23
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    $\begingroup$ A hint: You could try using $\sin^2\theta = \frac{1}{2}-\frac{1}{2}\cos 2\theta$ in $(*)$. $\endgroup$ – Minus One-Twelfth Jun 3 at 8:24
  • $\begingroup$ @kaviramamurthy oh okay, that's a mistake.. $\endgroup$ – Abhas Kumar Sinha Jun 3 at 8:24
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    $\begingroup$ Correct your question. In title sine is squared in question body it is only sine. $\endgroup$ – Vineet Jun 3 at 8:31
  • $\begingroup$ See math.stackexchange.com/questions/2667559/… $\endgroup$ – lab bhattacharjee Jun 3 at 8:34
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This problem is equivalent to

$$ \min(\max) a x^2+b x y + c y^2 \ \ \mbox{s. t.}\ \ x^2+y^2=1 $$

this is an homogeneous problem so calling $y = \lambda x$ and substituting we have equivalently

$$ \min(\max) f(\lambda) = \frac{a+\lambda b+\lambda^2c}{1+\lambda^2} $$

and the extremals condition is

$$ f'(\lambda) = 0\Rightarrow 2 \lambda (c-a)-b \lambda ^2+b = 0 $$

giving

$$ \lambda = \frac{c-a\pm\sqrt{(a-c)^2+b^2}}{b} $$

now substituting into $f(\lambda)$ we have

$$ \frac{1}{2} \left(-\sqrt{(a-c)^2+b^2}+a+c\right)\le f(\lambda)\le \frac{1}{2} \left(\sqrt{(a-c)^2+b^2}+a+c\right) $$

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By the double angle formulas, the expression is equivalent to

$$\frac12\left(a+c+(a-c)\cos2\theta+b\sin2\theta\right).$$

Now the expression $(a-c)\cos2\theta+b\sin2\theta$ can be seen as the dot product of a vector with a rotating unit vector, which takes its extreme values when the vectors are parallel or antiparallel, giving

$$\pm\|(a-c,b)\|=\pm\sqrt{(a-c)^2+b^2}.$$


The same result can be obtained by differentiation, or by reducing to the sine addition formula.


Yet another way is by finding the extrema of $(a-c)x+by$ under the constraint $x^2+y^2=1$. Using a Lagrange multiplier, the equations are

$$\begin{cases}x^2+y^2&=1,\\a-c&=2\lambda x,\\b&=2\lambda y,\end{cases}$$

easily giving

$$x=\pm\frac{a-c}{\sqrt{(a-c)^2+b^2}},\\y=\pm\frac{b}{\sqrt{(a-c)^2+b^2}}.$$

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  • $\begingroup$ Hey, I've not read Lagrange Multipliers yet, but as a pre high school student, I'm interested in it. If you can point some good resources, I'd be grateful :) $\endgroup$ – Abhas Kumar Sinha Jun 3 at 8:48
  • $\begingroup$ @AbhasKumarSinha: lookup "Lagrange multipliers". I mentioned three alternatives. $\endgroup$ – Yves Daoust Jun 3 at 9:00
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    $\begingroup$ I can't understand, there is lambda like symbol and what your doing is completely alien to me. $\endgroup$ – Abhas Kumar Sinha Jun 3 at 9:04
  • $\begingroup$ @AbhasKumarSinha: if you are not willing to do a Web search, just ignore this approach and stick to differentiation. $\endgroup$ – Yves Daoust Jun 3 at 9:09
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We can use a trick to express the function as something easier to deal with. We claim that we can write the function as $(d \cos \theta + e \sin \theta)^2 + f$ or $-(d \cos \theta - e \sin \theta)^2 + f$ for some constants $d, e, f$. From here, converting $d \cos \theta + e \sin \theta$ into a single trigonometric function is simple, and we can find the bounds exactly.

Note that for our expression to work, we must have $de = \frac{b}{2}$. So we expand $(d \cos \theta + \frac{b}{2d} \sin \theta)^2$ to get $d^2 \cos^2 \theta + b \sin \theta \cos \theta + \frac{b^2}{4d^2} \sin^2 \theta$. If we can find a $d$ satisfying $d^2 - \frac{b}{4d^2} = a^2 - c^2$, then the expression will be exactly $a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta - f(\sin^2 \theta + \cos^2 \theta)$ for some $f$. But solving for $d$ is now a quadratic in $d^2$. A similar approach can be taken with $-(d \cos \theta - e \sin \theta)^2$; one of the two will always work, and yield the exact bounds required.

