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## Section1.3Trigonometry

In this section we review the definitions of trigonometric functions.

### Subsection1.3.1Angles 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:

\begin{equation*} \pi~\mbox{rad} =180^\circ\text{.} \end{equation*}

Using this formula, some common angles can be derived:

\begin{equation*} \begin{array}{l|ccccccccccc} \mbox{Degrees} \amp 0^\circ \amp 30^\circ \amp 45^\circ \amp 60^\circ \amp 90^\circ \amp 120^\circ \amp 135^\circ \amp 150^\circ \amp 180^\circ \amp 270^\circ \amp 360^\circ \\ \hline \mbox{Radians} \amp 0 \amp \ds{\frac{\pi}{6}} \amp \ds{\frac{\pi}{4}} \amp \ds{\frac{\pi}{3}} \amp \ds{\frac{\pi}{2}} \amp \ds{\frac{2\pi}{3}} \amp \ds{\frac{3\pi}{4}} \amp \ds{\frac{5\pi}{6}} \amp \ds{\pi} \amp \ds{\frac{3\pi}{2}} \amp 2\pi \end{array} \end{equation*}
###### Example1.32. Degrees to Radians.

To convert $45^\circ$ to radians, multiply by $\ds{\frac{\pi}{180^\circ}}$ to get $\ds{\frac{\pi}{4}}\text{.}$

###### Example1.33. Radians to Degrees.

To convert $\ds{\frac{5\pi}{6}}$ radians to degrees, multiply by $\ds{\frac{180^\circ}{\pi}}$ to get $150^\circ\text{.}$

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 $r\text{:}$

\begin{equation*} \theta=\frac{s}{r}\mbox{ or } s=r\theta\text{.} \end{equation*}
###### Sector Area.

The area of the sector is equal to:

\begin{equation*} \mbox{Sector Area} =\frac{1}{2}r^2\theta\text{.} \end{equation*}
###### Example1.34. Angle Subtended by Arc.

If a circle has radius $3$ cm, then an angle of $2$ rad is subtended by an arc of $6$ cm ($s=r\theta=3\cdot 2=6$).

###### Example1.35. Area of Circle.

If we substitute $\theta=2\pi$ (a complete revolution) into the sector area formula we get the area of a circle:

\begin{equation*} A=\frac{1}{2}r^2(2\pi)=\pi r^2 \end{equation*}

### Subsection1.3.2Trigonometric 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:

###### Definition1.36. Basic Six Trigonometric Functions.
\begin{equation*} \begin{array}{ccc} \ds\sin\theta=\frac{ opp}{ hyp}\amp \qquad\amp \ds\csc\theta=\frac{ hyp}{ opp}\\ \\ \ds\cos\theta=\frac{ adj}{ hyp}\amp \qquad\amp \ds\sec\theta=\frac{ hyp}{ adj}\\ \\ \ds\tan\theta=\frac{ opp}{ adj}\amp \qquad\amp \ds\cot\theta=\frac{ adj}{ opp} \end{array} \end{equation*}

where opp stands for opposite, adj for adjacent, and hyp for hypotenuse.

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.

###### Mnemonic.

The mnemonic SOH CAH TOA is useful in remembering how trigonometric functions of acute angles relate to the sides of a right triangle.

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 $2\pi\text{:}$

\begin{equation*} \begin{array}{cc} \sin\theta = \sin\left(\theta + 2\pi k \right) \amp \cos\theta = \cos\left(\theta + 2\pi k \right) \end{array} \end{equation*}

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 $\cos\theta\text{:}$

\begin{equation*} \begin{array}{cc} \tan\theta = \dfrac{\sin\theta}{\cos\theta} \amp \sec\theta = \dfrac{1}{\cos\theta} \\[1ex] \csc\theta = \dfrac{1}{\sin\theta} \amp \cot\theta = \dfrac{\cos\theta}{\sin\theta} \end{array} \end{equation*}

### Subsection1.3.3Computing 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. ###### Mnemonic.

The first triangle should be easy to remember. To remember the second triangle, place the largest number ($2$) across from the largest angle ($90^\circ=\pi/2$). Place the smallest number ($1$) across from the smallest angle ($30^\circ=\pi/6$). Place the middle number ($\sqrt 3\approx 1.73$) across from the middle angle ($60^\circ=\pi/3$). Double check using the Pythagorean Theorem that the sides satisfy $a^2+b^2=c^2\text{.}$

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.

