11.2: Trigonometric Functions
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Another set of useful functions are the trigonometric functions. These functions have probably plagued you since high school. They have their origins as far back as the building of the pyramids. Typical applications in your introductory math classes probably have included finding the heights of trees, flag poles, or buildings. It was recognized a long time ago that similar right triangles have fixed ratios of any pair of sides of the two similar triangles. These ratios only change when the non-right angles change.
Thus, the ratio of two sides of a right triangle only depends upon the angle. Since there are six possible ratios (think about it!), then there are six possible functions. These are designated as sine, cosine, tangent and their reciprocals (cosecant, secant and cotangent). In your introductory physics class, you really only needed the first three. You also learned that they are represented as the ratios of the opposite to hypotenuse, adjacent to hypotenuse, etc. Hopefully, you have this down by now.
You should also know the exact values of these basic trigonometric functions for the special angles \(θ = 0, \frac{π}{6} , \frac{π}{3} , \frac{π}{4} ,\frac{π}{2} \), and their corresponding angles in the second, third and fourth quadrants. This becomes internalized after much use, but we provide these values in Table \(\PageIndex{1}\) just in case you need a reminder.
Table \(\PageIndex{1}\): Table of Trigonometric Values
| \(\theta\) | \(\cos \theta\) | \(\sin \theta\) | \(\tan \theta\) |
|---|---|---|---|
| 0 | 1 | 0 | 0 |
| \(\frac{\pi}{6}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{3}\) |
| \(\frac{\pi}{3}\) | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\sqrt{3}\) |
| \(\frac{\pi}{4}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{2}}{2}\) | 1 |
| \(\frac{\pi}{2}\) | 0 | 1 | \(\text{undefined}\) |
The problems students often have using trigonometric functions in later courses stem from using, or recalling, identities. We will have many an occasion to do so in this class as well. What is an identity? It is a relation that holds true all of the time. For example, the most common identity for trigonometric functions is the Pythagorean identity \[ \sin^2 \theta + \cos^2 \theta =1 \label{eq:1}\] This holds true for every angle \(θ\)! An even simpler identity is \[ \tan \theta = \dfrac{\sin \theta}{\cos \theta} \label{eq:2}\]
Other simple identities can be derived from the Pythagorean identity. Dividing the identity by \(\cos^2 θ\), or \(sin^2 θ\), yields \[ \begin{align} \tan^2 \theta +1 \quad &= \quad \sec^2 \theta\label{eq:3} \\ 1+\cot^2 \theta \quad &= \quad \csc^2 \theta\label{eq:4} \end{align} \]
Several other useful identities stem from the use of the sine and cosine of the sum and difference of two angles. Namely, we have that \[ \begin{align} \sin (A \pm B) \quad &= \quad \sin A \cos B \pm \sin B \cos A\label{eq:5} \\ \cos (A \pm B) \quad &= \quad \cos A \cos B \pm \sin A \sin B\label{eq:6} \end{align} \] Note that the upper (lower) signs are taken together.
Evaluate \(\sin \frac{π}{12} \).
Solution
\[ \begin{align} \sin \dfrac{\pi}{12} \quad &= \quad \sin \left( \dfrac{\pi}{3} - \dfrac{\pi}{4} \right) \nonumber \\ &= \quad \sin \dfrac{\pi}{3} \cos \dfrac{\pi}{4} - \sin \dfrac{\pi}{4} \cos \dfrac{\pi}{3} \nonumber \\ &= \quad \dfrac{\sqrt{3}}{2} \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} \dfrac{1}{2} \nonumber \\ &= \quad \dfrac{\sqrt{2}}{4} \left( \sqrt{3} -1 \right)\label{eq:7} \end{align} \]
The double angle formulae are found by setting \(A = B \): \[ \begin{align} \sin (2A) \quad &= \quad 2\sin A \cos B\label{eq:8} \\ \cos (2A) \quad &= \quad \cos^2 A - \sin^2 A\label{eq:9} \end{align} \]
Using Equation \(\eqref{eq:13}\), we can rewrite \(\eqref{eq:21}\) as \[ \begin{align} \cos (2A) \quad &= \quad 2\cos^2 A-1\label{eq:10} \\ &= \quad 1-2 \sin^2 A\label{eq:11} \end{align} \] These, in turn, lead to the half angle formulae. Solving for \(\cos^2 A\) and \(sin^2 A\), we find that \[ \begin{align} \sin^2 A \quad &= \quad \dfrac{ 1-\cos 2A}{2}\label{eq:12} \\ \cos^2 A \quad &= \quad \dfrac{1+\cos 2A}{2}\label{eq:13} \end{align} \]
Evaluate \(\cos \frac{π}{12} \). In the last example, we used the sum/difference identities to evaluate a similar expression. We could have also used a half angle identity.
