Trigonometric Ratios and Special Triangles Alternative Format
Lesson Part 1
Introduction
The Unit Circle and Trigonometric Ratios
Recall that a circle centred at the origin with radius \(1\) is called a unit circle and has equation \(x^2 + y^2 = 1.\)
We noticed in a previous module that any point on both the unit circle and the terminal arm of some standard position angle \(\theta\) can be written \(P\,(\cos(\theta), \sin(\theta))\).
Two observations can be made at this point.
We know that \(\tan(\theta) = \dfrac{y}{x}\), and, on the unit circle, \(x = \cos(\theta)\) and \(y = \sin(\theta)\), so \(\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}\).
This is true for all values of \(\theta\) in the domain of \(\tan(\theta)\), and is called a quotient identity.
Also, since every point is on the unit circle \(x^2 + y^2 = 1\), we can substitute \(x = \cos(\theta)\) and \(y = \sin(\theta)\).
Therefore, \((\cos(\theta))^2 + (\sin(\theta))^2 = 1\). This is usually written \(\cos^2 (\theta) + \sin^2 (\theta) = 1\).
This relationship is true for all values of \(\theta\) and is called the Pythagorean identity.
We will do more with identities later in this unit.
We have also defined three reciprocal trigonometric ratios, cosecant, secant, and cotangent:
\(\csc(\theta) = \dfrac{1}{\sin(\theta)} \qquad \sec(\theta) = \dfrac{1}{\cos(\theta)} \qquad \cot(\theta) = \dfrac{1}{\tan(\theta)}\)
Examples
Given any point \(P\,(x,y)\) on the terminal arm of a standard position angle, \(\theta\), we can determine the value of any or all of the six trigonometric ratios.
If that point is also on the unit circle, \(P\,(x,y)=P(\cos(\theta),\sin(\theta))\) for example, the process of determining the value of any or all of the six trigonometric ratios is easier.
Example 1
Point \(P\, \left( -\dfrac{3}{5} , -\dfrac{4}{5} \right)\) is on the terminal arm of standard position angle \(\theta\). Determine the exact values of each of the six trigonometric ratios.
Solution
A sketch is generally helpful. We discover that the point \(P(-3/5, -4/5)\) is in quadrant \(3\).
First, we calculate the value of \(r\):
\[ \begin{align*} r^2 = x^2 + y^2 &= \left( -\dfrac{3}{5} \right)^2 + \left( -\dfrac{4}{5} \right)^2 \\ &= \dfrac{9}{25} + \dfrac{16}{25} \\ &= 1 \end{align*} \]
and \(r = 1\) since \(r \gt 0\).
Since \(r = 1\), the point \(P\,\left( -\dfrac{3}{5}, -\dfrac{4}{5} \right)\) is on the unit circle and \((x,y) = (\cos(\theta), \sin(\theta))\).
Then,
\[\sin(\theta) = -\dfrac{4}{5}\]\[\csc(\theta) = \dfrac{1}{\sin(\theta)} = -\dfrac{5}{4}\]
\[\cos(\theta) = -\dfrac{3}{5}\]\[\sec(\theta) = \dfrac{1}{\cos(\theta)} = -\dfrac{5}{3}\]
\[\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{4}{3}\]\[\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{3}{4}\]
Check Your Understanding A
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Lesson Part 2
Special Triangles
In a previous course, two special triangles were introduced.
The first of the triangles is a \(30^{\circ}\)-\(60^{\circ}\)-\(90^{\circ}\) triangle. The ratio of the corresponding opposite side lengths is \(1 : \sqrt{3} : 2\).
The second triangle is a \(45^{\circ}\)-\(45^{\circ}\)-\(90^{\circ}\) triangle. The ratio of the corresponding opposite side lengths is \(1 : 1 : \sqrt{2}\).
Using radian measure, the first triangle is a \(\dfrac{\pi}{6}\)-\(\dfrac{\pi}{3}\)-\(\dfrac{\pi}{2}\) triangle and the second triangle is a \(\dfrac{\pi}{4}\)-\(\dfrac{\pi}{4}\)-\(\dfrac{\pi}{2}\) triangle.
