Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

Subsection 1.3.1 Angles and Sectors of Circles

Mathematicians tend to deal mostly with radians and we will see later that some formulas are more elegant when using radians (rather than degrees). The relationship between degrees and radians is:

Using this formula, some common angles can be derived:

From now on, unless otherwise indicated, we will always use radian measure.

In the diagram below is a sector of a circle with central angle \(\theta\) and radius \(r\) subtending an arc with length \(s\text{.}\)

When \(\theta\) is measured in radians, we have the following formula relating \(\theta\text{,}\) \(s\) and \(r\text{:}\)

Subsection 1.3.2 Trigonometric Functions

There are six basic trigonometric functions:

• Sine (abbreviated by \(\sin\))

• Cosecant (abbreviated by \(\csc\))

• Cosine (abbreviated by \(\cos\))

• Secant (abbreviated by \(\sec\))

• Tangent (abbreviated by \(\tan\))

• Cotangent (abbreviated by \(\cot\))

We first describe trigonometric functions in terms of ratios of two sides of a right angle triangle containing the angle \(\theta\text{.}\)

With reference to the above triangle, for an acute angle \(\theta\) (that is, \(0\leq\theta\lt \pi/2\)), the six trigonometric functions can be described as follows:

Notice that sine is the ratio of the opposite and hypotenuse. We use the mnemonic SOH to remember this ratio. Similarly, CAH and TOA remind us of the \(\cos\) and \(\tan\) ratios.

This description does not apply to obtuse or negative angles. To define the six basic trigonometric functions we first define sine and cosine as the lengths of various line segments from a unit circle, and then we define the remaining four basic trigonometric functions in terms of sine and cosine.

Take a line originating at the origin (making an angle of \(\theta\) with the positive half of the \(x\)-axis) and suppose this line intersects the unit circle at the point \((x,y)\text{.}\) The \(x\)- and \(y\)-coordinates of this point of intersection are equal to \(\cos\theta\) and \(\sin\theta\text{,}\) respectively.

Interactive Demonstration. Move the point on the circle clockwise or counterclockwise to investigate the sine curve (blue) and cosine curve (green) .

For angles greater than \(2\pi\) or less than \(-2\pi\text{,}\) simply continue to rotate around the circle. In this way, sine and cosine become periodic functions with period \(2\pi\text{:}\)

for any angle \(\theta\) and any integer \(k\text{.}\)

Above, only sine and cosine were defined directly by the circle. We now define the remaining four basic trigonometric functions in terms of the functions \(\sin\theta\) and \(\cos\theta\text{:}\)

Subsection 1.3.3 Computing Exact Trigonometric Ratios

The unit circle is often used to determine the exact value of a particular trigonometric function.

Reading from the unit circle one can see that \(\cos 5\pi/6=-\sqrt 3/2\) and \(\sin 5\pi/6=1/2\) (remember the that the \(x\)-coordinate is \(\cos\theta\) and the \(y\)-coordinate is \(\sin\theta\)). However, we don't always have access to the unit circle. In this case, we can compute the exact trigonometric ratios for \(\theta=5\pi/6\) by using special triangles and the CAST Rule described below.

The first special triangle has angles of \(45^\circ,45^\circ,90^\circ\) (i.e., \(\pi/4,\pi/4,\pi/2\)) with side lengths \(1,1,\sqrt 2\text{,}\) while the second special triangle has angles of \(30^\circ,60^\circ,90^\circ\) (i.e., \(\pi/6,\pi/3,\pi/2\)) with side lengths \(1,2,\sqrt 3\text{.}\) They are classically referred to as the \(1\)-\(1\)-\(\sqrt{2}\) triangle, and the \(1\)-\(2\)-\(\sqrt{3}\) triangle, respectively, shown below.

The special triangles allow us to compute the exact value (excluding the sign) of trigonometric ratios, but to determine the sign, we can use the CAST Rule.

Example 1.37. Determining Trigonometric Ratios Without Unit Circle.

Determine \(\sin 5\pi/6\text{,}\) \(\cos 5\pi/6\text{,}\) \(\tan 5\pi/6\text{,}\) \(\sec 5\pi/6\text{,}\) \(\csc 5\pi/6\) and \(\cot 5\pi/6\) exactly by using the special triangles and CAST Rule.

Solution

We start by drawing the \(x\)-\(y\)-plane and indicating our angle of \(5\pi/6\) in standard position (positive angles rotate counterclockwise while negative angles rotate clockwise). Next, we drop a perpendicular to the \(x\)-axis (never drop it to the \(y\)-axis!).

