Reference Angles and the Fundamental Identity

- Trigonometry Angles

In the previous article in the series on trigonometry, the focus was on solving right triangles using their unique properties. But what about angles that don't fit nicely within quadrant I of the cartesian coordinate plane? This article will demonstrate how to solve these types of angles/triangles using reference angles. Finally, it will wrap up by explaining an interesting relationship between sine and cosine, also known as the Fundamental Identiy.

Reference Angles

Right triangles have interesting properties that make it possible solve them, or find the lengths of all sides and angles; all one has to do is draw a line straight down from any point on the terminal side, as shown in Figure 1. In reality, this is exactly what a reference angle is: drawing a perpendicular line from any point on the terminal side of an angle to the x-axis. It just happens that, for angles in quadrant I, the actual angle and the reference angle are the same.

Quadrant I angle

A simple example demonstrates how this can be applied to angles with terminal sides in quadrants II, III, and IV. In this case, take the 135° angle \( \alpha \) shown in Figure 2.

Quadrant II angle

Creating a reference angle can be accomplished then by dropping a perpendicular line down from the terminal side to the \( x \)-axis. This creates the 45° angle \( \alpha\text{'} \) shown in Figure 3.

Quadrant II reference angle

Now, finding the sine and cosine is as simple as recalling the values from the previous article. The final step is determining the appropriate sign to apply to the results. This can be found by identifying the quadrant that the terminal side of the angle resides.

Trigonometric signs

In the example of the 45° reference angle \( \alpha\text{'} \), the terminal side lies in quadrant II, so only the sine will be positive; the cosine and tangent are both negative. Therefore, the values of the trigonometric functions for 135° are:

\( \text{sin}\ 135° = \frac{\sqrt{2}}{2} \) \( \text{csc}\ 135° = \sqrt{2} \)
\( \text{cos}\ 135° = -\frac{\sqrt{2}}{2} \) \( \text{sec}\ 135° = -\sqrt{2} \)
\( \text{tan}\ 135° = -1 \) \( \text{cot}\ 135° = -1 \)

The Fundamental Identity

Finally, an interesting relationship exists between the sine and cosine of an angle. Recall trigonometric ratios for sine and cosine: \( \text{sin}\ \alpha = \frac{y}{r} \) and \( \text{cos}\ \alpha = \frac{x}{r} \). Also, recall that \( r = \sqrt{x^{2} + y^{2}} \). See the previous articles in this series on trigonometry for more information about these values.

With these values, if the squares of the sine and cosine are added together, the Fundamental Identity is found.

\[ \begin{equation} \begin{split} \text{sin}^{2}\ \alpha + \text{cos}^{2}\ \alpha &= \frac{y^{2}}{r^{2}} + \frac{x^{2}}{r^{2}} \\ &= \frac{y^{2} + x^{2}}{r^{2}} \end{split} \end{equation} \]

Note that the numerator of the answer, \( y^{2} + x^{2} \), is the square of \( r \), and so it can be replaced, allowing for the equation to be further simplified.

\[ \begin{equation} \begin{split} \text{sin}^{2}\ \alpha + \text{cos}^{2}\ \alpha &= \frac{r^{2}}{r^{2}} \\ &= 1 \end{split} \end{equation} \]

This means that \( \text{sin}^{2}\ \alpha + \text{cos}^{2}\ \alpha = 1 \).

Conclusion

This article showed how right triangles can be created from any type of angle in standard position and explained the Fundamental Identity. Next up in the series will take a closer look at the unit circle and graphing the trigonomentric functions.