5.2 – Trig Functions of Real Numbers

5.2 Video 1 of 2: Trigonometric Functions of Real Numbers.

5.2 Video 2 of 2: Pythagorean Identities.

Section Notes

Trigonometric Functions

A function is a relationship between two sets. It assigns an “output” for each “input”. For trigonometric functions, the “input” is the number $t$ , the distance traveled around the unit circle. The output is a value involving the $x$ and $y$ coordinates of the terminal point of $t$ , as defined below.

The Trigonometric Functions
There are six trigonometric functions. For any real number $t$ with terminal point $P(x,y)$ they are: $\begin{align} \sin t&=y & \cos t&=x & \tan t&=\dfrac{y}{x} \ \ \ (x\not=0)\\ \csc t&=\dfrac{1}{y} \ \ \ (y\not=0) & \sec t&=\dfrac{1}{x} \ \ \ (x\not=0) & \cot t&=\dfrac{x}{y} \ \ \ (y\not=0)\end{align}$

The Trigonometric Functions

There are six trigonometric functions. For any real number $t$ with terminal point $P(x,y)$ they are:

$\begin{align*} \sin t&=y & \cos t&=x & \tan t&=\dfrac{y}{x} \ \ \ (x\not=0)\\ \csc t&=\dfrac{1}{y} \ \ \ (y\not=0) & \sec t&=\dfrac{1}{x} \ \ \ (x\not=0) & \cot t&=\dfrac{x}{y} \ \ \ (y\not=0)\end{align*}$

Find the values of the six trigonometric functions for $t=\dfrac{\pi}{4}$
First, note that the terminal point of $t=\dfrac{\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$ . Thus, the six trigonometric functions are: $\begin{align} \sin\dfrac{\pi}{4}&=\dfrac{\sqrt{2}}{2} & \cos\dfrac{\pi}{4}&=\dfrac{\sqrt{2}}{2} & \tan\dfrac{\pi}{4}&=1\\ \csc\dfrac{\pi}{4}&=\sqrt{2} & \sec\dfrac{\pi}{4}&=\sqrt{2} & \cot\dfrac{\pi}{4}&=1\end{align}$

Find the values of the six trigonometric functions for $t=\dfrac{\pi}{4}$

First, note that the terminal point of $t=\dfrac{\pi}{4}$ is $\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$ . Thus, the six trigonometric functions are:

$\begin{align*} \sin\dfrac{\pi}{4}&=\dfrac{\sqrt{2}}{2} & \cos\dfrac{\pi}{4}&=\dfrac{\sqrt{2}}{2} & \tan\dfrac{\pi}{4}&=1\\ \csc\dfrac{\pi}{4}&=\sqrt{2} & \sec\dfrac{\pi}{4}&=\sqrt{2} & \cot\dfrac{\pi}{4}&=1\end{align*}$

Find the values of the six trigonometric functions for $t=\dfrac{2\pi}{3}$
The terminal point of $t=\dfrac{2\pi}{3}$ is $\left(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ , so the six trigonometric functions are: $\begin{align} \sin\dfrac{2\pi}{3}&=\dfrac{\sqrt{3}}{2} & \cos\dfrac{2\pi}{3}&=-\dfrac{1}{2} & \tan\dfrac{2\pi}{3}&=-\sqrt{3}\\ \csc\dfrac{2\pi}{3}&=\dfrac{2\sqrt{3}}{3} & \sec\dfrac{2\pi}{3}&=-2 & \cot\dfrac{2\pi}{3}&=-\dfrac{\sqrt{3}}{3}\end{align}$

Find the values of the six trigonometric functions for $t=\dfrac{2\pi}{3}$

The terminal point of $t=\dfrac{2\pi}{3}$ is $\left(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$ , so the six trigonometric functions are:

$\begin{align*} \sin\dfrac{2\pi}{3}&=\dfrac{\sqrt{3}}{2} & \cos\dfrac{2\pi}{3}&=-\dfrac{1}{2} & \tan\dfrac{2\pi}{3}&=-\sqrt{3}\\ \csc\dfrac{2\pi}{3}&=\dfrac{2\sqrt{3}}{3} & \sec\dfrac{2\pi}{3}&=-2 & \cot\dfrac{2\pi}{3}&=-\dfrac{\sqrt{3}}{3}\end{align*}$

Find the values of the six trigonometric functions for $t=\dfrac{\pi}{2}$
The terminal point of $t=\dfrac{\pi}{2}$ is $(0,1)$ , so the six trigonometric functions are: $\begin{align} \sin\dfrac{\pi}{2}&=1 & \cos\dfrac{\pi}{2}&=0 & \tan\dfrac{\pi}{2}&=\varnothing\\ \csc\dfrac{\pi}{2}&=1 & \sec\dfrac{\pi}{2}&=\varnothing & \cot\dfrac{\pi}{2}&=0\end{align}$ The functions $\tan$ and $\sec$ are undefined, as they have zeros in the denominators.

Using the definitions of the trigonometric functions the chart below shows the six trigonometric functions for $t=0,\dfrac{\pi}{6},\dfrac{\pi}{4},\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ :

$\bgroup\def\arraystretch{1.5}\begin{tabular}{|c|c|c|c|c|c|c|}\hline \cellcolor{blue!25!green!25}$t$ & \cellcolor{blue!25}$\sin t$ & \cellcolor{blue!25}$\cos t$ & \cellcolor{blue!25}$\tan t$ & \cellcolor{blue!25}$\csc t$ & \cellcolor{blue!25}$\sec t$ & \cellcolor{blue!25}$\cot t$ \\ \hline \cellcolor{green!25}$0$ & $0$ & $1$ & $0$ & $\varnothing$ & $1$ & $\varnothing$ \\ \hline \cellcolor{green!25}$\frac{\pi}{6}$ & $\frac{1}{2}$ & $\frac{\sqrt{3}}{2}$ & $\frac{\sqrt{3}}{3}$ & $2$ & $\frac{2\sqrt{3}}{3}$ & $\sqrt{3}$\\ \hline \cellcolor{green!25}$\frac{\pi}{4}$ & $\frac{\sqrt{2}}{2}$ & $\frac{\sqrt{2}}{2}$ & $1$ & $\sqrt{2}$ & $\sqrt{2}$ & $1$ \\ \hline \cellcolor{green!25}$\frac{\pi}{3}$ & $\frac{\sqrt{3}}{2}$ & $\frac{1}{2}$ & $\sqrt{3}$ & $\frac{2\sqrt{3}}{3}$ & $2$ & $\frac{\sqrt{3}}{3}$ \\ \hline \cellcolor{green!25}$\frac{\pi}{2}$ & $1$ & $0$ & $\varnothing$ & $1$ & $\varnothing$ & $0$\\ \hline\end{tabular}\egroup$

We can make note of the numbers for which each function is undefined by defining the domain of each function. The domain of $\sin t$ and $\cos t$ is $\mathbb{R}$ , all real numbers.

The domain of $\tan t$ and $\sec t$ is all real numbers except $\dfrac{\pi}{2}+n\pi$ for $n$ , an integer.

The domain of $\cot t$ and $\csc t$ is all real numbers except $n\pi$ for $n$ an integer.

The signs of each trigonometric function is determined by the quadrant for which $t$ lies.

In Quadrant I all trigonometric functions are positive
In Quadrant II $\sin$ and $\cos$ are positive. The other four are negative.
In Quadrant III $\tan$ and $\cot$ are positive. The other four are negative.
In Quadrant IV $\cos$ and $\sec$ are positive. The other four are negative.