###### The CAST Rule.

The CAST Rule says that in quadrant I all three of $\sin\theta\text{,}$ $\cos\theta\text{,}$ $\tan\theta$ are positive. In quadrant II, only $\sin\theta$ is positive, while $\cos\theta\text{,}$ $\tan\theta$ are negative. In quadrant III, only $\tan\theta$ is positive, while $\sin\theta\text{,}$ $\cos\theta$ are negative. In quadrant IV, only $\cos\theta$ is positive, while $\sin\theta\text{,}$ $\tan\theta$ are negative. To remember this, simply label the quadrants by the letters C-A-S-T starting in the bottom right and labelling counter-clockwise. ###### Example1.37. 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*}
###### Example1.38. 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*}

### Subsection1.3.4Graphs 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,

\begin{equation*} -1\leq\sin x\leq 1\qquad -1\leq\cos x\leq 1\text{.} \end{equation*}

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 $2\pi\text{.}$

### Subsection1.3.5Trigonometric 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.

###### Shifts and Periodicity.

$\begin{array}{lll} ~~\ds{\sin (\theta + 2\pi)}=\ds{\sin \theta} ~~\amp ~~ ~~\ds{\cos (\theta + 2\pi)}=\ds{\cos \theta}~~\amp ~~ ~~\ds{\tan (\theta + 2\pi)}=\ds{\tan \theta} \\[1ex] ~~\ds{\sin (\theta + \pi)}=\ds{-\sin \theta} ~~\amp ~~ ~~\ds{\cos (\theta + \pi)}=\ds{-\cos \theta}~~ \amp ~~ ~~\ds{\tan (\theta + \pi)}=\ds{\tan \theta} \\[1ex] ~~\ds{\sin (-\theta)}=\ds{-\sin \theta} ~~\amp ~~ ~~\ds{\cos (-\theta)}=\ds{\cos \theta}~~\amp ~~ ~~\ds{\tan (-\theta)}=\ds{-\tan \theta} \\[1ex] ~~\ds{\sin \left(\frac{\pi}{2}-\theta\right)}=\ds{\cos \theta} ~~\amp ~~ ~~\ds{\cos \left(\frac{\pi}{2}-\theta\right)}=\ds{\sin \theta} ~~\amp ~~ ~~\ds{\tan \left(\frac{\pi}{2}-\theta\right)} =\ds{\cot \theta} \end{array}$

###### Pythagoras Type Formulas.
\begin{equation*} \begin{array}{l} \sin ^2 \theta + \cos ^2 \theta = 1\\ \tan^2 \theta + 1 = \sec^2 \theta\\ 1 + \cot ^2 \theta = \csc^2 \theta\\ \end{array} \end{equation*}
###### Double-angle Formulas.
\begin{equation*} \begin{array}{l} \sin(2\theta) = 2\sin\theta\cos\theta\\ \cos(2\theta)=\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1= 1-2\sin^2\theta\\ \end{array} \end{equation*}
###### Half-angle Formulas.
\begin{equation*} \begin{array}{l} \ds{\cos^2\theta = \frac{1+\cos(2\theta)}{2}}\\ \ds{\sin^2\theta = \frac{1-\cos(2\theta)}{2}}\\ \end{array} \end{equation*}
###### Addition Formulas.
\begin{equation*} \begin{array}{l} \sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi\\ \cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi\\ \ds{\tan(\theta+\phi) = \frac{\tan\theta + \tan\phi}{1 - \tan\theta\tan\phi}}\\ \sin(\theta-\phi) = \sin\theta\cos\phi - \cos\theta\sin\phi\\ \cos(\theta-\phi) = \cos\theta\cos\phi + \sin\theta\sin\phi \end{array} \end{equation*}
###### Example1.39. Double Angle.

