B.2 Trigonometry

Angles — Radians vs Degrees

For mathematics, and especially in calculus, it is much better to measure angles in units called radians rather than degrees. By definition, an arc of length $\theta$ on a circle of radius one subtends an angle of $\theta$ radians at the centre of the circle.

The circle on the left has radius 1, and the arc swept out by an angle of $\theta$ radians has length $\theta\text{.}$ Because a circle of radius one has circumference $2\pi$ we have

$$\begin{align*} 2\pi\text{ radians }&=360^\circ & \pi\text{ radians }&=180^\circ & \frac{\pi}{2}\text{ radians }&=90^\circ\\ \frac{\pi}{3}\text{ radians }&=60^\circ & \frac{\pi}{4}\text{ radians }&=45^\circ & \frac{\pi}{6}\text{ radians }&=30^\circ \end{align*}$$

More generally, consider a circle of radius $r\text{.}$ Let $L(\theta)$ denote the length of the arc swept out by an angle of $\theta$ radians and let $A(\theta)$ denote the area of the sector (or wedge) swept out by the same angle. Since the angle sweeps out the fraction $\frac{\theta}{2\pi}$ of a whole circle, we have

$$\begin{align*} L(\theta) &= 2\pi r \cdot \frac{\theta}{2\pi} = \theta r & \text{and}\\ A(\theta) &= \pi r^2 \cdot \frac{\theta}{2\pi} = \frac{\theta}{2} r^2 \end{align*}$$

Trig Function Definitions

The trigonometric functions are defined as ratios of the lengths of the sides of a right-angle triangle as shown in the left of the diagram below . These ratios depend only on the angle $\theta\text{.}$

The trigonometric functions sine, cosine and tangent are defined as ratios of the lengths of the sides

$$\begin{align*} \sin\theta &= \frac{\text{opposite}}{\text{hypotenuse}} & \cos\theta &= \frac{\text{adjacent}}{\text{hypotenuse}} & \tan\theta &= \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin \theta}{\cos \theta}.\\ \end{align*}$$

These are frequently abbreviated as

$$\begin{align*} \sin\theta &= \frac{\text{o}}{\text{h}} & \cos\theta &= \frac{\text{a}}{\text{h}} & \tan\theta &= \frac{\text{o}}{\text{a}} \end{align*}$$

which gives rise to the mnemonic

$$\begin{align*} \text{SOH} && \text{CAH} && \text{TOA} \end{align*}$$

If we scale the triangle so that they hypotenuse has length $1$ then we obtain the diagram on the right. In that case, $\sin \theta$ is the height of the triangle, $\cos \theta$ the length of its base and $\tan \theta$ is the length of the line tangent to the circle of radius 1 as shown.

Since the angle $2\pi$ sweeps out a full circle, the angles $\theta$ and $\theta+2\pi$ are really the same.

Hence all the trigonometric functions are periodic with period $2\pi\text{.}$ That is

$$\begin{align*} \sin(\theta+2\pi) &= \sin(\theta) & \cos(\theta+2\pi) &= \cos(\theta) & \tan(\theta+2\pi) &= \tan(\theta) \end{align*}$$

The plots of these functions are shown below

$$\sin \theta \nonumber$$

$$\cos \theta \nonumber$$

$$\tan \theta \nonumber$$

The reciprocals (cosecant, secant and cotangent) of these functions also play important roles in trigonometry and calculus:

$$\begin{align*} \csc \theta &= \frac{1}{\sin\theta} = \frac{\text{h}}{\text{o}} & \sec\theta &= \frac{1}{\cos \theta} = \frac{\text{h}}{\text{a}} & \cot\theta &= \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin\theta} = \frac{\text{a}}{\text{o}} \end{align*}$$

The plots of these functions are shown below

$$\csc \theta \nonumber$$

$$\sec \theta \nonumber$$

$$\cot \theta \nonumber$$

These reciprocal functions also have geometric interpretations:

Since these are all right-angled triangles we can use Pythagoras to obtain the following identities:

$$\begin{align*} \sin^2\theta + \cos^2 \theta &=1 & \tan^2\theta + 1 &= \sec^2\theta & 1 + \cot^2 \theta &=\csc^2\theta \end{align*}$$

Of these it is only necessary to remember the first

$$\begin{align*} \sin^2\theta + \cos^2 \theta &=1 \end{align*}$$

The second can then be obtained by dividing this by $\cos^2\theta$ and the third by dividing by $\sin^2\theta\text{.}$

Important Triangles

Computing sine and cosine is non-trivial for general angles — we need Taylor series (or similar tools) to do this. However there are some special angles (usually small integer fractions of $\pi$) for which we can use a little geometry to help. Consider the following two triangles.

