The town of Avery lies 48 miles due east of Baker, and Clio is 34 miles in the bearing \( \mathrm \, 35^ < \circ >\, \mathrm \) from Baker. How far is it from Avery to Clio?
We know how to solve right triangles using the trigonometric ratios; however, the triangle formed by the three towns is not a right triangle because it includes an obtuse angle of \(125^\) at \(B\), as shown in Figure \( \PageIndex < 1 >\).
This type of triangle is called an oblique triangle. Later in this course, we learn how to solve oblique triangles, but first, we must be able to find the values of the trigonometric functions for obtuse angles.
We now introduce an essential tool for evaluating trigonometric functions and solving trigonometric equations involving non-acute angles.
The reference angle for an angle \( \theta \) in standard form is the acute angle formed between the terminal side of \( \theta \) and the \( x \)-axis. By convention, the reference angle is always considered positive (even if it opens in a clockwise direction). It is denoted with the "hat" notation, \( \hat \).
Any acute angle \(\theta\) is the reference angle for four angles between \(0^\) and \(360^\), one in each quadrant. The figure below shows the four angles in standard position whose reference angle is \(35^\). Note that each angle is found by measuring \(35^\) from the \(x\)-axis in the appropriate quadrant and that the four angles together make a "bow-tie" shape.
Find and label the reference angle for the given angle.
While you should strive for understanding in mathematics over memorization, some students initially find the following guidelines helpful until they get used to reference angles. Trust me when I say you want to use the following guideline only for a while - it's too bulky.
Let \( \theta \) be an angle in standard position.
By definition, reference angles are acute angles. This means they are not \( 0^ < \circ >\) or \( 90^ < \circ >\). In fact, quadrantal angles do not have reference angles.
Draw the angle and state its reference angle.
We can use reference angles to help us evaluate a trigonometric function at a non-acute angle \( \theta \). To do so, we construct a reference triangle for \( \theta \) using the following method:
The figure below shows angles \(\theta\) in all four quadrants and the reference angle, \(\hat\), for each. Study the figures, and make sure you understand how to find the reference angle in each quadrant.
Figures \( \PageIndex < 1a >\) (left) and \( \PageIndex < 1b >\) (right): Reference triangles for Quadrant I and Quadrant II angles, respectively.
Figures \( \PageIndex < 1c >\) (left) and \( \PageIndex < 1d >\) (right): Reference triangles for Quadrant III and Quadrant IV angles, respectively.
These observations may help you remember the formulas:
The following activity reveals the usefulness of reference angles and reference triangles when evaluating trigonometric functions of non-acute angles.
In all four versions of Figure \( \PageIndex < 1 >\), coordinate trigonometry states that\[ \begin
\sin\left( \theta \right) & = & \dfrac & \quad & \csc\left( \theta \right) & = & \dfrac \\
\cos\left( \theta \right) & = & \dfrac & \quad & \sec\left( \theta \right) & = & \dfrac \\
\tan\left( \theta \right) & = & \dfrac & \quad & \cot\left( \theta \right) & = & \dfrac \\
\end \nonumber \]However, it is also true that in all four versions of Figure \( 1 \), the side of the reference triangle opposite \( \hat \) has length \( |y| \), the side of the reference triangle adjacent to \( \hat \) has length \( |x| \), and the hypotenuse of the reference triangle has a length of \( r \). Therefore, by right triangle trigonometry,\[ \begin
\sin\left( \hat \right) & = & \dfrac<|y|> & \quad & \csc\left( \hat \right) & = & \dfrac <|y|>\\
\cos\left( \hat \right) & = & \dfrac<|x|> & \quad & \sec\left( \hat \right) & = & \dfrac <|x|>\\
\tan\left( \hat \right) & = & \dfrac<|y|> <|x|>& \quad & \cot\left( \hat \right) & = & \dfrac<|x|> <|y|>\\
\end \nonumber \]Thus, the values of the trigonometric functions of the angle \( \theta \) match the values of the trigonometric functions of \( \hat \), except, possibly, for a difference in sign. The sign of the trigonometric function's value can be determined by the quadrant in which the terminal side of the angle lies. This result is summarized in the following theorem.
The value of a trigonometric function of any angle is equal to that of the function at its reference angle, except for sign. The quadrant determines the sign of the function.
In the previous chapter, we used a calculator to approximate the values of the trigonometric functions at acute angles. Your calculator can easily find the value of a trigonometric function at any angle. We showcase this in the following example.