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  • $\begingroup$ Beautiful, nice creativity :) $\endgroup$ – Abhas Kumar Sinha Jun 3 at 8:48
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    $\begingroup$ An Awesome Script from @auscrypt +1 $\endgroup$ – Ak19 Jun 3 at 9:41
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By C-S we obtain: $$a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta=a\cdot\frac{1-\cos2\theta}{2}+b\cdot\frac{\sin2\theta}{2}+c\cdot\frac{1+\cos2\theta}{2}=$$ $$=\frac{1}{2}\left(a+c+b\sin2\theta-(a-c)\cos2\theta\right)\leq\frac{1}{2}\left(a+c+\sqrt{(b^2+(-a+c)^2)\left(\sin^22\theta+\cos^22\theta\right)}\right)=$$ $$=\frac{1}{2}\left(a+c+\sqrt{b^2+(a-c)^2}\right)$$ and by C-S again we obtain: $$a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta=\frac{1}{2}\left(a+c+b\sin2\theta-(a-c)\cos2\theta\right)\geq$$ $$\geq \frac{1}{2}\left(a+c-\sqrt{(b^2+(-a+c)^2)\left(\sin^22\theta+\cos^22\theta\right)}\right)=$$ $$=\frac{1}{2}\left(a+c-\sqrt{b^2+(a-c)^2}\right).$$ The equality in the both cases occurs for $$(\sin2\theta,\cos2\theta)||(b,-a+c),$$ which says that we got a minimal and the maximal value of the expression.

Now, since $f$ is a continuous function, we obtain that the range of $f$ it's: $$\left[\frac{1}{2}\left(a+c-\sqrt{b^2+(a-c)^2}\right),\frac{1}{2}\left(a+c+\sqrt{b^2+(a-c)^2}\right)\right]$$

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    $\begingroup$ You should add a word about the tightness of the bounds. $\endgroup$ – Yves Daoust Jun 3 at 10:13
  • $\begingroup$ @Yves Daoust I added something for you. See now. $\endgroup$ – Michael Rozenberg Jun 3 at 10:33
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It's not actually a full answer but, in my view, interesting generalization. Note that the problem is a special case of the following statement.

Proposition. Let $\beta$ be a symmetric bilinear form in $n$-dimensional euclidean space $\left(V, \langle\cdot, \cdot\rangle\right)$ and $\lambda_1\geq \lambda_2\geq\ldots\geq\lambda_n$ are eigenvalues of linear operator $A$, which is defined by equality $\beta(x, y)=\langle Ax, y\rangle$. For vector subspace $L\subset V$ define $$ \underline{\lambda}(L):=\min\{\beta(v,v)| v\in L,\|v\|=1\}, \\ \overline{\lambda}(L):=\max\{\beta(v,v)| v\in L,\|v\|=1\}. $$ Then, for all $1\leq k\leq n$ the following equlaity holds $$ \lambda_k=\max\{\underline{\lambda}(L)|\dim L= k\}=\min\{\overline{\lambda}(L)|\dim L= n+1-k\}. $$

In our case $n=2$ and operator $A$ (and symmetric bilinear form $\beta$) has matrix \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} Characteristic equation of this matrix is $(a-\lambda)(c-\lambda)-\frac{b^2}{4}=0$, so we have eigenvalues $\lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$ and $\lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2}$.

Using proposition above we obtain that $$ \min_{\|v\|=1}\beta(v, v)=\lambda_2, \max_{\|v\|=1}\beta(v, v)=\lambda_1. $$ Finally, note that $\|v\|=1$ is equivalent to $v=(\sin\theta, \cos\theta)^T$ for some $\theta\in [0,2\pi)$. Thus, $$ \min_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_2=\dfrac{a+c}{2}-\dfrac{1}{2}\sqrt{b^2+(a-c)^2}, \\ \max_{\theta} (a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta) = \lambda_1=\dfrac{a+c}{2}+\dfrac{1}{2}\sqrt{b^2+(a-c)^2}, $$ as desired.

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  • $\begingroup$ For god's sake, that math is monstrous! I didn't get that very first line, is $\beta$ is a vector? Does it uses Lagrange multipliers? $\endgroup$ – Abhas Kumar Sinha Jun 4 at 3:16
  • $\begingroup$ Actually, as I mentioned before it's just a generalization of your problem. $\beta$ is a symmetric bilinear form in vector space $V$, which means that $\beta$ is a function defined on $V\times V$ and satisfying the following conditions. Firstly, $\beta$ is symmetric which means that $\beta (x,y)=\beta (y,x)$ for all $x,y\in V$. Secondly, $\beta$ is bilinear, so for all scalars $\lambda, \mu$ and $x,y,z\in V$ we have $\beta (\lambda x+\mu y,z)=\lambda\beta (x,z)+\mu\beta (y,z)$ and similar relation for the second argument. $\endgroup$ – richrow Jun 4 at 8:02
  • $\begingroup$ You can learn more about bilinear forms in any course of linear algebra. Andthis solution does not use Lagrange multipliers (which are more connected with multivariate calculus). $\endgroup$ – richrow Jun 4 at 8:06

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