The ratios of the side lengths remain the same at \(1 : \sqrt{3} : 2\) and \(1 : 1 : \sqrt{2}\), respectively.
Using these triangles, we are able to determine the exact value of each of the six trigonometric ratios corresponding to angles \(\dfrac{\pi}{6}\), \(\dfrac{\pi}{4}\), and \(\dfrac{\pi}{3}\) or \(30^{\circ}\), \(45^{\circ}\), and \(60^{\circ}\).
Determine the exact values of the six trigonometric ratios for each of the angles \(\dfrac{\pi}{6}\), \(\dfrac{\pi}{4}\), and \(\dfrac{\pi}{3}\).
| Angle \((\theta)\) |
\(\sin(\theta)\) |
\(\cos(\theta)\) |
\(\tan(\theta)\) |
| \(\dfrac{\pi}{6}\) or \(30^{\circ}\) |
\(\dfrac{1}{2}\) |
\(\dfrac{\sqrt{3}}{2}\) |
\(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\) |
| \(\dfrac{\pi}{4}\) or \(45^{\circ}\) |
\(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\) |
\(\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\) |
\(1\) |
| \(\dfrac{\pi}{3}\) or \(60^{\circ}\) |
\(\dfrac{\sqrt{3}}{2}\) |
\(\dfrac{1}{2}\) |
\(\sqrt{3}\) |
| Angle \((\theta)\) |
\(\csc(\theta)\) |
\(\sec(\theta)\) |
\(\cot(\theta)\) |
| \(\dfrac{\pi}{6}\) or \(30^{\circ}\) |
\(2\) |
\(\dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\) |
\(\sqrt{3}\) |
| \(\dfrac{\pi}{4}\) or \(45^{\circ}\) |
\(\sqrt{2}\) |
\(\sqrt{2}\) |
\(1\) |
| \(\dfrac{\pi}{3}\) or \(60^{\circ}\) |
\(\dfrac{2}{\sqrt{3}} =\dfrac{2\sqrt{3}}{3}\) |
\(2\) |
\(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\) |
I wouldn't think that you would memorize these, but the triangles are excellent resources for recovering this information quickly.
Check Your Understanding B
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Examples
Example 2
A unit circle is shown with \(OP\) on the terminal arm of a \(\dfrac{\pi}{3}\) radian standard position angle, where \(O\) is the center of the circle and \(P(x,y)\) is a point on the unit circle. Determine the coordinates of \(P\).
Solution
Drop a perpendicular from \(P\) to the \(x\)-axis at \(A\).
We know that \(\sin \left( \dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{2}\), \(\cos \left( \dfrac{\pi}{3} \right) = \dfrac{1}{2}\), and \(\tan \left( \dfrac{\pi}{3} \right) = \sqrt{3}\).
We also know that, on the unit circle, each point has coordinates \((\cos(\theta), \sin(\theta))\).
Therefore, \(P\) has coordinates \(\left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\) and it follows that \(OA = \dfrac{1}{2}\), \(AP = \dfrac{\sqrt{3}}{2}\), and, since \(P\) is on the unit circle, \(OP = 1\).
Finding More Points on the Unit Circle
We're going to use this to generate a total of four points on the unit circle, each of which has a related angle or a reference angle (interchangeable terms), of \(\dfrac{\pi}{3}\) radians.
Suppose we reflect \(\triangle OAP\) about the \(y\)-axis, forming congruent \(\triangle O A' P'\), which lies in the second quadrant.
Using properties of reflections, \(P'\) has coordinates \(\left( -\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)\) and \(\angle P'OA' = \angle POA = \dfrac{\pi}{3}\).
Then \(OP'\) is along the terminal arm of a \(\dfrac{2 \pi}{3}\) standard position angle.
It follows that
\(\sin\left(\dfrac{2 \pi}{3}\right) = \dfrac{\sqrt{3}}{2}\), \(\cos\left(\dfrac{2 \pi}{3}\right) = -\dfrac{1}{2}\), and \(\tan \left( \dfrac{2 \pi}{3} \right) = \dfrac{\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}} = -\sqrt{3}\).
