Posts Tagged ‘trigonometry’

A calculator is broken so that the only

April 27, 2013

A calculator is broken so that the only keys that still work are the sin, cos, tan, arcsin, arccos, and arctan buttons. The display initially shows 0. Given any positive rational number q, show that pressing some finite sequence of buttons will yield q. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

I’ve started reading Zeitz’s The Art and Craft of Problem Solving. This one took me about 90 minutes, though as usual, once I had a solution it seemed obvious. Originally from USAMO 1995. Who comes up with these problems? How? You sit down and say, “Okay, it’s time to invent a problem that can be solved with elementary math, but only if you see some diabolical trick”, and then you do what?

Visualizing Elementary Calculus: Trigonometry

March 26, 2011

Here we’ll find the derivatives of trigonometric functions. The goal is to reinforce the idea of \textrm{d} as a thing that means “a little bit of” and grant some new insight into why these derivatives are what they are. The first argument is based on the preface of Tristan Needham’s Visual Complex Analysis. I haven’t read the bulk of it, but the preface is good.

This series
I – Introduction
II – Trigonometry


The Sine Function

Let’s find \textrm{d}(\sin\theta) / \textrm{d}\theta. The sine function is the height of a right triangle in the unit circle. We’ll draw it, and add a little change in \theta. This induces a change in \sin\theta. The change in \theta is called \textrm{d}\theta and the change in \sin\theta is called \textrm{d}(\sin\theta).

We show the sine of an angle as the dark blue line. The change in the sine when we change the angle slightly is the light blue line.

The interesting part is \textrm{d}\sin\theta, so we’ll zoom in there in the next picture. Before we do, remember that the arc length along a piece of the unit circle is equal to the angle it subtends. This will tell us the length of the little piece of the circumference near \textrm{d}\sin\theta. Also remember that we’re imagining \textrm{d}\theta to get smaller and smaller, until the two radii in the picture are parallel. We get this:

The interesting region is blown up to large size. The black line d(theta) is part of the edge of the circle. The angles marked are congruent to theta.

The section of the circle is \textrm{d}\theta long. It looks like a straight line because we are zoomed in close, like the horizon at the beach. You can use some geometry to show that the angles marked are congruent to \theta.

Looking at the right triangle formed, we can use the definition of the cosine function to read off

\frac{\textrm{d}(\sin\theta)}{\textrm{d}\theta} = \cos\theta

which is the derivative of the sine function.

Motion on the Unit Circle

Another way to view these derivatives is to imagine a point moving around the outside edge of the unit circle with speed one. Its location as a function of time is (\cos t, \sin t).

Its velocity is tangent to the circle and length one. Let’s draw the velocity vector right at the point, and then also translate it to the origin.

The position of the point is the red vector r. Its velocity is the green tangent v, which has also been copied to the origin.

We want to know the coordinates of \vec{v}. That’s not too hard; \vec{v} is a quarter-circle rotation of \vec{r}. Draw in the components of \vec{r}, and rotate those components to get \vec{v}. The x-component of the position becomes the y-component of the velocity, and the y-component of the position becomes minus one times the x-component of the velocity.

The components of the position get rotated a quarter turn to make the components of the velocity.

The derivative of position is velocity, and so comparing components between the position and velocity vectors, we get

\frac{\textrm{d}(\cos\theta)}{\textrm{d}\theta} = -\sin\theta

\frac{\textrm{d}(\sin\theta)}{\textrm{d}\theta} = \cos\theta

Exercises

  • Look back at the first derivation we gave that \textrm{d}(\sin\theta)/\textrm{d}\theta = \cos\theta. Rework it to find derivatives of the other five trig functions. You might want to note that one way to interpret \tan\theta and \sec\theta is

The tangent and secant of an angle are side lengths of a right triangle with "adjacent" side length one.


(Answers)

  • Look back at the argument about a dot moving around a circle. Consider a larger circle to find the derivative of 5\sin\theta with respect to \theta. (Answer: 5\cos\theta)
  • Suppose the dot moving around the edge of the circle is going three times as fast. What does this mean for the derivative of \sin(3 t) and \cos(3 t) with respect to t? Remember that the velocity must still be perpendicular to the position, but not necessarily unit length and more. (Answer: the derivative of \sin(3 t) with respect to t is 3\cos(3 t).
  • Suppose the dot is moving at a variable speed v(t) = t, so that it keeps getting faster. Then the y-coordinate of the position is \sin(\frac{1}{2}t^2). Again, the velocity is perpendicular to position, but its length is changing. What is the derivative of \sin(\frac{1}{2}t^2) with respect to t? (Answer: t\cos(\frac{1}{2}t^2)

Viete’s Formula and Spinning Pizza

September 17, 2010

Have you seen Viete’s formula?:

It’s a special case of a trig identity found by Euler:

If you plug in \pi/2 to the trig identity and use the half-angle formula for cosine over and over, you get Viete’s formula.

But why would you want to consecutively cut angles in half and multiply their cosines? Well, you might be eating pizza.

You have a slice of pizza that is too hot to hold, so you want to balance it on your fork and gnaw at it instead. There’s a precise spot on the underside of the slice where the fork should go.

Balancing pizza on a fork.

This point is called the slice’s center of mass, and we’re going to find it. By symmetry, it must be on the line cutting the slice in half lengthwise, but we don’t yet know how far down. It depends on the shape of the slice, which we measure by \theta, the angle its edges make.

The center of mass of the slice of pizza is a green dot. It lies on the line cutting the slice in half vertically.