Note that the results in Example \( \PageIndex < 2 >\) align with what we said previously - the sign of the trigonometric function's value can be determined by the quadrant in which the terminal side of the angle lies. Thus, to compute \( \sin\left( 200^ < \circ >\right) \), we could note that \( \theta \in \mathrm \) and the reference angle is \( 20^ < \circ >\). Since the sine is negative in \( \mathrm \), we get\[ \begin
\sin\left( 200^ < \circ >\right) & = & \fbox \sin\left( 20^ < \circ >\right) & \quad & \left( \sin\left( \theta \right) \text< is equal to >\sin\left( \hat \right), \text < except for, possibly, the sign>\right) \\
& = & - \sin\left( 20^ < \circ >\right) & \quad & \left( \text200^ < \circ >\in \mathrm, \text < sine is negative>\right) \\
\end \nonumber \]From this point forward in Trigonometry, the following Trigonometry Mantra will be helpful.
Trigonometry is about two things - reference angles and quadrants.
While this mantra oversimplifies the subject, most of Trigonometry is easier once you know the reference angle and quadrant for a given angle.
When combined with the Reference Angle Theorem, the following corollary improves the speed with which we can evaluate trigonometric functions.
For any integer \( n \),\[ \begin
\sin\left( \theta + 360^< \circ >n \right) & = & \sin\left( \theta \right) & \quad & \csc\left( \theta + 360^< \circ >n \right) & = & \csc\left( \theta \right) \\
\cos\left( \theta + 360^< \circ >n \right) & = & \cos\left( \theta \right) & \quad & \sec\left( \theta + 360^< \circ >n \right) & = & \sec\left( \theta \right) \\
\tan\left( \theta + 360^< \circ >n \right) & = & \tan\left( \theta \right) & \quad & \cot\left( \theta + 360^< \circ >n \right) & = & \cot\left( \theta \right) \\
\end \nonumber \]
This result is from our previous discussion that the value of a trigonometric function is the same for all angles coterminal with the original given angle.
As we will investigate later, it turns out that\[ \tan\left( \theta + 180^< \circ >n \right) = \tan\left( \theta \right) \quad \text \quad \cot\left( \theta + 180^< \circ >n \right) = \cot\left( \theta \right) \nonumber \]for any integer \( n \); however, this is not needed for our current discussion.
Let's see how this corollary, along with focusing on reference angles and quadrants, speeds up the evaluation of trigonometric functions whose reference angles are the special angles.
Find the exact value of each trigonometric function.
Find the exact value of each trigonometric function (without touching a calculator).
We now turn our attention back to finding angles given values of trigonometric functions. Last chapter, we were able to find \( \theta \) such that \( \tan\left( \theta \right) = 0.5662 \). At that time, we used the \( \mathrm^ \) button on our calculator to find \( \tan^\left( 0.5662 \right) \approx 29.5185^ < \circ >\). At that time, however, we were only working with right triangle trigonometry, and the angles we were interested in were all acute. When working with non-acute angles, the inverse trigonometric functions can get tricky.
To avoid the trickiness associated with inverse trigonometric functions (and to give us more time with Trigonometry before diving into the more profound theory of the inverse trigonometric functions), it's best to think of them as returning reference angles. Since reference angles are always acute, this interpretation works with what we did in the last chapter; however, as we have seen in this section, we are now dealing with angles that are not necessarily acute.
For now, think of the inverse trigonometric functions as returning reference angles.
We must evaluate the inverse trigonometric function at a positive value to return a reference angle. The reason for this cannot be fully justified at this time. Still, we can say that, for example, \( \sin^< -1 >\left( -0.31 \right) \) returns an angle \( \theta \) so that \( \sin\left( \theta \right) = -0.31\); however, we know that the values of all trigonometric functions are positive in \( \mathrm \). Since we want the returned angle to be a reference angle, it must be acute (and, therefore, in the first quadrant). Hence, there is no possibility for a reference angle to be returned if the argument of our inverse trigonometric function is negative.
If we agree that inverse trigonometric functions evaluated at positive numbers return reference angles, we can ask for solutions to equations involving trigonometric functions.
For example, find \( \theta \) if \[ \sin\left( \theta \right) = -0.31, \nonumber \]where \( \theta \in \mathrm \).
First, it should make sense that \( \theta \) could be in the third quadrant because the sine at \( \theta \) is negative, and the sine is negative in the third quadrant (as well as in the fourth quadrant). Remember, Trigonometry is about two things - reference angles and quadrants. We have already been given a quadrant. We evaluate the inverse sine function at \( |-0.31| = 0.31 \) to find the reference angle. Hence, the reference angle is\[ \hat = \sin^\left( |-0.31| \right) = \sin^\left( 0.31 \right) \approx 18.1^< \circ >. \nonumber \]The angle in \( \mathrm \) with reference angle \( \hat \approx 18.1^ < \circ >\) is \( \theta = 180^ < \circ >+ \hat \approx 180^ < \circ >+ 18.1^ < \circ >= 198.1^< \circ >. \)
To find the angle \( \theta \) so that the trigonometric function of \( \theta \) is a given ratio:
The second step in the summary above assumes you are given extra information to help determine the quadrant in which \( \theta \) terminates. While this will be true in this section, we will eventually shed that requirement. We will begin to find all angles that make a given trigonometric equation valid.
