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6.5 Circular Functions: Graphs and Properties

Given the coordinates of a point on the unit circle, find its reflections across the x-axis, the y-axis, and the origin.
Determine the six trigonometric function values for a real number when the coordinates of the point on the unit circle determined by that real number are given.
Find trigonometric function values for any real number using a calculator.
Graph the six circular functions and state their properties.

The domains of the trigonometric functions, defined in Sections 6.1 and 6.3, have been sets of angles or rotations measured in a real number of degree units. We can also consider the domains to be sets of real numbers, or radians, introduced in Section 6.4. Many applications in calculus that use the trigonometric functions refer only to radians.

Let’s again consider radian measure and the unit circle. We defined radian measure for $θ$ $θ$ as

θ = \frac{s}{r} . θ is a real number without units .

$θ = \frac{s}{r} . θ is a real number without units .$

When $r = 1$ $r = 1$ ,

θ = \frac{s}{1}, or θ = s .

$θ = \frac{s}{1}, or θ = s .$

The arc length s on the unit circle is the same as the radian measure of the angle $θ$ $θ$

In the figure above, the point $(x, y)$ $(x, y)$ is the point where the terminal side of the angle with radian measure $s$ $s$ intersects the unit circle. We can now extend our definitions of the trigonometric functions using domains composed of real numbers, or radians. Trigonometric functions with domains composed of real numbers are called circular functions.

In the definitions,

s can be considered the radian measure of an angle or
the measure of an arc length on the unit circle.

Either way, s is a real number. To each real number $s$ $s$ , there corresponds an arc length $s$ $s$ on the unit circle.

We can consider the domains of trigonometric functions to be real numbers rather than angles. We can determine these values for a specific real number if we know the coordinates of the point on the unit circle determined by that number. As with degree measure, we can also find these function values directly using a calculator.

Reflections on the Unit Circle

Let’s consider the unit circle and a few of its points. For any point $(x, y)$ $(x, y)$ on the unit circle, $x^{2} + y^{2} = 1$ $x^{2} + y^{2} = 1$ , we know that $- 1 \leq x \leq 1$ $- 1 \leq x \leq 1$ and $- 1 \leq y \leq 1.$ $- 1 \leq y \leq 1.$ If we know the x- or y-coordinate of a point on the unit circle, we can find the other coordinate. If $x = \frac{3}{5}$ $x = \frac{3}{5}$ , then

\begin{matrix} {(\frac{3}{5})}^{2} + y^{2} & = & 1 \\ y^{2} & = & 1 - \frac{9}{25} = \frac{16}{25} \\ y & = & \pm \frac{4}{5} . \end{matrix}

$\begin{matrix} {(\frac{3}{5})}^{2} + y^{2} & = & 1 \\ y^{2} & = & 1 - \frac{9}{25} = \frac{16}{25} \\ y & = & \pm \frac{4}{5} . \end{matrix}$

Thus, $(\frac{3}{5}, \frac{4}{5})$ $(\frac{3}{5}, \frac{4}{5})$ and $(\frac{3}{5}, - \frac{4}{5})$ $(\frac{3}{5}, - \frac{4}{5})$ are points on the unit circle. There are two points with an x-coordinate of $\frac{3}{5}$ $\frac{3}{5}$ .

Now let’s consider the radian measure $π / 3$ $π / 3$ and determine the coordinates of the point on the unit circle determined by $π / 3.$ $π / 3.$ We construct a right triangle by dropping a perpendicular segment from the point to the x-axis.

Since $π / 3 = 60 °$ $π / 3 = 60 °$ , we have a $30 ° - 60 °$ $30 ° - 60 °$ right triangle in which the side opposite the $30 °$ $30 °$ angle is one half of the hypotenuse. The hypotenuse, or radius, is 1, so the side opposite the $30 °$ $30 °$ angle is $\frac{1}{2} \cdot 1$ $\frac{1}{2} \cdot 1$ , or $\frac{1}{2}$ $\frac{1}{2}$ . Using the Pythagorean equation, we can find the other side:

\begin{matrix} {(\frac{1}{2})}^{2} + y^{2} & = & 1 \\ y^{2} & = & 1 - \frac{1}{4} = \frac{3}{4} \\ y & = & \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} . \end{matrix}

$\begin{matrix} {(\frac{1}{2})}^{2} + y^{2} & = & 1 \\ y^{2} & = & 1 - \frac{1}{4} = \frac{3}{4} \\ y & = & \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} . \end{matrix}$

We know that y is positive since the point is in the first quadrant. Thus the coordinates of the point determined by $π / 3$ $π / 3$ are $x = 1 / 2$ $x = 1 / 2$ and $y = \sqrt{3} / 2$ $y = \sqrt{3} / 2$ , or $(1 / 2, \sqrt{3} / 2)$ $(1 / 2, \sqrt{3} / 2)$ . We can always check to see if a point is on the unit circle by substituting into the equation $x^{2} + y^{2} = 1$ $x^{2} + y^{2} = 1$ :

{(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} = \frac{1}{4} + \frac{3}{4} = 1.

${(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} = \frac{1}{4} + \frac{3}{4} = 1.$

Because a unit circle is symmetric with respect to the x-axis, the y-axis, and the origin, we can use the coordinates of one point on the unit circle to find coordinates of its reflections.

Example 1

Each of the following points lies on the unit circle. Find their reflections across the x-axis, the y-axis, and the origin.

