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1.6: Applying Trig Functions to Angles of Rotation

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

  • Find the values of the six trigonometric functions for angles of rotation.
  • Recognize angles of the unit circle.

Trigonometric Functions of Angles in Standard Position

In section 1.3, we defined the six trigonometric functions for angles in right triangles. We can also define the same functions in terms of angles of rotation. Consider an angle in standard position, whose terminal side intersects a circle of radius \begin{align*}r\end{align*}r. We can think of the radius as the hypotenuse of a right triangle:

The point \begin{align*}(x, y)\end{align*}(x,y) where the terminal side of the angle intersects the circle tells us the lengths of the two legs of the triangle. Now, we can define the trigonometric functions in terms of \begin{align*}x, y\end{align*}x,y, and \begin{align*}r\end{align*}r:

\begin{align*}\cos \theta = \frac{x}{r} && \sec \theta = \frac{r}{x}\\ \sin \theta = \frac{y}{r} && \csc \theta = \frac{r}{y}\\ \tan \theta = \frac{y}{x} && \cot \theta = \frac{x}{y}\end{align*}cosθ=xrsinθ=yrtanθ=yxsecθ=rxcscθ=rycotθ=xy

And, we can extend these functions to include non-acute angles.

Example 1: The point (-3, 4) is a point on the terminal side of an angle in standard position. Determine the values of the six trigonometric functions of the angle.

Solution:

Notice that the angle is more than 90 degrees, and that the terminal side of the angle lies in the second quadrant. This will influence the signs of the trigonometric functions.

\begin{align*}\cos \theta & = \frac{-3}{5} && \sec \theta = \frac{5}{-3}\\ \sin \theta & = \frac{4}{5} && \csc \theta = \frac{5}{4}\\ \tan \theta & = \frac{4}{-3} && \cot \theta = \frac{-3}{4}\end{align*}cosθsinθtanθ=35=45=43secθ=53cscθ=54cotθ=34

Notice that the value of \begin{align*}r\end{align*}r depends on the coordinates of the given point. You can always find the value of \begin{align*}r\end{align*}r using the Pythagorean Theorem. However, often we look at angles in a circle with radius 1. As you will see next, doing this allows us to simplify the definitions of the trig functions.

The Unit Circle

Consider an angle in standard position, such that the point \begin{align*}(x, y)\end{align*}(x,y) on the terminal side of the angle is a point on a circle with radius 1.

This circle is called the unit circle. With \begin{align*}r = 1\end{align*}r=1, we can define the trigonometric functions in the unit circle:

\begin{align*}\cos \theta & = \frac{x}{r} = \frac{x}{1} = x && \sec \theta = \frac{r}{x} = \frac{1}{x}\\ \sin \theta & = \frac{y}{r} = \frac{y}{1} = y && \csc \theta = \frac{r}{y} = \frac{1}{y}\\ \tan \theta & = \frac{y}{x} && \cot \theta = \frac{x}{y}\end{align*}cosθsinθtanθ=xr=x1=x=yr=y1=y=yxsecθ=rx=1xcscθ=ry=1ycotθ=xy

Notice that in the unit circle, the sine and cosine of an angle are the \begin{align*}x\end{align*}x and \begin{align*}y\end{align*}y coordinates of the point on the terminal side of the angle. Now we can find the values of the trigonometric functions of any angle of rotation, even the quadrantal angles, which are not angles in triangles.

We can use the figure above to determine values of the trig functions for the quadrantal angles. For example, \begin{align*}\sin 90^\circ = y = 1\end{align*}sin90=y=1.

Example 2: Use the unit circle above to find each value:

a. \begin{align*}\cos 90^\circ\end{align*}cos90

b. \begin{align*}\cot 180^\circ\end{align*}cot180

c. \begin{align*}\sec 0^\circ\end{align*}sec0

Solution:

a. \begin{align*}\cos 90^\circ = 0\end{align*}cos90=0

The ordered pair for this angle is (0, 1). The cosine value is the \begin{align*}x\end{align*}x coordinate, \begin{align*}0\end{align*}0.

b. \begin{align*}\cot 180^\circ\end{align*}cot180 is undefined

The ordered pair for this angle is (-1, 0). The ratio \begin{align*}\frac{x}{y}\end{align*}xy is \begin{align*}\frac{-1}{0}\end{align*}10, which is undefined.

c. \begin{align*}\sec 0^\circ = 1\end{align*}sec0=1

The ordered pair for this angle is (1, 0). The ratio \begin{align*}\frac{1}{x}\end{align*}1x is \begin{align*}\frac{1}{1} = 1\end{align*}11=1.

