Now that we have learned how to work with angles in standard position and to convert between degrees and radians, we can now extend this to find the trigonometric ratios of these angles. While it can still be helpful to think of the sides of the triangle in terms of opposite, adjacent, and hypotenuse, we are also going to see what these values would correspond to on the coordinate plane. This will also allow us to find the trig values of angles that have a terminal side on one of the axes where you wouldn’t be able to draw a triangle.
Below are each of the six trig functions listed in terms of \(x\), \(y\), and \(r\).
\(\sin\theta=\large\frac{y}{r}\) \(\csc\theta=\large\frac{r}{y}\)
\(\cos\theta=\large\frac{x}{r}\) \(\sec\theta=\large\frac{x}{r}\)
\(\tan\theta=\large\frac{y}{x}\) \(\cot\theta=\large\frac{x}{y}\)
\(\sin\theta=\large\frac{y}{r}\) \(\csc\theta=\large\frac{r}{y}\)
\(\cos\theta=\large\frac{x}{r}\) \(\sec\theta=\large\frac{x}{r}\)
\(\tan\theta=\large\frac{y}{x}\) \(\cot\theta=\large\frac{x}{y}\)
As we work on these trig ratios, we will use the two special right triangles, \(30^{\circ}-60^{\circ}-90^{\circ}\) and \(45^{\circ}-45^{\circ}-90^{\circ}\), quite often so these triangles are shown below for reference. Keep in mind that when we have angles with terminal sides in different quadrants, we will be working with \(x\) and \(y\) values that may be negative. The radius (or hypotenuse) will always be non-negative since that refers to the distance from the origin or vertex of the angle.
Let’s look at some examples. If you need any help with the converting between radians and degrees, be sure to go back and look at the Guided Learning for Target B.
Example 1: Find the exact value of each expression.
a) \(\sin\left(\frac{4\pi}{3}\right)\)
We will first convert \( \frac{4 \pi}{3} \) to degrees which is \(240^{\circ}\). This angle is in the third quadrant with a reference angle of \(60^{\circ}\) so we will graph that and make a triangle based on that reference angle. Note that both the \(x\) and \(y\) values will be negative since we are in the third quadrant, and the radius will always be non-negative regardless of the quadrant.
\(\sin\left(\frac{4\pi}{3}\right)=\frac{y}{r}=-\frac{\sqrt{3}}{2}\)
b) \(\cos\left(-225^{\circ}\right)\) c) \(\tan\left(\frac{11\pi}{6}\right)\)
Now let’s say we are asked to find \(\sin\left(\frac{3\pi}{2}\right)\). We would first think about converting the angle to degrees which would be \(270^{\circ}\). Well, if we graph that then we get the terminal side being on the y-axis which doesn’t allow us to draw a triangle to the x-axis like we could in the examples above (you essentially have a “triangle” where one of the sides is \(0\). However since we now know how to find trig ratios based on the \(x\), \(y\), and \(r\), we can use those to find these trig ratios. Any point on an axis will work for \(x\) and \(y\) values, but the easiest points to work with are shown below since the \(r\) value for each of those would be \(1\). So, in thinking about the point and the radius, we can get any of the trig ratios for those angles. You will see a lot of answers involving \(1\), \(-1\), \(0\), and undefined (which comes when you try to divide a number by \(0\).
\(\sin\left(\frac{3\pi}{2}\right)=-\frac{1}{1}=-1\) since the \(y\) value is \(-1\) and the radius is \(1\).
\(\cot\left(-\pi\right)=-\frac{1}{0}\) which is undefined. Be careful to write out “undefined” if that is the case.
However, \(\tan\left(-\pi\right)=\frac{0}{-1}=0\).
Example 2: Find the exact value of each expression.
a) \(\cos\left(-2\pi\right)\) b) \(\csc\left(\frac{\pi}{2}\right)\)
Let’s now look at a few other applications that we can now work with. We talked about the fact that you will get \(x\) and \(y\) values that can now be negative depending on what quadrant the terminal side of an angle is located in. This is shown below. You will notice that all of the trig ratios are positive in the first quadrant, only sine/cosecant are positive in the second quadrant, only tangent/cotangent are positive in the third quadrant, and only cosine/secant are positive in the fourth quadrant. We can use this to help us identify what quadrant a particular angle should be located in based on given information.
Example 3: Given the following conditions, identify the quadrant in which the terminal side of must lie.
a) \(\sin\theta<0,\ \tan\theta>0\)
Given \(\sin\theta<0\), \(\theta\) can be located in Quadrants III or IV. Given \(\tan\theta>0\), \(\theta\) can be located in Quadrants I or III. Therefore, \(\theta\) must be located in Quadrant III to meet both conditions.
b) \(\sec\theta<0,\ \cot\theta<0\)
Given \(\sec\theta<0\), \(\theta\) can be located in Quadrants II or III (since cosine is also negative in those quadrants). Given \(\cot\theta<0\), \(\theta\) can be located in Quadrants II or IV (since tangent is also negative in those quadrants. Therefore, \(\theta\) must be located in Quadrant II to meet both conditions.
Example 4: Given that \(\sin\theta=-\frac{7}{25}\) and \(\tan\theta<0\), find the remaining five trig ratios of \(\theta\).
We know that \(\theta\) must have its terminal side in Quadrant IV since that is the only quadrant where sine and tangent are both negative. So we can draw a triangle in that quadrant and label the opposite side (or \(y\)-value) \(-7\) and the hypotenuse (or radius) with \(25\) as shown below. We will then use the Pythagorean Theorem to find the other side of the triangle which is \(24\). Given the location of the triangle, the side that is \(24\) should be positive and not negative. From there, we can list the remaining trig ratios (also included is sine even though we were given that to start with).
\(\sin\theta=\large-\frac{7}{25}\) \(\csc\theta=\large-\frac{25}{7}\)
\(\cos\theta=\large\frac{24}{25}\) \(\sec\theta=\large\frac{25}{24}\)
\(\tan\theta=\large-\frac{7}{24}\) \(\cot\theta=\large-\frac{24}{7}\)
Example 5: Let \(\left(-12,-5\right)\) be a point on the terminal side of an angle \(\theta\) in standard position on a circle with radius \(r\). Find the values of the six trig ratios.