Notice that we can now figure out the angles in the triangle. Since \(180^\circ=\pi\text{,}\) we have an interior angle of \(\pi-5\pi/6=\pi/6\) inside the triangle. As the angles of a triangle add up to \(180^\circ=\pi\), the other angle must be \(\pi/3\text{.}\) This gives one of our special triangles. We label it accordingly and add the CAST Rule to our diagram.

From the above figure we see that \(5\pi/6\) lies in quadrant II where \(\sin\theta\) is positive and \(\cos\theta\) and \(\tan\theta\) are negative. This gives us the sign of \(\sin\theta\text{,}\) \(\cos\theta\) and \(\tan\theta\text{.}\) To determine the value we use the special triangle and SOH CAH TOA.

Using \(\sin\theta=opp/hyp\) we find a value of \(1/2\text{.}\) Since \(\sin\theta\) is positive in quadrant II, we have

\begin{equation*} \sin \frac{5\pi}{6}=+\frac{1}{2}\text{.} \end{equation*}

Using \(\cos\theta=adj/hyp\) we find a value of \(\sqrt 3/2\text{.}\) But \(\cos\theta\) is negative in quadrant II, therefore,

\begin{equation*} \cos \frac{5\pi}{6}=-\frac{\sqrt 3}{2}\text{.} \end{equation*}

Using \(\tan\theta=opp/adj\) we find a value of \(1/\sqrt 3\text{.}\) But \(\tan\theta\) is negative in quadrant II, therefore,

\begin{equation*} \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}\text{.} \end{equation*}

To determine \(\sec\theta\text{,}\) \(\csc\theta\) and \(\cot\theta\) we use the definitions:

\begin{equation*} \csc \frac{5\pi}{6} = \ds\frac{1}{\sin \frac{5\pi}{6}} = +2, \qquad \sec \frac{5\pi}{6} = \ds\frac{1}{\cos \frac{5\pi}{6}} = -\frac{2}{\sqrt 3}, \qquad\cot \frac{5\pi}{6} = \ds\frac{1}{\tan \frac{5\pi}{6}} = -\sqrt 3\text{.} \end{equation*}

Example 1.38. CAST Rule.

If \(\cos\theta=3/7\) and \(3\pi/2\lt \theta\lt 2\pi\text{,}\) then find \(\cot\theta\text{.}\)

Solution

We first draw a right angle triangle. Since \(\cos\theta=adj/hyp=3/7\text{,}\) we let the adjacent side have length \(3\) and the hypotenuse have length \(7\text{.}\)

Using the Pythagorean Theorem, we have \(3^2+(\mbox{opp} )^2=7^2\text{.}\) Thus, the opposite side has length \(\sqrt{40}\text{.}\)

To find \(\cot\theta\) we use the definition:

\begin{equation*} \cot\theta=\frac{1}{\tan\theta}\text{.} \end{equation*}

Since we are given \(3\pi/2\lt \theta\lt 2\pi\text{,}\) we are in the fourth quadrant. By the CAST Rule, \(\tan\theta\) is negative in this quadrant. As \(\tan\theta=opp/adj\text{,}\) it has a value of \(\sqrt{40}/3\text{,}\) but by the CAST Rule it is negative, that is,

\begin{equation*} \tan\theta=-\frac{\sqrt{40}}{3}\text{.} \end{equation*}

Therefore,

\begin{equation*} \cot\theta=-\frac{3}{\sqrt{40}}\text{.} \end{equation*}

Subsection 1.3.4 Graphs of Trigonometric Functions

The graph of the functions \(\sin x\) and \(\cos x\) can be visually represented as:

Both \(\sin x\) and \(\cos x\) have domain \((-\infty,\infty)\) and range \([-1,1]\text{.}\) That is,

The zeros of \(\sin x\) occur at the integer multiples of \(\pi\text{,}\) that is, \(\sin x=0\) whenever \(x=n\pi\text{,}\) where \(n\) is an integer. Similarly, \(\cos x=0\) whenever \(x=\pi/2+n\pi\text{,}\) where \(n\) is an integer.

The six basic trigonometric functions can be visually represented as:

Both tangent and cotangent have range \((-\infty,\infty)\text{,}\) whereas cosecant and secant have range \((-\infty,-1]\cup[1,\infty)\text{.}\) Each of these functions is periodic. Tangent and cotangent have period \(\pi\text{,}\) whereas sine, cosine, cosecant and secant have period \(2\pi\text{.}\)

Subsection 1.3.5 Trigonometric Identities

There are numerous trigonometric identities, including those relating to shift/periodicity, Pythagoras type identities, double-angle formulas, half-angle formulas and addition formulas. We list these below.