There are several important angles in the unit circle that you will work with extensively in your study of trigonometry, primarily \begin{align*}30^\circ\end{align*}30, \begin{align*}45^\circ\end{align*}45, and \begin{align*}60^\circ\end{align*}60. Recall section 1.2 to find the values of the trigonometric functions of these angles. First, we need to know the ordered pairs. Let’s begin with \begin{align*}30^\circ\end{align*}30.

This triangle is identical to #8 from 1.3. If you look back at this problem, you will recall that you found the sine, cosine and tangent of \begin{align*}30^\circ\end{align*}30 and \begin{align*}60^\circ\end{align*}60. It is no coincidence that the endpoint on the unit circle is the same as your answer from #8.

The terminal side of the angle intersects the unit circle at the point \begin{align*}\left ( \frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}. Therefore we can find the values of any of the trig functions of \begin{align*}30^\circ\end{align*}. For example, the cosine value is the \begin{align*}x-\end{align*}coordinate, so \begin{align*}\cos (30^\circ) = \frac{\sqrt{3}}{2}\end{align*}. Because the coordinates are fractions, we have to do a bit more work in order to find the tangent value:

\begin{align*}\tan 30^\circ = \frac{y}{x} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{2} \times \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}.\end{align*}

In the review exercises you will find the values of the remaining four trig functions of this angle. The table below summarizes the ordered pairs for \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, and \begin{align*}60^\circ\end{align*} on the unit circle.

Angle \begin{align*}x-\end{align*}coordinate \begin{align*}y-\end{align*}coordinate
\begin{align*}30^\circ\end{align*} \begin{align*}\frac{\sqrt{3}}{2}\end{align*} \begin{align*}\frac{1}{2}\end{align*}
\begin{align*}45^\circ\end{align*} \begin{align*}\frac{\sqrt{2}}{2}\end{align*} \begin{align*}\frac{\sqrt{2}}{2}\end{align*}
\begin{align*}60^\circ\end{align*} \begin{align*}\frac{1}{2}\end{align*} \begin{align*}\frac{\sqrt{3}}{2}\end{align*}

We can use these values to find the values of any of the six trig functions of these angles.

Example 3: Find the value of each function.

a. \begin{align*}\cos 45^\circ\end{align*}

b. \begin{align*}\sin 60^\circ\end{align*}

c. \begin{align*}\tan 45^\circ\end{align*}

Solution:

a. \begin{align*}\cos 45^\circ = \frac{\sqrt{2}}{2}\end{align*} The cosine value is the \begin{align*}x-\end{align*} coordinate of the point.

b. \begin{align*}\sin 60^\circ = \frac{\sqrt{3}}{2}\end{align*} The sine value is the \begin{align*}y-\end{align*} coordinate of the point.

c. \begin{align*}\tan 45^\circ = 1\end{align*} The tangent value is the ratio of the \begin{align*}y-\end{align*} coordinate to the \begin{align*}x-\end{align*} coordinate. Because the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*} coordinates are the same for this angle, the tangent ratio is 1.

Points to Consider

  • How can some values of the trig functions be negative? How is it that some are undefined?
  • Why is the unit circle and the trig functions defined on it useful, even when the hypotenuses of triangles in the problem are not 1?