A bigger slice will have its center of mass closer to the tip. We would like to know r(\theta), the distance from the tip to the center of mass as a function of \theta.

We want to know the distance to the center of mass.

There are two limiting cases – a very skinny slice and a whole pie. A very skinny slice is basically an isosceles triangle. Its center of mass is 2/3 the way from the tip to the edge1, so

\lim_{\theta \to 0} r(\theta) = \frac{2}{3}R .

Let’s choose the radius of the pizza as unit of length, so R = 1 from here on.

In the other limiting case, an entire pizza has its center of mass right at the tip (i.e. center), so

r(2\pi) = 0 .

To investigate intermediate cases, we start with a slice of angle \theta and imagine cutting it in half lengthwise, creating two skinny pieces of angle \theta/2. These have their own centers of mass at r(\theta/2).

The center of mass of the big piece is on the line connecting the smaller pieces’ centers of mass.

A bit of trigonometry tells us

r(\theta) = r(\theta/2)\cos(\theta/4)

If we take this formula and divide all angles by 2, we get a formula for r(\theta/2). We substitute this for where r(\theta/2) appeared in the original. We obtain

r(\theta) = \left[r(\theta/4)\cos(\theta/8)\right]\cos(\theta/4)

Repeat the process ad infinitum. Rearranging the order of the terms and substituting the limiting value of r for small \theta, we get

r(\theta) =\frac{2}{3} \cos(\theta/4)\cos(\theta/8)\cos(\theta/16)\ldots

It involves one half of Euler’s trig identity. If we find r(\theta) by a different method and get a different expression for it, we can set our two expressions for r(\theta) equal to each other, and prove Euler’s identity. We’ll do this by invoking some physics ideas.

Suppose you’re spinning some pizza dough in the air. You know, like this:

If the pizza is spinning, each little bit of dough undergoes centripetal acceleration. Where there’s acceleration, there’s force. The pizza isn’t touching anything, so the force on any one piece of pizza must be coming from the rest of the pizza.

Let’s again examine a slice of size \theta, this time still attached to the spinning pizza. It has two forces of size F acting on it; one force is exerted by the slice to its left and one by the slice to its right.

There are two forces on the slice - one from the pizza to the left and one from the pizza to the right. They're both drawn originating from the center of mass. The slice is accelerating towards its tip (red arrow).

The sum of these forces is the mass of the slice times the acceleration of its center of mass. That acceleration is \omega^2 r(\theta). Hence, if we determine the forces we can deduce r(\theta).

Some trigonometry shows that the net force is 2F\sin(\theta/2).

Equating this to mass times acceleration, we get

2F\sin(\theta/2) = \frac{m \theta}{2\pi} \omega^2 r(\theta)

We might as well let m = \omega = 1 and solve for r to get

r(\theta) = 2F\sin(\theta/2)\frac{2 \pi}{\theta} .

We still need to determine F, but we can do that because we know r(\theta) \to 2/3 as \theta \to 0. After a little algebra, we get

r(\theta) = \frac{4}{3} \frac{sin(\theta/2)}{\theta}

This gives us the sought two expressions for r. We can now equate them and simplify to

\frac{sin(\theta)}{\theta} = \cos(\theta/2)\cos(\theta/4)\cos(\theta/8)\ldots


1) To see why an isosceles triangle’s center of mass is 2/3 up the altitude, first show it’s true for an equilateral triangle. Then explain why all isosceles triangles have their center of mass the same fraction of the way down the altitude.

A Physical Trig Identity

September 6, 2010

Can a basic physics problem give you insight into math?

For example, mathematically

\cos(x) - \sin(x) = \sqrt{2}\cos(x + \pi/4)

which is easy to verify using the angle addition formula.

I came across this formula while solving a simple problem in statics.

Imagine the classic “block on an inclined plane”. Gravity (F_g) pulls the block down, and you push (F_p) on it sideways, like this:

Gravity pulls down on the block and you push on it to the right.

What is the minimum coefficient of static friction to keep the block stationary? In order to calculate this, we need to know the component of force parallel to the plane.

First look at gravity. We want to find the green component F_{g1}.

The force of gravity can be decomposed into two components. One points along the plane and the other is normal to it.

Let’s say the positive direction is to the right. Then gravity is pulling backwards some, so F_{g1} is negative. I know it’s either a sine or cosine of \theta, and in the limit as \theta \to 0, I see that F_{g1} = 0, so

F_{g1} = -F_g \sin\theta .

Then we look at the pushing force.

The force from pushing is likewise decomposed.

A similar procedure gives

F_{p1} = F_p \cos\theta .

So the total force in the direction of the ramp is

-F_g \sin\theta + F_p \cos\theta .

In the special case where F_g = F_p = 1 the force is

\cos\theta - \sin\theta .

Now we will find this component of the force another way. We start by tip-to-tail adding the force of gravity and the force of the push. They’re at right angles, and assuming they’re equal in magnitude we get a resultant force with length \sqrt{2} that bisects the angle between the gravity and pushing forces.

We can also find the component of force along the plane by first adding the two vectors...

The angle between this resultant force and the plane is \pi/4 + \theta.

...and then finding a component of the sum.

The component of this force along the plane is then the cosine of \pi/4 + \theta, so the force along the direction of the plane is

\sqrt{2} \cos(\pi/4 + \theta)

and since it’s the same quantity we calculated before, we have

\cos\theta - \sin\theta = \sqrt{2}\cos(\pi/4 + \theta) .

There is no physics in this calculation, but if you had simply asked me to write \cos\theta-\sin\theta as a single trig function, I wouldn’t have thought to approach it like this.