Find all values of $x$ with $0\leq x\leq \pi$ such that $\sin 2x=\sin x\text{.}$

Solution

Using the double-angle formula $\sin 2x = 2\sin x\cos x$ we have:

\begin{equation*} 2\sin x\cos x = \sin x \end{equation*}
\begin{equation*} 2\sin x\cos x- \sin x = 0 \end{equation*}
\begin{equation*} \sin x(2\cos x - 1)=0 \end{equation*}

Thus, either $\sin x=0$ or $\cos x = 1/2\text{.}$ For the first case when $\sin x = 0\text{,}$ we get $x=0$ or $x=\pi\text{.}$ For the second case when $\cos x=1/2\text{,}$ we get $x=\pi/3$ (use the special triangles and CAST rule to get this). Thus, we have three solutions: $x=0\text{,}$ $x=\pi/3\text{,}$ $x=\pi\text{.}$

##### Exercises for Section 1.3.

Find all values of $\theta$ such that $\sin(\theta) = -1\text{;}$ give your answer in radians.

Answer
$2n\pi-\pi/2\text{,}$ any integer $n$

Find all values of $\theta$ such that $\cos(2\theta) = 1/2\text{;}$ give your answer in radians.

Answer
$n\pi\pm\pi/6\text{,}$ any integer $n$

Compute the following:

1. $\sin(3\pi)$

2. $\sec(5\pi/6)$

3. $\cos(-\pi/3)$

4. $\csc(4\pi/3)$

5. $\tan(7\pi/4)$

6. $\cot(13\pi/4)$

If $\sin\theta=\frac{3}{5}$ and $\frac{\pi}{2}\lt \theta\lt \pi\text{,}$ then find $\sec\theta\text{.}$

Answer

$-\frac{5}{4}$

Suppose that $\tan\theta=x$ and $\pi\lt \theta\lt \frac{3\pi}{2}\text{,}$ find $\sin\theta$ and $\cos\theta$ in terms of $x\text{.}$

Answer
$\sin\theta=-x/\sqrt{x^{2}+1}\text{,}$ $\cos\theta=-1/\sqrt{x^{2}+1}\text{.}$

Find an angle $\theta$ such that $-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}$ and $\sin\theta=\sin\frac{23\pi}{7}\text{.}$

Answer
$-\frac{2\pi}{7}$ is the unique answer.

Use an angle sum identity to compute $\cos(\pi/12)\text{.}$

Answer
$(\sqrt{2}+\sqrt{6})/4$

Use an angle sum identity to compute $\tan(5\pi/12)\text{.}$

Answer
$-(1+\sqrt3)/(1-\sqrt3)=2+\sqrt3$

Verify the following identities

1. $\ds \cos^2(t)/(1-\sin(t)) = 1+\sin(t)$

2. $2\csc(2\theta)=\sec(\theta)\csc(\theta)$

3. $\sin(3\theta) - \sin(\theta) = 2\cos(2\theta)\sin(\theta)$

Sketch the following functions:

1. $y=2\sin(x)$

2. $y=\sin(3x)$

3. $y=\sin(-x)$

Find all of the solutions of $\ds 2\sin(t) -1 -\sin^2(t) =0$ in the interval $[0,2\pi]\text{.}$

Answer
$t=\pi/2$
Solution

We want to find all solutions to the equation

\begin{equation*} 2\sin t - 1 -\sin^{2}t = 0 \ \ \ t \in \left[0,2\pi\right] \end{equation*}

We first make the substitution, $w = \sin t\text{.}$ Then our equation becomes,

\begin{equation*} \begin{split} 2w - 1 - w^{2} \amp = 0 \\ w^{2} - 2w + 1 \amp = 0 \\ (w-1)^{2} \amp = 0 \\ w = 1. \end{split} \end{equation*}

We have thus reduced the problem to solving $\sin t = 1$ on $\left[0,2\pi\right]\text{.}$ Therefore, $t = \frac{\pi}{2}\text{.}$