Evaluate $\cos\dfrac{3\pi}{4}$
Because $\dfrac{3\pi}{4}$ is in quadrant II, the $\cos$ value will be negative. The reference number for $\dfrac{3\pi}{4}$ is $\dfrac{\pi}{4}$ which has $x-$ coordinate $\dfrac{\sqrt{2}}{2}$ , so then $\cos\dfrac{3\pi}{4}=-\dfrac{\sqrt{2}}{2}$

Evaluate $\tan\left(-\dfrac{5\pi}{3}\right)$
$-\dfrac{5\pi}{3}$ is in quadrant I, so the $\tan$ value will be positive. The reference angle is $\dfrac{\pi}{3}$ and so we find that $\tan\left(-\frac{5\pi}{3}\right)=\tan\frac{\pi}{3}=\sqrt{3}$

Evaluate $\sin\dfrac{19\pi}{4}$
$\dfrac{19\pi}{4}-4\pi=\dfrac{3\pi}{4}$ , so the terminal point of $\dfrac{19\pi}{4}$ is the same as the terminal point of $\dfrac{3\pi}{4}$ , which lies in quadrant II, making $\sin$ positive. The reference angle for $\dfrac{3\pi}{4}$ is $\dfrac{\pi}{4}$ , thus, $\sin\dfrac{19\pi}{4}=\sin\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$

While we have begun a list of trigonometric values for certain $t$ values, there are many times we will need to evaluate the trig functions for other $t$ values. For example, how would we estimate $\sin 1.8$ ? Engineers and mathematicians used to use books filled with tables to estimate such values. Slide rulers were also used to be able to estimate such values. Luckily, today, modern calculators have programs to evaluate such values. The calculator must be set to radian mode (as opposed to degrees). Inputting $\sin 1.8$ into a calculator and rounding to six decimal places gives 0.973848.

Use a calculator to find the value of $\sin 2.6$ . Round to six decimal places.
Inputting this into a calculator yields $\sin 2.6=0.515501$

Use a calculator to find the value of $\cos 5.2$ . Round to six decimal places.
Inputting this into a calculator yields $\cos 5.2=0.468517$

Use a calculator to find the value of $\tan 2.5$ . Round to six decimal places.
Inputting this into a calculator yields $\tan 2.5=-0.747022$

Some calculators have the functions $\csc, \sec,$ and $\cot$ and some do not. If your calculator does not have these functions, you can easily use the reciprocal relations to evaluate them:

Reciprocal relations
$\csc t=\dfrac{1}{\sin t}, \ \ \ \ \ \ \sec t=\dfrac{1}{\cos t}, \ \ \ \ \ \ \cot t=\dfrac{1}{\tan t}$

Use a calculator to find the value of $\sec 4.7$ . Round to six decimal places.
$\sec 4.7=\dfrac{1}{\cos 4.7}=-80.718958$

Use a calculator to find the value of $\csc 0.5$ . Round to six decimal places.
$\csc 0.5=\dfrac{1}{\sec 0.5}=2.085830$

When we view the relationship of the terminal point of $t$ and $-t$ , as pictured below:

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It can be seen that the terminal point of $t$ is $(x,y)$ , while the terminal point of $-t$ is $(x,y)$ . Thus, the $\sin$ value of $-t$ is the opposite of the $\sin$ value of $t$ . The $\cos$ is unchanged. This gives way to the even and odd properties of trigonometric functions:

Even-Odd Properties of Trig Functions
$\begin{align} \sin (-t)&=-\sin(t) & \cos(-t)&=\cos(t) & \tan(-t)&=-\tan(t)\\ \csc (-t)&=-\csc(t) & \sec(-t)&=\sec(t) & \cot(-t)&=-\cot(t)\end{align}$

Use the even-odd properties of the trigonometric functions to determine the value of $\cos\left(\dfrac{-\pi}{3}\right)$
$\cos$ is a n even function, so we have $\cos\left(\dfrac{-\pi}{3}\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$