If we now reflect both \(\triangle O A' P'\) and \(\triangle OAP\) about the \(x\)-axis, we get \(\triangle O A' P''\) and \(\triangle OAP'''\), respectively.
The four triangles we have seen are congruent and each contains a \(\dfrac{\pi}{3}\) radian reference angle.
We also obtain the following:
- \(P''\left( -\dfrac{1}{2}, -\dfrac{\sqrt{3}}{2} \right)\) in the third quadrant and is on the terminal arm of a \(\dfrac{4\pi}{3}\) standard position angle.
- \(P'''~\left(\dfrac{1}{2}, - \dfrac{\sqrt{3}}{2}\right)\) in the fourth quadrant and is on the terminal arm of a \(\dfrac{5\pi}{3}\) standard position angle.
Each of the four points, \(P\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\), \(P'\left(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\), \(P''\left(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2}\right)\), and \(P'''\left(\dfrac{1}{2},-\dfrac{\sqrt{3}}{2}\right)\), lie on the unit circle.
The standard position angles drawn to \(P, P', P'',\) and \(P'''\) all have the same reference angle: \(\dfrac{\pi}{3}\), or \(60^{\circ}\), in this case.
If we disregard signs, all of the four points have the same \(x\) and \(y\)-coordinates.
So for each of the four related standard position angles, each trigonometric ratio, without regard for the sign, will be equal.
The sign of the trigonometric ratio depends on the quadrant in which the terminal arm lies.
Lesson Part 3
Determining the Sign of a Trigonometric Ratio
In previous courses, you may have determined the sign of a particular trigonometric ratio by analyzing the signs of \(x\) and \(y\) in the quadrant containing the terminal arm of the standard position angle.
Recall:
- In the first quadrant, \(x\gt 0\) and \(y\gt 0\);
- in the second quadrant, \(x\lt 0\) and \(y\gt 0\)
- in the third quadrant, \(x\lt 0\) and \(y\lt 0\); and
- in the fourth quadrant, \(x\gt 0\) and\(y\lt 0\).
Others may have used a memory device known as the CAST rule. Each letter in the rule tells which trigonometric ratio is positive in a particular quadrant.
In quadrant 1, all trigonometric ratios are positive.
In quadrant 2, sin\((\theta)\) is positive. (\(\csc(\theta)\) is also positive.)
In quadrant 3, tan\((\theta)\) is positive. (\(\cot(\theta)\) is also positive.)
In quadrant 4, cos\((\theta)\) is positive. (\(\sec(\theta)\) is also positive.)
Note: The trigonometric ratios associated with any reference angle are positive. (A reference angle is an acute angle. The terms related acute angle and reference angle mean the same thing. In this course, both terms are used.)
Check Your Understanding C
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Examples
Example 3
Determine the exact value of \(\sin \left( \dfrac{7 \pi}{4} \right)\).
Solution
The standard position angle \(\dfrac{7 \pi}{4}\) has its terminal arm in quadrant \(4\). So, the point \(P\left(\cos\left(\dfrac{7\pi}{4}\right),\sin\left(\dfrac{7\pi}{4}\right)\right)\) lies on the unit circle in the fourth quadrant.
The reference angle is \(2\pi-\dfrac{7\pi}{4}=\dfrac{\pi}{4}\) and we know that \(\sin \left( \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}}\).
The sine function is negative in quadrant \(4\).
Therefore, \(\sin \left( \dfrac{7 \pi}{4} \right) = -\sin \left( \dfrac{\pi}{4} \right) = -\dfrac{1}{\sqrt{2}}\).
This answer, \(-\dfrac{1}{\sqrt{2}}\), is referred to as the exact value of \(\sin \left( \dfrac{7 \pi}{4} \right)\).
Most calculators will be able to give the approximate value of a trig ratio but not the exact value.
When \(-\dfrac{1}{\sqrt{2}}\) is converted to a decimal, it is \(-0.707\), rounded to three decimals.
This second answer is an approximate answer.