$(\frac{3}{5}, \frac{4}{5})$ $(\frac{3}{5}, \frac{4}{5})$
$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$
$(\frac{1}{2}, \frac{\sqrt{3}}{2})$ $(\frac{1}{2}, \frac{\sqrt{3}}{2})$

Solution

Now Try Exercise 1.

Finding Function Values

Knowing the coordinates of only a few points on the unit circle along with their reflections allows us to find trigonometric function values of the most frequently used real numbers, or radians.

Example 2

Find each of the following function values.

a) $tan \frac{π}{3}$ $tan \frac{π}{3}$
b) $cos \frac{3 π}{4}$ $cos \frac{3 π}{4}$
c) $sin (- \frac{π}{6})$ $sin (- \frac{π}{6})$
d) $cos \frac{4 π}{3}$ $cos \frac{4 π}{3}$
e) $cot π$ $cot π$
f) $csc (- \frac{7 π}{2})$ $csc (- \frac{7 π}{2})$

Solution

We locate the point on the unit circle determined by the rotation, and then find its coordinates using reflection if necessary.

a) The coordinates of the point determined by $π / 3$ $π / 3$ are $(1 / 2, \sqrt{3} / 2)$ $(1 / 2, \sqrt{3} / 2)$ .

Thus, $tan \frac{π}{3} = \frac{y}{x} = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}$ $tan \frac{π}{3} = \frac{y}{x} = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}$ .
b) The reflection of $(\sqrt{2} / 2, \sqrt{2} / 2)$ $(\sqrt{2} / 2, \sqrt{2} / 2)$ across the y-axis is $(- \sqrt{2} / 2, \sqrt{2} / 2)$ $(- \sqrt{2} / 2, \sqrt{2} / 2)$ .

Thus, $cos \frac{3 π}{4} = x = - \frac{\sqrt{2}}{2}$ $cos \frac{3 π}{4} = x = - \frac{\sqrt{2}}{2}$ .
c) The reflection of $(\sqrt{3} / 2, 1 / 2)$ $(\sqrt{3} / 2, 1 / 2)$ across the x-axis is $(\sqrt{3} / 2, - 1 / 2)$ $(\sqrt{3} / 2, - 1 / 2)$ .

Thus, $sin (- \frac{π}{6}) = y = - \frac{1}{2}$ $sin (- \frac{π}{6}) = y = - \frac{1}{2}$ .
d) The reflection of $(1 / 2, \sqrt{3} / 2)$ $(1 / 2, \sqrt{3} / 2)$ across the origin is $(- 1 / 2, - \sqrt{3} / 2)$ $(- 1 / 2, - \sqrt{3} / 2)$ .

Thus, $cos \frac{4 π}{3} = x = - \frac{1}{2}$ $cos \frac{4 π}{3} = x = - \frac{1}{2}$ .
d) The coordinates of the point determined by $π$ $π$ are $(- 1,0)$ $(- 1,0)$ .

Thus, $cot π = \frac{x}{y} = \frac{- 1}{0}$ $cot π = \frac{x}{y} = \frac{- 1}{0}$ , which is not defined.
e) The coordinates of the point determined by $- 7 π / 2$ $- 7 π / 2$ are $(0, 1)$ $(0, 1)$ .

Thus, $csc (- \frac{7 π}{2}) = \frac{1}{y} = \frac{1}{1} = 1.$ $csc (- \frac{7 π}{2}) = \frac{1}{y} = \frac{1}{1} = 1.$

We can also think of cot $π$ $π$ as the reciprocal of tan $π$ $π$ . Since $tan π = y / x = 0 / - 1 = 0$ $tan π = y / x = 0 / - 1 = 0$ and the reciprocal of 0 is not defined, we know that $cot π$ $cot π$ is not defined.

Now Try Exercises 9 and 11.

Using a calculator, we can find trigonometric function values of any real number without knowing the coordinates of the point that it determines on the unit circle. Most calculators have both degree mode and radian mode. When finding function values of radian measures, or real numbers, we must set the calculator in RADIAN mode.

Example 3

Find each of the following function values of radian measures using a calculator. Round the answer to four decimal places.

$cos \frac{2 π}{5}$ $cos \frac{2 π}{5}$
$tan (- 3)$ $tan (- 3)$
sin 24.9
$sec \frac{π}{7}$ $sec \frac{π}{7}$

Solution

Using a calculator set in RADIAN mode, we find the values.

$cos \frac{2 π}{5} \approx 0.3090$ $cos \frac{2 π}{5} \approx 0.3090$
$tan (- 3) \approx 0.1425$ $tan (- 3) \approx 0.1425$
$sin 24.9 \approx - 0.2306$ $sin 24.9 \approx - 0.2306$
$sec \frac{π}{7} = \frac{1}{cos \frac{π}{7}} \approx 1.1099$ $sec \frac{π}{7} = \frac{1}{cos \frac{π}{7}} \approx 1.1099$

Note in part (d) that the secant function value can be found by taking the reciprocal of the cosine value. Thus we can enter $cos π / 7$ $cos π / 7$ and use the reciprocal key.

Now Try Exercises 25 and 33.

From the definitions on p. 449, we can relabel any point $(x, y)$ $(x, y)$ on the unit circle as ( $cos s$ $cos s$ , $sin s$ $sin s$ ), where s is any real number.

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Table of Contents for 6.5 Circular Functions: Graphs and Properties

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Table of Contents for
6.5 Circular Functions: Graphs and Properties