Use the even-odd properties of the trigonometric functions to determine the value of $\sin\left(\dfrac{-\pi}{4}\right)$
$\sin$ is an odd function, so we have $\sin\left(\dfrac{-\pi}{4}\right)=-\sin\left(\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}$

It is also worth noting that for any terminal point, this point is on the unit circle, and thus, the $x$ and $y$ values must solve the equation of the unit circle. If we replace $x$ and $y$ with the expression $\cos t$ and $\sin t$ , respectively, we obtain the first of three Pythagorean Identities:

$x^2+y^2=1$
$\cos^2t+sin^2t=1$ There are two other Pythagorean identities, and they can be created from the first one (above) by dividing by either $\sin^2t$ or $\cos^2t$ and simplifying. The results of these calculations are listed here:

Pythagorean Identities
$\sin^2t+\cos^2t=1 \ \ \ \ \ \ \tan^2t+1=\sec^2t \ \ \ \ \ \ 1+\cot^2t=\csc^2y$

These identities can often aid in finding the other trigonometric values when one of them is known

If $\cos t=\dfrac{4}{5}$ and $t$ is in quadrant IV, find the values of all of the trigonometric functions at $t$
Using the first Pythagorean identity, we have $\sin^2t+\cos^2t=1$ $\sin^2t+\left(\dfrac{4}{5}\right)^2=1$ $\sin^2t+\dfrac{16}{25}=1$ $\sin^2t=\dfrac{9}{25}$ $\sin t=\pm\sqrt{\dfrac{9}{25}}$ $\sin t=-\dfrac{3}{5}$ The negative square root is chosen because $t$ is in Quadrant IV, so $\sin t$ is a negative value. Knowing both the $\sin$ and $\cos$ values at $t$ is enough to find the remaining trigonometric functions, yielding: $\begin{align}\cos t & = \dfrac{4}{5} & \sin t & = -\dfrac{3}{5} & \tan t & = \dfrac{\sin t}{\cos t}=\dfrac{-\frac{3}{5}}{\frac{4}{5}}=-\dfrac{3}{4}\\ \csc t&=\frac{1}{\sin t} = -\dfrac{5}{3} & \sec t & = \dfrac{1}{\cos t} = \dfrac{5}{4} & \cot t & = \frac{1}{\tan t} = -\dfrac{4}{3}\end{align}$

If $\cos t=\dfrac{4}{5}$ and $t$ is in quadrant IV, find the values of all of the trigonometric functions at $t$

Using the first Pythagorean identity, we have

$\sin^2t+\cos^2t=1$
$\sin^2t+\left(\dfrac{4}{5}\right)^2=1$
$\sin^2t+\dfrac{16}{25}=1$
$\sin^2t=\dfrac{9}{25}$

$\sin t=\pm\sqrt{\dfrac{9}{25}}$

$\sin t=-\dfrac{3}{5}$

The negative square root is chosen because $t$ is in Quadrant IV, so $\sin t$ is a negative value. Knowing both the $\sin$ and $\cos$ values at $t$ is enough to find the remaining trigonometric functions, yielding:

$\begin{align*}\cos t & = \dfrac{4}{5} & \sin t & = -\dfrac{3}{5} & \tan t & = \dfrac{\sin t}{\cos t}=\dfrac{-\frac{3}{5}}{\frac{4}{5}}=-\dfrac{3}{4}\\ \csc t&=\frac{1}{\sin t} = -\dfrac{5}{3} & \sec t & = \dfrac{1}{\cos t} = \dfrac{5}{4} & \cot t & = \frac{1}{\tan t} = -\dfrac{4}{3}\end{align*}$

PreCalculus I

5.2 – Trig Functions of Real Numbers

5.2 Video 1 of 2: Trigonometric Functions of Real Numbers.

5.2 Video 2 of 2: Pythagorean Identities.

Section Notes

Trigonometric Functions

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