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.

  • $\begingroup$ 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

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

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


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

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

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


By the double angle formulas, the expression is equivalent to


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


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


  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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

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.

  • $\begingroup$ 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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