The first results from cutting a square along its diagonal, while the second is obtained by cutting an equilateral triangle from one corner to the middle of the opposite side. These, together with the angles $0,\frac{\pi}{2}$ and $\pi$ give the following table of values

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\csc\theta$	$\sec\theta$	$\cot\theta$
$0$ rad	0	1	0	DNE	1	DNE
$\tfrac{\pi}{2}$ rad	1	0	DNE	1	DNE	0
$\pi$ rad	0	-1	0	DNE	-1	DNE
$\tfrac{\pi}{4}$ rad	$\tfrac{1}{\sqrt{2}}$	$\tfrac{1}{\sqrt{2}}$	1	$\sqrt{2}$	$\sqrt{2}$	1
$\tfrac{\pi}{6}$ rad	$\tfrac{1}{2}$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{\sqrt{3}}$	2	$\tfrac{2}{\sqrt{3}}$	$\sqrt{3}$
$\tfrac{\pi}{3}$ rad	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{2}$	$\sqrt{3}$	$\tfrac{2}{\sqrt{3}}$	2	$\tfrac{1}{\sqrt{3}}$

Some More Simple Identities

Consider the figure below

The pair triangles on the left shows that there is a simple relationship between trigonometric functions evaluated at $\theta$ and at $-\theta\text{:}$

$$\begin{align*} \sin(-\theta)&=-\sin(\theta) & \cos(-\theta) &=\cos(\theta) \end{align*}$$

That is — sine is an odd function, while cosine is even. Since the other trigonometric functions can be expressed in terms of sine and cosine we obtain

$$\begin{align*} \tan(-\theta) &=-\tan(\theta) & \csc(-\theta) &=-\csc(\theta) & \sec(-\theta) &=\sec(\theta) & \cot(-\theta) &=-\cot(\theta) \end{align*}$$

Now consider the triangle on the right — if we consider the angle $\frac{\pi}{2}-\theta$ the side-lengths of the triangle remain unchanged, but the roles of “opposite” and “adjacent” are swapped. Hence we have

$$\begin{align*} \sin\left(\tfrac{\pi}{2}-\theta\right)&=\cos\theta & \cos\left(\tfrac{\pi}{2}-\theta\right)&=\sin\theta \end{align*}$$

Again these imply that

$$\begin{align*} \tan\left(\tfrac{\pi}{2}-\theta\right)&=\cot\theta & \csc\left(\tfrac{\pi}{2}-\theta\right)&=\sec\theta & \sec\left(\tfrac{\pi}{2}-\theta\right)&=\csc\theta & \cot\left(\tfrac{\pi}{2}-\theta\right)&=\tan\theta \end{align*}$$

We can go further. Consider the following diagram:

This implies that

$$\begin{align*} \sin(\pi-\theta)&=\sin(\theta) & \cos(\pi-\theta) &= -\cos(\theta)\\ \sin(\pi+\theta)&=-\sin(\theta) & \cos(\pi+\theta) &=-\cos(\theta) \end{align*}$$

From which we can get the rules for the other four trigonometric functions.

Identities — Adding Angles

We wish to explain the origins of the identity

$$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta). \end{align*}$$

A very geometric demonstration uses the figure below and an observation about areas.

• The left-most figure shows two right-angled triangles with angles $\alpha$ and $\beta$ and both with hypotenuse length $1\text{.}$
• The next figure simply rearranges the triangles — translating and rotating the lower triangle so that it lies adjacent to the top of the upper triangle.
• Now scale the lower triangle by a factor of $q$ so that edges opposite the angles $\alpha$ and $\beta$ are flush. This means that $q \cos \beta = \cos \alpha\text{.}$ ie

  $$\begin{align*} q &= \frac{\cos\alpha}{\cos\beta} \end{align*}$$

  Now compute the areas of these (blue and red) triangles

  $$\begin{align*} A_\text{red} &= \frac{1}{2} q^2 \sin\beta \cos \beta\\ A_\text{blue} &= \frac{1}{2} \sin \alpha \cos \alpha\\ \end{align*}$$

  So twice the total area is

  $$\begin{align*} 2 A_\text{total} &= \sin \alpha \cos \alpha + q^2 \sin\beta \cos \beta \end{align*}$$
• But we can also compute the total area using the rightmost triangle:

  $$\begin{align*} 2 A_\text{total} &= q \sin(\alpha+\beta) \end{align*}$$

Since the total area must be the same no matter how we compute it we have

$$\begin{align*} q \sin(\alpha+\beta) &= \sin \alpha \cos \alpha + q^2 \sin\beta \cos \beta\\ \sin(\alpha+\beta) &= \frac{1}{q} \sin \alpha \cos \alpha + q \sin\beta \cos \beta\\ &= \frac{\cos \beta}{\cos \alpha} \sin \alpha \cos \alpha + \frac{\cos \alpha}{\cos \beta} \sin\beta \cos \beta\\ &= \sin \alpha \cos \beta + \cos \alpha \sin\beta \end{align*}$$

as required.