Lesson Part 4
The Reference Angle \(\dfrac{\pi}{4}\)
Since \(\sin \left( \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}}\) and \(\cos \left( \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}}\), the point \(\left( \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \right)\) is on the unit circle in the first quadrant and on the terminal arm of a standard position angle \(\dfrac{\pi}{4}\).
The point \(\left( \dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}} \right)\) is also on the unit circle but in the fourth quadrant and would be on the terminal arm of a standard position angle, \(\dfrac{7 \pi}{4}\).
Using symmetry arguments, we also get the point \(\left( -\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \right)\) is on the unit circle in the second quadrant corresponding to a standard position angle, \(\dfrac{3 \pi}{4}\).
The point \(\left( -\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}} \right)\) is on the unit circle in the third quadrant corresponding to a standard position angle, \(\dfrac{5 \pi}{4}\).
The Reference Angle \(\dfrac{\pi}{6}\)
We looked at the exact trigonometric ratios for angles between \(0\) and \(2\pi\) with reference angles \(\dfrac{\pi}{3}\) and \(\dfrac{\pi}{4}\).
We have also determined the corresponding points on the unit circle.
Determine the coordinates of the four points on the unit circle corresponding to angles between \(0\) and \(2 \pi\) associated with reference angle \(\dfrac{\pi}{6}\).
Since \(\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\) and \(\sin\left( \dfrac{\pi}{6} \right) = \dfrac{1}{2}\), the point \(P_1 \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\) is on the unit circle in the first quadrant and the terminal arm of standard position angle is \(\dfrac{\pi}{6}\). The other three points are easily found. The four points are shown in the chart with the corresponding standard position angle.
| Quadrant the Terminal Arm is in |
Standard Position Angle |
Point on Unit Circle |
| \(1\) |
\(\dfrac{\pi}{6}\) |
\(P_1\left( \dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right)\) |
| \(2\) |
\(\pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}\) |
\(P_2\left( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\) |
| \(3\) |
\(\pi + \dfrac{\pi}{6} = \dfrac{7 \pi}{6}\) |
\(P_3\left( -\dfrac{\sqrt{3}}{2}, -\dfrac{1}{2} \right)\) |
| \(4\) |
\(2\pi - \dfrac{\pi}{6} = \dfrac{11 \pi}{6}\) |
\(P_4\left( \dfrac{\sqrt{3}}{2}, -\dfrac{1}{2} \right)\) |
The points, \(P_1, P_2, P_3,\) and \(P_4\), are also illustrated on the diagram.
Lesson Part 5
Examples
Example 4
Without using a calculator, determine the exact value of \(\cot(\pi)\).
Solution
We discover that \(\pi\) radians is a standard position angle with the terminal arm along the negative \(x\)-axis. In particular, the point \((-1,0)\) is on the unit circle and the terminal arm of the standard position angle of \(\pi\) radians.
We know that \(\tan(\theta) = \dfrac{y}{x}\) so \(\tan(\pi) = \dfrac{0}{-1}=0\).
Therefore, \(\cot(\pi) = \dfrac{-1}{0}\) and is undefined.
For any integer multiple of \(\dfrac{\pi}{2}\) radians or \(90^{\circ}\), the exact value of any trigonometric ratio can be found by selecting a point on the appropriate axis.
Example 5
If \(\tan (\theta) = -\sqrt{3}\), determine the possible values of \(\theta\) such that \(0 \leq \theta \leq 2 \pi\).
Solution
Some observations are helpful here. First, the ratio \(\dfrac{\sqrt{3}}{1}\) is familiar. We know that \(\tan\left( \dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{1}\). It follows that \(\dfrac{\pi}{3}\) is the reference angle. Second, \(\tan(\theta) \lt 0\). Therefore, the terminal arm can be in quadrant \(2\) or \(4\).
That is, there are two possible answers in the required domain.
In quadrant \(2\), \(\theta_1 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\).
In quadrant \(4\), \(\theta_2 = 2 \pi - \dfrac{\pi}{3} = \dfrac{5 \pi}{3}\).
If there had been no restriction on the domain, we would have to add integer multiples of \(2 \pi\) to the answers.