Review Questions

  1. The point (3, -4) is a point on the terminal side of an angle \begin{align*}\theta\end{align*} in standard position.
    1. Determine the radius of the circle.
    2. Determine the values of the six trigonometric functions of the angle.
  2. The point (-5, -12) is a point on the terminal side of an angle \begin{align*}\theta\end{align*} in standard position.
    1. Determine the radius of the circle.
    2. Determine the values of the six trigonometric functions of the angle.
  3. \begin{align*}\tan \theta = -\frac{2}{3}\end{align*} and \begin{align*}\cos \theta > 0\end{align*}. Find \begin{align*}\sin \theta\end{align*}.
  4. \begin{align*}\csc \theta = -4\end{align*} and \begin{align*}\tan \theta > 0\end{align*}. Find the exact values of the remaining trigonometric functions.
  5. (2, 6) is a point on the terminal side of \begin{align*}\theta\end{align*}. Find the exact values of the six trigonometric functions.
  6. The terminal side of the angle \begin{align*}270^\circ\end{align*} intersects the unit circle at (0, -1). Use this ordered pair to find the six trig functions of \begin{align*}270^\circ\end{align*}.
  7. In the lesson you learned that the terminal side of the angle \begin{align*}30^\circ\end{align*} intersects the unit circle at the point \begin{align*}\left ( \frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}. Here you will prove that this is true.
    1. Explain why Triangle \begin{align*}ABD\end{align*} is an equiangular triangle. What is the measure of angle \begin{align*}DAB\end{align*}?
    2. What is the length of \begin{align*}BD\end{align*}? How do you know?
    3. What is the length of \begin{align*}BC\end{align*} and \begin{align*}CD\end{align*}? How do you know?
    4. Now explain why the ordered pair is \begin{align*}\left ( \frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}.
    5. Why does this tell you that the ordered pair for \begin{align*}60^\circ\end{align*} is \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}?
  8. In the lesson you learned that the terminal side of the angle \begin{align*}45^\circ\end{align*} is \begin{align*}\left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right )\end{align*}. Use the figure below and the Pythagorean Theorem to show that this is true.
  9. In what quadrants will an angle in standard position have a positive tangent value? Explain your thinking.
  10. Sketch the angle \begin{align*}150^\circ\end{align*} on the unit circle is. How is this angle related to \begin{align*}30^\circ\end{align*}? What do you think the ordered pair is?
  11. We now know that \begin{align*}\sin \theta = y\end{align*}, \begin{align*}\cos \theta = x\end{align*}, and \begin{align*}\tan \theta = \frac{y}{x}\end{align*}. First, explain how it looks as though sine, cosine, and tangent are related. Second, can you rewrite tangent in terms of sine and cosine?