Find \( \theta \) to the nearest tenth of a degree, if \( 0^ < \circ >\lt \theta \lt 360^ < \circ >\).
Of course, if given a "special" trigonometric ratio, you should not reach for a calculator.
Find \( \alpha \) if \( \cos\left( \alpha \right) = -\frac> \) and \( \alpha \in \mathrm \), where \( 0^ < \circ >\lt \alpha \lt 360^ < \circ >\).
Solution Reference Angle. It is crucial that you recognize the special ratios. Sketching the \( 30^ < \circ >\)-\( 60^ < \circ >\)-\( 90^ < \circ >\) triangle, we see that the reference angle for which \( \cos\left( \alpha \right) = \left|-\frac>\right| = \frac> \) is \( \hat = 30^ < \circ >\).
Quadrant. \( \alpha \in \mathrm \)
Actual Angle. Thus,\[ \alpha = 180^ < \circ >- \hat = 180^ < \circ >- 30^ < \circ >= 150^< \circ >. \nonumber \]
Solve \( \tan\left( \beta \right) = -\sqrt \) given \( 0^ < \circ >\lt \beta \lt 360^ < \circ >\) and \( \cos\left( \beta \right) \gt 0 \).
Answer
True or False? For Problems 3 - 4, determine if the statement is true or false. If true, cite the definition or theorem stated in the text supporting your claim. If false, explain why it is false and, if possible, correct the statement.
For Problems 7 - 14, find the reference angle for the given angle. Make a sketch showing the angle, the reference angle, and the reference triangle.
For Problems 15 - 20, find three angles between \(90^\) and \(360^\) having the given reference angle, and sketch all four angles on the same grid.
For Problems 21 - 35, find the exact value of the trigonometric function using reference angles and quadrants (do not use a calculator).
For Problems 36 - 43, use the values below to find the value of the trigonometric function. Do not use a calculator!
\(\cos \left(23^\right) \approx 0.9205 \quad \sin \left(46^\right) \approx 0.7193 \quad \tan \left(78^\right) \approx 4.7046\)
\(\theta\) | \(30^\) | \(45^\) | \(60^\) | \(120^\) | \(135^\) | \(150^\) | \(210^\) | \(225^\) | \(240^\) | \(300^\) | \(315^\) | \(330^\) |
\(\sin \left(\theta\right)\) | ||||||||||||
\(\cos \left(\theta\right)\) | ||||||||||||
\(\tan \left(\theta\right)\) | ||||||||||||
\(\csc \left(\theta\right)\) | ||||||||||||
\(\sec \left(\theta\right)\) | ||||||||||||
\(\cot \left(\theta\right)\) |
For Problems 49 - 54, find a second angle between \(0^\) and \(360^\) with the given trigonometric ratio.
For Problems 55 - 69, use the given information to find an approximation of \( \theta \) to the nearest degree if \(0^ \le \theta \lt 360^\).
For Problems 70 - 84, use the given information to find the exact value of \( \theta \) if \(0^ \le \theta \lt 360^\). Do not use a calculator!
For Problems 85 - 90, find exact values for all solutions between \(0^\) and \(360^\).
For Problems 93 - 102,
(a) Sketch an angle in standard position with the given properties.
(b) Find \(\cos \left(\theta\right)\), \(\sin \left(\theta\right)\), and \(\tan \left(\theta\right)\).
(c) Find the angle \(\theta\), rounded to tenths of a degree.
For Problems 103 - 106,
(a) Find the sine and cosine of the angle.
(b) Sketch the supplement of the angle in standard position. (Use congruent triangles.)
(c) Find the sine and cosine of the supplement.
(d) Find the angle and its supplement, rounded to the nearest degree.
Problems 107 and 108 refer to the following table of values.
\(\theta\) | \(15^\) | \(25^\) | \(65^\) | \(75^\) | \(105^\) | \(115^\) | \(155^\) | \(165^\) |
\(\cos \left(\theta\right)\) | ||||||||
\(\sin \left(\theta\right)\) | ||||||||
\(\tan \left(\theta\right)\) |
Problems 109 - 114 preview the concept of solving trigonometric equations (a topic reserved for later in the course). For each equation, find two different angles between \( 0^ < \circ >\) and \( 360^ < \circ >\) that satisfy the equation. Round your answers to the nearest \(0.1^\).
For Problems 115 - 118, fill in the blanks with complements or supplements.
For Problems 121 and 122,
(a) Find the coordinates of point \(P\). Round to two decimal places.
(b) Find the sides \(B C\) and \(P C\) of \(\triangle P C B\).
(c) Find side \(P B\).
This page titled 4.1: Reference Angles is shared under a CC BY-NC 12 license and was authored, remixed, and/or curated by Roy Simpson.