We can obtain the angle addition formula for cosine by substituting $\alpha \mapsto \pi/2-\alpha$ and $\beta \mapsto -\beta$ into our sine formula:

$$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) & \text{becomes}\\ \underbrace{\sin(\pi/2-\alpha-\beta)}_{\cos(\alpha+\beta)} &= \underbrace{\sin(\pi/2-\alpha)}_{\cos(\alpha)}\cos(-\beta) + \underbrace{\cos(\pi/2-\alpha)}_{\sin(\alpha)}\sin(-\beta)\\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \end{align*}$$

where we have used $\sin(\pi/2-\theta)=\cos(\theta)$ and $\cos(\pi/2-\theta)=\sin(\theta)\text{.}$

It is then a small step to the formulas for the difference of angles. From the relation

$$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \end{align*}$$

we can substitute $\beta \mapsto -\beta$ and so obtain

$$\begin{align*} \sin(\alpha - \beta) &= \sin(\alpha)\cos(-\beta) + \cos(\alpha)\sin(-\beta)\\ &= \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \end{align*}$$

The formula for cosine can be obtained in a similar manner. To summarise

$$\begin{align*} \sin(\alpha \pm \beta) &= \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta)\\ \cos(\alpha \pm \beta) &= \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta) \end{align*}$$

The formulas for tangent are a bit more work, but

$$\begin{align*} \tan(\alpha + \beta) &= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\\ &= \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) }\\ &= \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) } \cdot \frac{\sec(\alpha) \sec(\beta)}{\sec(\alpha) \sec(\beta)}\\ &= \frac{\sin(\alpha)\sec(\alpha) + \sin(\beta)\sec(\beta)}{1 - \sin(\alpha)\sec(\alpha)\sin(\beta)\sec(\beta) }\\ &= \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta) }\\ \end{align*}$$

and similarly we get

$$\begin{align*} \tan(\alpha - \beta) &= \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta) } \end{align*}$$

Identities — Double-angle Formulas

If we set $\beta=\alpha$ in the angle-addition formulas we get

$$\begin{align*} \sin(2\alpha) &= 2\sin(\alpha)\cos(\alpha)\\ \cos(2\alpha) &= \cos^2(\alpha)-\sin^2(\alpha)\\ &= 2\cos^2(\alpha)-1 & \text{since } \sin^2\theta =1-\cos^2\theta\\ &= 1-2\sin^2(\alpha) & \text{since } \cos^2\theta =1-\sin^2\theta\\ \tan(2\alpha) &= \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}\\ &= \frac{2}{\cot(\alpha)-\tan(\alpha)} &\text{divide top and bottom by $\tan(\alpha)$} \end{align*}$$

Identities — Extras

Sums to Products

Consider the identities

$$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha) \sin(\beta) & \sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta) - \cos(\alpha) \sin(\beta) \end{align*}$$

If we add them together some terms on the right-hand side cancel:

$$\begin{align*} \sin(\alpha+\beta) + \sin(\alpha-\beta) &= 2\sin(\alpha)\cos(\beta). \end{align*}$$

If we now set $u=\alpha+\beta$ and $v = \alpha-\beta$ (i.e. $\alpha=\frac{u+v}{2}, \beta=\frac{u-v}{2}$) then

$$\begin{align*} \sin(u) + \sin(v) &= 2\sin\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right) \end{align*}$$

This transforms a sum into a product. Similarly:

$$\begin{align*} \sin(u) - \sin(v) &= 2\sin\left(\frac{u - v}{2}\right)\cos\left(\frac{u + v}{2}\right)\\ \cos(u) + \cos(v) &= 2\cos\left(\frac{u + v}{2}\right)\cos\left(\frac{u - v}{2}\right)\\ \cos(u) - \cos(v) &= -2\sin\left(\frac{u + v}{2}\right)\sin\left(\frac{u - v}{2}\right) \end{align*}$$

Products to sums

Again consider the identities

$$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha) \sin(\beta) & \sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta) - \cos(\alpha) \sin(\beta) \end{align*}$$

and add them together:

$$\begin{align*} \sin(\alpha+\beta) + \sin(\alpha-\beta) &= 2\sin(\alpha)\cos(\beta). \end{align*}$$

Then rearrange:

$$\begin{align*} \sin(\alpha)\cos(\beta)&= \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2} \end{align*}$$

In a similar way, start with the identities

$$\begin{align*} \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) & \cos(\alpha-\beta) &= \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \end{align*}$$

If we add these together we get

$$\begin{align*} 2\cos(\alpha)\cos(\beta) &= \cos(\alpha+\beta) + \cos(\alpha-\beta)\\ \end{align*}$$

while taking their difference gives

$$\begin{align*} 2\sin(\alpha)\sin(\beta) &= \cos(\alpha-\beta) - \cos(\alpha+\beta) \end{align*}$$

Hence

$$\begin{align*} \sin(\alpha)\sin(\beta)&= \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}\\ \cos(\alpha)\cos(\beta)&= \frac{\cos(\alpha-\beta) + \cos(\alpha+\beta)}{2} \end{align*}$$