That is, if we wanted all possible values of \(\theta\) in a general sense, \(\theta = \left\{ \dfrac{2\pi}{3} + 2\pi n, \dfrac{5 \pi}{3} + 2\pi n, n \in \mathbb{Z} \right\}\).
Check to see if any coterminal angles are in the given domain.
In this case, there are no additional coterminal angles since we're only looking for angles between \(0\) and \(2\pi\).
Check Your Understanding D
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Example 6
If \(\cos (\theta) = - \dfrac{1}{4}\), determine the possible values of \(\theta\) such that \(-180^{\circ} \leq \theta \leq 180^{\circ}\).
Solution
Some observations are helpful here.
First, \(\cos(\theta) \lt 0\). Therefore, the terminal arm can be in quadrant \(2\) or \(3\).
Second, the ratio \(\dfrac{1}{4}\) is unfamiliar.
We can use a calculator to determine the approximate value of the reference angle.
Let \(\alpha\) be the reference angle. Then \(\cos (\alpha) = + \dfrac{1}{4}\).
(The trig ratio of any angle between \(0^{\circ}\) and \(90^{\circ}\), is greater than zero.)
Using your calculator, \(\alpha = \cos^{-1}\left( \dfrac{1}{4} \right) \approx 75.5^{\circ}\).
Be sure your calculator is set in degree mode for this example.
In quadrant \(2\), \(\theta_1 = 180^{\circ} - \alpha \approx 104.5^{\circ}\).
In quadrant \(3\), \(\theta_2 = 180^{\circ} + \alpha \approx 255.5^{\circ}\). But note that this answer is not in the required domain, i.e., between \(-180^{\circ}\) and \(180^{\circ}\). An angle coterminal with \(255.5^{\circ}\) in the required domain is \(255.5^{\circ}-360^{\circ}\) or \(-104.5^{\circ}\).
Therefore, \(\theta \approx \left\{ -104.5^{\circ}, 104.5^{\circ} \right\}\).
Check Your Understanding E
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Example 7
If \(\sin (\theta) = \dfrac{1}{2}\), determine the possible values of the other five trigonometric ratios.
Solution
Since \(\sin(\theta) \gt 0\), the terminal arm can be in quadrant \(1\) or \(2\). We are looking for two sets of trigonometric ratios.
Second, if we use the unit circle, the points have coordinates \((\cos(\theta), \sin(\theta)) = \left( \cos(\theta), \dfrac{1}{2} \right)\).
We know that \(x^2 + y^2 = 1\), so \(\cos^2(\theta) + \dfrac{1}{4} = 1\) and \(\cos^2(\theta) = \dfrac{3}{4}\) follows.
Therefore, \(\cos(\theta) = \pm \dfrac{\sqrt{3}}{2}\). The two points on the unit circle are \(P_1 \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\) in the first quadrant and \(P_2\left( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\) in the second quadrant.
Using the definitions with \(P_1\left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\),
\(\sin(\theta) = \dfrac{1}{2}\), \(\cos(\theta) = \dfrac{\sqrt{3}}{2}\), \(\tan(\theta) = \dfrac{1}{\sqrt{3}}\), \(\csc(\theta) = 2\), \(\sec(\theta) = \dfrac{2}{\sqrt{3}}\), and \(\cot(\theta) = \sqrt{3}\).
Using the definitions with \(P_2\left( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)\),
\(\sin(\theta) = \dfrac{1}{2}\), \(\cos(\theta) = -\dfrac{\sqrt{3}}{2}\), \(\tan(\theta) = -\dfrac{1}{\sqrt{3}}\), \(\csc(\theta) = 2\), \(\sec(\theta) = -\dfrac{2}{\sqrt{3}}\), and \(\cot(\theta) = -\sqrt{3}\).
In this lesson, you'll notice that there are a variety of examples. There are many different kinds of questions that can be formulated.
Example 8
The point \(P\,(5,-3)\) is on the terminal arm of some standard position angle \(\theta\), such that \(0 \leq \theta \leq 2 \pi\). Determine the measure of \(\theta\), correct to two decimal places.
Solution
Notice that \(P\,(5,-3)\) is a point in the fourth quadrant.
