# Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$

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.

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

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

giving

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

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

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

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

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

• 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 :) – Abhas Kumar Sinha Jun 3 at 8:48
• @AbhasKumarSinha: lookup "Lagrange multipliers". I mentioned three alternatives. – Yves Daoust Jun 3 at 9:00
• I can't understand, there is lambda like symbol and what your doing is completely alien to me. – Abhas Kumar Sinha Jun 3 at 9:04
• @AbhasKumarSinha: if you are not willing to do a Web search, just ignore this approach and stick to differentiation. – 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.

• Beautiful, nice creativity :) – Abhas Kumar Sinha Jun 3 at 8:48
• An Awesome Script from @auscrypt +1 – 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: $$\left[\frac{1}{2}\left(a+c-\sqrt{b^2+(a-c)^2}\right),\frac{1}{2}\left(a+c+\sqrt{b^2+(a-c)^2}\right)\right]$$

• You should add a word about the tightness of the bounds. – Yves Daoust Jun 3 at 10:13
• @Yves Daoust I added something for you. See now. – 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.

• 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? – Abhas Kumar Sinha Jun 4 at 3:16
• 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. – richrow Jun 4 at 8:02
• 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). – richrow Jun 4 at 8:06