Section B.2 Trigonometry
Subsection B.2.1 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*}
Subsection B.2.2 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
\begin{equation*}
\sin \theta
\end{equation*}
\begin{equation*}
\cos \theta
\end{equation*}
\begin{equation*}
\tan \theta
\end{equation*}
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
\begin{equation*}
\csc \theta
\end{equation*}
\begin{equation*}
\sec \theta
\end{equation*}
\begin{equation*}
\cot \theta
\end{equation*}
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 \(\sin^2\theta\text{.}\)
Subsection B.2.3 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 | \(\tfrac{1}{\sqrt{3}}\) |
Subsection B.2.4 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.
Subsection B.2.5 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*}Subsection B.2.6 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*}
Subsection B.2.7 Identities — Extras
Subsubsection B.2.7.1 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*}
Subsubsection B.2.7.2 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*}
