# 7.2 Trigonometry and derivatives and addition theorems

Introduction

In 7.1, we introduced lots of trigonometry without actually mentioning it.

Trigonometry is a long and off-putting name for what is really a fun subject. A Trigon is a fancy name for a triangle; analogous to the words octagon or pentagon, Metry refers to measurement. So trigonometry means either measuring triangles, or using triangles to measure other things, I’m not sure which; maybe both.

We will describe the remaining important trigonometric results. One of the mysteries of trigonometry is: Why does every one of the six ratios of side lengths in a right triangle have its own special name? Why for example does $$\frac{1}{\sin x}$$ have a name of its own? When I studied trigonometry in school, (in prehistoric times) we were confronted with all six names, and quizzed on and expected to memorize which means which without any clues at all. This turned many of us off to trigonometry.

Suppose our angle $$theta$$, as in the picture here lies between the $$x$$ axis and the line $$0B$$. The ancients drew a line segment that extends from the point $$B$$ tangent to the unit circle to the $$x$$ axis at point $$C$$. The length of this segment they called the tangent of the angle $$\theta$$. (When the line has a positive slope the tangent is taken to be negative.) Tangent is a Latin word that means 'touching', and that is what this line does to the circle, at point $$B$$.

The $$x$$ coordinate of the point $$A$$ where the tangent line meets the $$x$$ axis, is called the secant of $$\theta$$ (we are assuming that the origin is at the center of the unit circle.) Secant is a Latin word meaning 'cutting' which is what this line does to the circle.

They also defined the complement of an angle that is less than a right angle to be the difference between a right angle and it. This got them to define the cosine, cotangent and cosecant as the sine, tangent and secant of the complement of the original angle.

Fortunately for us, all of these six functions are easily related to the sine function, which means that we need only really become familiar with the sine, and we can then figure out what the others are.

Here are the relations between these functions, all of which follow from the definitions from the fact that corresponding angles of similar triangles are equal.

By definition, $$\cos\theta$$ is $$\sin\left(\frac{\pi}{2}-\theta\right)$$.

From triangle $$BCD$$ in which the hypotenuse is $$\tan \theta$$ and the side not opposite $$\theta$$ is $$\sin θ$$, we get

$(\cos \theta)(\tan \theta) = \sin \theta$

which means

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\sin(\pi/2-\theta)}$

The complementary version of this is:

$\cot θ = \frac{\cos \theta}{\sin θ} = \frac{\sin(\pi/2-\theta)}{\sin \theta}$

From the triangle $$BCO$$

we similarly get

$(\sec \theta)(\sin \theta) = \tan \theta$

which means

$\sec \theta = \frac{1}{\cos \theta}$

and the complementary version is

$\csc \theta = \frac{1}{\sin \theta}$

So all this explains why every ratio of side lengths of a right triangle has a name of its own.

I like this picture so much, here it is again but as a mathlet.

Exercises:

7.21. Draw a this picture yourself, without looking at this one, for an angle $$\theta$$ that is less than $$\frac{\pi}{2}$$ showing all of these entities.

7.22. How many similar triangles do you see? Remember that the two angles other than a right angle in a triangle with a right angle are complementary.

7.23. What strange relations do you get if you use the definition of sine in the triangle $$OAC$$?

We have, in the previous section, already discussed the three basic theorems of trigonometry that you should know. There are also useful "addition theorems" of trigonometry, which describe what sine and cosine of sums of arguments are. We also describe the derivatives of the sine and cosine, and their relation to exponentials.

What were the basic Theorems of Trigonometry?

1.The Pythagorean Theorem: This famous result states that the square of the hypotenuse of a right triangle is the sum of the squares of its other two sides. Translated to our definitions it says that for any angle, we have

$(\sin\theta)^2 + (\cos\theta)^2 = 1$

which implies that, up to sign we have

$\cos\theta = \sqrt{1-(\sin\theta)^2}$

2. The Law of Sines: This states that in any triangle $$ABC$$ the ratio of the sines of its angle at $$A$$ to its angle at $$B$$ is the ratio of the lengths of the side opposite $$A$$ to the side opposite $$B$$. If we describe these lengths as $$l(BC)$$ and $$l(AC)$$ respectively, we have

$\frac{\sin A}{\sin B} = \frac{|BC|}{|AC|}$

We proved this in section 7.1C

3. The Law of Cosines: This statement gives the length of the side $$BC$$ of a triangle in terms of the lengths of $$AB$$ and $$AC$$ and its angle at $$A$$

$|BC|^2 = |AB|^2 + |AC|^2 - 2 |AB||AC|\cos A$

Also proven in 7.1C

Derivatives of Sines and Cosines

Consider a point $$P$$ on the unit circle which circle is centered at the point $$C$$. Let $$\theta$$ be the angle clockwise from the line segment $$CP$$ to the $$x$$ axis.

We then have $$x = \cos\theta$$ and $$y = \sin\theta$$. We know that $$(\cos\theta)^2 + (\sin\theta)^2$$ is $$1$$ for any $$P$$ on the unit circle since this is the statement that $$x^2 + y^2$$ for such $$P$$ is $$1$$, which is the definition of the unit circle.

This means that the derivative of $$(\cos\theta)^2 + (\sin\theta)^2$$ is $$0$$ as we move around the unit circle is $$0$$. This tells us

$2\cos\theta\left(\frac{d\cos\theta}{d\theta}\right) + 2\sin\theta\left(\frac{d\sin\theta}{d\theta}\right) = 0$

which means that the vector with components $$\cos\theta$$ and $$\sin\theta$$ has $$0$$ dot product with the vector whose components are their corresponding derivatives.