Let \(\alpha\) represent the reference angle.
If \(O\) is located at the origin and \(A\) at the point \((5,0)\), then, as illustrated in the diagram, \(OA = 5\) and \(AP = 3\).
Then \(\tan(\alpha) = \dfrac{3}{5}\) and \(\alpha = \tan^{-1} \left(\dfrac{3}{5} \right) \approx 0.54\) radians.
It follows that \(\theta = 2 \pi - \alpha \approx 5.74\) radians.
It should be noted that you could go directly from the definition \(\tan(\theta) = \dfrac{y}{x} = -\dfrac{3}{5}\).
When using your calculator to determine an angle in radians, your calculator will always return an answer between \(-\pi\) and \(\pi\).
You must interpret the result in light of the problem.
Finding the reference angle and placing the terminal arm in the correct quadrant removes any doubt.
Check Your Understanding F
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Lesson Part 6
Examples
Example 9
Solve \(2 \sin\left( x + \dfrac{\pi}{2} \right) = -\sqrt{3}, 0 \leq x \leq 3 \pi\). State the exact answers.
Solution
This example is a little bit more challenging. Note that it asks to state exact answers, which probably is giving you a little bit of a hint.
When you look at the equation, the angle looks a little bit messy. So a common trick is to replace the messy angle with a simpler angle.
Let \(\theta = x + \dfrac{\pi}{2}\). The equation simplifies to \(2 \sin \left( \theta \right) = -\sqrt{3}\).
Dividing both sides by \(2\), we get \(\sin (\theta) = -\dfrac{\sqrt{3}}{2}\).
Since \(\sin (\theta) \lt 0\), the terminal arm of the standard position angle is in either quadrant \(3\) or quadrant \(4\).
If \(\alpha\) is the reference angle, then \(\sin \alpha = \dfrac{\sqrt{3}}{2}\) and, from our special triangles, \(\alpha = \dfrac{\pi}{3}\).
In quadrant \(3\), \(\theta_1 = \pi + \alpha = \dfrac{4 \pi}{3}\).
But \(\theta = x + \dfrac{\pi}{2}\). So,
\begin{align*} x_1 &= \theta_1 - \dfrac{\pi}{2}\\ &= \dfrac{4 \pi}{3} - \dfrac{\pi}{2}\\ &= \dfrac{8 \pi}{6} - \dfrac{3 \pi}{6}\\ &= \dfrac{5 \pi}{6} \end{align*}
In quadrant \(4\), \(\theta_2 = 2\pi - \alpha = \dfrac{5 \pi}{3}\).
But \(\theta = x + \dfrac{\pi}{2}\). So,
\begin{align*} x_2 &= \theta_2 - \dfrac{\pi}{2}\\ &= \dfrac{5 \pi}{3} - \dfrac{\pi}{2}\\ &= \dfrac{10 \pi}{6} - \dfrac{3 \pi}{6}\\ &= \dfrac{7 \pi}{6} \end{align*}
Both \(x = \dfrac{5 \pi}{6}\) and \(x = \dfrac{7 \pi}{6}\) are in the domain \(0 \leq x \leq 3 \pi\).
We must check for coterminal angles in the domain.
Subtracting \(2\pi\) from either angle produces a negative angle outside of the domain.
However, \(\dfrac{5\pi}{6} + 2 \pi = \dfrac{17\pi}{6}\), which is in the domain.
Adding \(2\pi\) to either \(\dfrac{17\pi}{6}\) or \(\dfrac{7 \pi}{6}\) produces an angle outside of the domain.
Therefore, \(x = \left\{ \dfrac{5 \pi}{6}, \dfrac{7 \pi}{6}, \dfrac{17 \pi}{6} \right\}\).
We're going to encounter problems where the equations are messy like in this example. In such situations, the technique we used here is really useful.
Summary
In this module,
- we determined the exact and approximate values of various trigonometric ratios, and
- we worked with special triangles and solved a variety of equations.
In future modules, we will use methods developed in this module to assist in the graphing of trigonometric functions and in the solving of more difficult, but maybe more practical, applications.