Review Answers

  1. The radius of the circle is 5. \begin{align*}\cos \theta & = \frac{3}{5} && \sec \theta = \frac{5}{3}\\ \sin \theta & = \frac{-4}{5} && \csc \theta = \frac{5}{-4}\\ \tan \theta & = \frac{-4}{3} && \cot \theta = \frac{3}{-4}\end{align*}
  2. The radius of the circle is 13. \begin{align*}\cos \theta & = \frac{-5}{13} && \sec \theta = \frac{13}{-5}\\ \sin \theta & = \frac{-12}{13} && \csc \theta = \frac{13}{-12}\\ \tan \theta & = \frac{-12}{-5} = \frac{12}{5} && \cot \theta = \frac{-5}{-12} = \frac{5}{12}\end{align*}
  3. If \begin{align*}\tan \theta = -\frac{2}{3}\end{align*}, it must be in either Quadrant II or IV. Because \begin{align*}\cos \theta > 0\end{align*}, we can eliminate Quadrant II. So, this means that the 3 is negative. (All Students Take Calculus) From the Pythagorean Theorem, we find the hypotenuse: \begin{align*}2^2 + (-3^2) & = c^2\\ 4 + 9 & = c^2\\ 13 & = c^2\\ \sqrt{13} & = c\end{align*} Because we are in Quadrant IV, the sine is negative. So, \begin{align*}\sin \theta = -\frac{2}{\sqrt{13}}\end{align*} or \begin{align*}-\frac{2\sqrt{13}}{13}\end{align*} (Rationalize the denominator)
  4. If \begin{align*}\csc \theta = -4\end{align*}, then \begin{align*}\sin \theta = -\frac{1}{4}\end{align*}, sine is negative, so \begin{align*}\theta\end{align*} is in either Quadrant III or IV. Because \begin{align*}\tan \theta > 0\end{align*}, we can eliminate Quadrant IV, therefore \begin{align*}\theta\end{align*} is in Quadrant III. From the Pythagorean Theorem, we can find the other leg: \begin{align*}a^2 + (-1)^2 & = 4^2\\ a^2 + 1 & = 16\\ a^2 & = 15\\ a & = \sqrt{15}\\ \text{So}, \ \cos \theta & = -\frac{\sqrt{15}}{4}, \sec \theta = -\frac{4}{\sqrt{15}} \ \text{or} \ -\frac{4\sqrt{15}}{15}\\ \tan \theta & = -\frac{1}{\sqrt{15}} \ \text{or} \ \frac{\sqrt{15}}{15}, \cot \theta = \sqrt{15}\end{align*}
  5. If the terminal side of \begin{align*}\theta\end{align*} is on (2, 6) it means \begin{align*}\theta\end{align*} is in Quadrant I, so sine, cosine and tangent are all positive. From the Pythagorean Theorem, the hypotenuse is: \begin{align*}2^2 + 6^2 & = c^2\\ 4 + 36 & = c^2\\ 40 & = c^2\\ \sqrt{40} & = 2\sqrt{10} = c\end{align*} Therefore, \begin{align*}\sin \theta = \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}, \cos \theta = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}\end{align*} and \begin{align*}\tan \theta = \frac{6}{2} = 3\end{align*}
  6. \begin{align*}\cos 270 & = 0 && \sec 270 = undefined\\ \sin 270 & = -1 && \csc 270 = \frac{1}{-1} = -1\\ \tan 270 & = undefined && \cot 270 = 0\end{align*}
    1. The triangle is equiangular because all three angles measure 60 degrees. Angle \begin{align*}DAB\end{align*} measures 60 degrees because it is the sum of two 30 degree angles.
    2. \begin{align*}BD\end{align*} has length 1 because it is one side of an equiangular, and hence equilateral, triangle.
    3. \begin{align*}BC\end{align*} and \begin{align*}CD\end{align*} each have length \begin{align*}\frac{1}{2}\end{align*}, as they are each half of \begin{align*}BD\end{align*}. This is the case because Triangle \begin{align*}ABC\end{align*} and \begin{align*}ADC\end{align*} are congruent.
    4. We can use the Pythagorean theorem to show that the length of \begin{align*}AC\end{align*} is \begin{align*}\frac{\sqrt{3}}{2}\end{align*}. If we place angle \begin{align*}BAC\end{align*} as an angle in standard position, then \begin{align*}AC\end{align*} and \begin{align*}BC\end{align*} correspond to the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates where the terminal side of the angle intersects the unit circle. Therefore the ordered pair is \begin{align*}\left ( \frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}.
    5. If we draw the angle \begin{align*}60^\circ\end{align*} in standard position, we will also obtain a \begin{align*}30-60-90\end{align*} triangle, but the side lengths will be interchanged. So the ordered pair for \begin{align*}60^\circ\end{align*} is \begin{align*}\left ( \frac{1}{2}, \frac{\sqrt{3}}{2} \right )\end{align*}.

  7. \begin{align*}n^2 + n^2 & = 1^2\\ 2n^2 & = 1\\ n^2 & = \frac{1}{2}\\ n & = \pm \sqrt{\frac{1}{2}}\\ n & = \pm \frac{1}{\sqrt{2}}\\ n & = \pm \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}\end{align*} Because the angle is in the first quadrant, the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates are positive.
  8. An angle in the first quadrant, as the tangent is the ratio of two positive numbers. And, angle in the third quadrant, as the tangent in the ratio of two negative numbers, which will be positive.
  9. The terminal side of the angle is a reflection of the terminal side of \begin{align*}30^\circ\end{align*}. From this, students should see that the ordered pair is \begin{align*}\left ( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right )\end{align*}.
  10. Students should notice that tangent is the ratio of \begin{align*}\frac{\sin}{\cos}\end{align*}, which is \begin{align*}\frac{y}{x}\end{align*}, which is also slope.

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