It is, fortunately very easy to find all vectors in two dimensions that have $$0$$ dot product with a given one: you reverse the components, change one of their signs, and multiply by any constant $$c$$. Thus $$(a,b)$$ (which is a shorter way of writing $$a\hat{i} + b\hat{j}$$) has $$0$$ dot product with $$(cb, -ca)$$.

This tells us $$\frac{d\sin\theta}{d\theta} = c\cos\theta$$ and $$\frac{d\cos\theta}{d\theta} = -c\sin\theta$$, for some constant $$c$$.

We can determine the constant $$c$$ by examining these statements at the point for which $$\theta = 0$$.

If our angle is measured in radians, we have observed that on the unit circle, moving a distance $$d\theta$$ from angle $$0$$ changes the sine from $$0$$ to almost $$dθ$$. Thus the constant $$c$$ above is $$1$$ at angle $$0$$, and being a constant, is always $$1$$.

We conclude $$(\sin\theta)' = \cos\theta, and (\cos\theta)' = -\sin\theta$$. (The latter relation actually follows from the former from the fact that $$\cos\theta$$ is $$\sin(\frac{\pi}{2} - \theta)$$.

Exercise 7.25: Deduce the derivatives of the secant and tangent from these facts.

Why not tell us the answers?

If we did this you will have trouble remembering them. If you figure them out yourself you will have trouble forgetting them.

An interesting consequence is gotten by looking at the combination $$\cos x + i \sin x$$. ($$i$$ here is the square root of $$-1$$.) Notice that its derivative is $$i$$ times itself. And its value at $$x = 0$$ is $$1$$. We know what that means. A function whose derivative is $$q$$ times itself, whose value at $$x = 0$$ is $$1$$, is $$\exp(qx)$$.

We therefore find: $$\cos x + i \sin x = \exp(ix)$$.

In general we can divide any function $$f$$ into an odd part and an even part; the even part is $$\frac{f(x) + f(-x)}{2}$$ and the odd part is $$\frac{f(x) - f(-x)}{2}$$. The sum of the two parts is $$f(x)$$.

Since $$\cos x$$ is even and $$\sin x$$ is odd, we can identify $$\cos x$$ as the even part of $$\exp(ix)$$, and $$i\sin x$$ as its odd part.

The formal expressions are

$\cos x = \frac{\exp(ix) + \exp(-ix)}{2}$

and

$\sin x = \frac{\exp(ix) - \exp(-ix)}{2i}$

Power series expansions of Sines and Cosines

We have seen that $$\cos 0$$ is $$1$$. Since the derivative of the sine is the cosine, $$\sin x$$, when written as a sum of powers of $$x$$, must have a term in that sum whose derivative is $$1$$. That has to be $$x$$, so that the first term in the power series expansion for $$\sin x$$ is $$x$$. The cosine has derivative $$-\sin x$$, so it must have a term in its power series expansion whose derivative is $$-x$$, which term must be $$-\frac{x^2}{2}$$. $$\sin x$$ similarly must have a term whose derivative is that, namely $$-\frac{x^3}{3!}$$; this forces a term in the cosine series to be $$-\frac{x^4}{4!}$$, etc.

So what do we get?

The sine has contributions from all the odd powers and their signs alternate:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... + (-1)^k\frac{x^{2k+1}}{(2k+1)!} + ...$

while the cosine has similarly alternating sign terms from the even powers:

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + (-1)^k\frac{x^{2k}}{(2k)!} + ...$

By the way, $$\cosh x$$ and $$\sinh x$$ are the even and odd parts of $$\exp(x)$$. Their power series expansions are similar to those of the cosine and sine except that all terms are positive.

There are power series expressions for all the trigonometric functions, but you can figure them out yourself if you ever want to do so, from their relations to the sine and cosine of angle

We have already noticed that the standard measure of angle, in terms of degrees or radians is additive: this measure of the sum of two angles is the sum of the same measures of each summand. This statement is not true for sines or cosines. The sine of the sum of two angles is not the sum of their sines. The addition theorems tell us how to compute the sine and cosine of the sum of two angles in terms of the sines and cosines of the two angles that are summed.

The easiest way to find or prove addition theorems for sines and cosines is to use their relations to exponentials. We already know the addition theorem for exponentials:

$\exp(cx + cy) = \exp(cx)\exp(cy)$

Since $$\exp(ix)$$ is $$\cos x + i\sin x$$, we find that $$\exp(i(x+y))$$ is

$(\cos x + i\sin x)(\cos y + i\sin y)$

which is

$((\cos x)(\cos y) - (\sin x)(\sin y)) + i((\sin x)(\cos y) + (\cos x)(\sin y))$

The real part of this last expression is $$\cos(x+y)$$ and the imaginary part is $$\sin(x+y)$$ and these are our addition theorems.

For the hasty:

$\frac{d(\tan x)}{dx} = \frac{d(\sin x/\cos x)}{dx} = \frac{\cos x}{\cos x} - \frac{\sin x(-\sin x)}{(\cos x)^2} = 1 + (\tan x)^2$

From triangle $$OBC$$ in the picture at the top of section 7.2 we find that the last is $$(\sec x)^2$$.

$\frac{d(\sec x)}{dx} = \frac{d(1/\cos x)}{dx} = -\frac{(-\sin x)}{(\cos x)^2} = (\tan x)(\sec x)$