Different ways of solving a contest problem (Part 1)

The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:

If $3 \sin \theta = \cos \theta$ , what is $\sin \theta \cos \theta$ ?

When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.

Here’s the first solution that I received: draw the appropriate triangles for the angle $\theta$ :

$3 \sin \theta = \cos \theta$

$\tan \theta = \displaystyle \frac{1}{3}$

Therefore, the angle $\theta$ must lie in either the first or third quadrant, as shown. (Of course, $\theta$ could be coterminal with either displayed angle, but that wouldn’t affect the values of $\sin \theta$ or $\cos \theta$ .)

In Quadrant I, $\sin \theta = \displaystyle \frac{1}{\sqrt{10}}$ and $\cos \theta = \displaystyle \frac{3}{\sqrt{10}}$ . Therefore,

$\sin \theta \cos \theta = \displaystyle \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \displaystyle \frac{3}{10}$ .

In Quadrant III, $\sin \theta = \displaystyle -\frac{1}{\sqrt{10}}$ and $\cos \theta = -\displaystyle \frac{3}{\sqrt{10}}$ . Therefore,

$\sin \theta \cos \theta = \displaystyle \left( - \frac{1}{\sqrt{10}} \right) \times \left( -\frac{3}{\sqrt{10}} \right) = \displaystyle \frac{3}{10}$ .

Either way, we can be certain that $\sin \theta \cos \theta = \displaystyle \frac{3}{10}$ .

How I Impressed My Wife: Part 5h

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$ . In today’s post, I’ll start considering the case $|b| < 1$ .

Factoring the denominator is a bit more complicated if $|b| < 1$ . Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$ , the right-hand side is now a complex number. So, To solve for $u$ , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$ .

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$ , or $|u| = 1$ . This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$ .

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$ ,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$ ,

$\sin 2 \phi = 2 \sin\phi \cos \phi$ ,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$ ,

$u = \pm( \sqrt{1-b^2} + i |b|)$ .

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$ .

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

$= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

To double-check my work, I can directly multiply this product:

$([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

$= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)$

$= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})$

$= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2$

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$ .

So, at last, I can rewrite the integral $Q$ as

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$ , in tomorrow’s post.

How I Impressed My Wife: Part 4b

Previously in this series, I have used two different techniques to show that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

Let me backtrack to a point in the middle of the previous solution:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$ ,

where $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ (and $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

Earlier in this series, I used the magic substitution $u = \tan \displaystyle \frac{\phi}{2}$ to evaluate this last integral. Now, I’ll instead use contour integration; see Wikipedia for more details. I will use Euler’s formula as a substitution (see here and here for more details):

$z = e^{i \phi} = \cos \phi + i \sin \phi$ ,

so that the integral $Q$ is transformed to a contour integral in the complex plane. Under this substitution, as discussed in yesterday’s post,

$\cos \phi = \displaystyle \frac{1}{2} \left[z + \displaystyle \frac{1}{z} \right]$

and

$d\phi = \displaystyle -\frac{i}{z} dz$

Employing this substitution, the region of integration changes from $0 \le \phi \le 2\pi$ to a the unit circle $C$ , a closed counterclockwise contour in the complex plane:

$Q = 2 \displaystyle \oint_C \frac{\displaystyle -\frac{i}{z} dz}{S + \displaystyle \frac{R}{2} \left[z + \displaystyle \frac{1}{z} \right]}$

$= -4i \displaystyle \oint_C \frac{dz}{Rz^2 + 2Sz + R}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}$

While this looks integral in the complex plane looks a lot more complicated than a regular integral, it’s actually a lot easier to compute using residues. I’ll discuss the computation of this contour integral in tomorrow’s post.

How I Impressed My Wife: Part 4a