Solution
In this example, we have \[ \begin{align} \cos^2 \dfrac{\pi}{12} \quad &= \quad \dfrac{1}{2} \left( 1+ \cos \dfrac{\pi}{6}\right) \nonumber \\ &= \quad \dfrac{1}{2} \left( 1 + \dfrac{\sqrt{3}}{2} \right) \nonumber \\ &= \dfrac{1}{4} \left( 2+\sqrt{3} \right)\label{eq:14} \end{align} \]
So, \(\cos \frac{π}{12} = \frac{1}{2}\sqrt{2 + \sqrt{3}}\). This is not the simplest form and is called a nested radical. In fact, if we proceeded using the difference identity for cosines, then we would obtain \[ \cos{\pi}{12} = \dfrac{\sqrt{2}}{4} (1+ \sqrt{3}) \nonumber \] So, how does one show that these answers are the same?
It is useful at times to know when one can reduce square roots of such radicals, called denesting. More generally, one seeks to write \(\sqrt{a + b\sqrt{q}} = c + d\sqrt{q}\). Following the procedure in this example, one has \(d = \frac{b}{2c}\) and \[ c^2 = \dfrac{1}{2}\left( a \pm \sqrt{a^2 − qb^2}\right)\nonumber\] As long as \(a^2 − qb^2\) is a perfect square, there is a chance to reduce the expression to a simpler form.
Let’s focus on the factor \(\sqrt{2 + \sqrt{3}}\). We seek to write this in the form \(c + d\sqrt{3}\). Equating the two expressions and squaring, we have \[\begin{align} 2+ \sqrt{3} \quad &= \quad (c+d\sqrt{3})^2 \nonumber \\ &= \quad c^2 +3d^2+2cd \sqrt{3}\label{eq:15} \end{align} \] In order to solve for \(c\) and \(d\), it would seem natural to equate the coefficients of \(\sqrt{3}\) and the remaining terms. We obtain a system of two nonlinear algebraic equations, \[ \begin{align} c^2 3d^2 \quad &= 2\label{eq:16}\\ 2cd \quad &= \quad 1\label{eq:17} \end{align} \]
Solving the second equation for \(d = 1/2c\), and substituting the result into the first equation, we find \[ 4c^4 - 8c^2 +3=0 \nonumber \] This fourth order equation has four solutions \[ c+ \pm \dfrac{\sqrt{2}}{2} , \pm \dfrac{\sqrt{6}}{2} \nonumber \] and \[ b=\pm \dfrac{\sqrt{2}}{2} , \pm \dfrac{\sqrt{6}}{6} \nonumber \] Thus, \[\begin{align} \cos{\pi}{12} \quad &= \quad \dfrac{1}{2} \sqrt{2+\sqrt{3}} \nonumber \\ &= \quad \pm \dfrac{1}{2} \left( \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \sqrt{3} \right) \nonumber \\ &= \quad \pm \dfrac{\sqrt{2}}{4} \left( 1 + \sqrt{3} \right)\label{eq:18} \end{align} \] and \[ \begin{align} \cos \dfrac{\pi}{12} \quad &= \quad \nonumber \dfrac{1}{2} \sqrt{2 + \sqrt{3}} \\ &= \quad \nonumber \pm \dfrac{1}{2} \left( \dfrac{\sqrt{6}}{2} + \dfrac{\sqrt{6}}{6} \sqrt{3} \right) \\ &= \quad \pm \dfrac{\sqrt{6}}{12} \left( 3 + \sqrt{3} \right)\label{eq:19} \end{align} \] Of the four solutions, two are negative and we know the value of the cosine for this angle has to be positive. The remaining two solutions are actually equal! A quick computation will verify this \[ \begin{align} \dfrac{\sqrt{6}}{12} \left( 3+\sqrt{3} \right) \quad &= \quad \nonumber \dfrac{\sqrt{3}\sqrt{2}}{12} \left( 3 +\sqrt{3} \right) \\ &= \quad \nonumber \dfrac{\sqrt{2}}{12} \left( 3\sqrt{3} + 3 \right) \\ &= \quad \dfrac{\sqrt{2}}{4} \left( \sqrt{3} +1 \right)\label{eq:20} \end{align} \] We could have bypassed this situation be requiring that the solutions for \(b\) and \(c\) were not simply proportional to \(\sqrt{3}\) like they are in the second case.
Finally, another useful set of identities are the product identities. For example, if we add the identities for \(\sin (A + B)\) and \(\sin (A − B)\), the second terms cancel and we have \[ \sin (A+B) + \sin (A-B) = 2\sin A\cos B\nonumber \] Thus, we have that \[ \sin A \cos B = \dfrac{1}{2} (\sin (A+B)+ \sin (A-B)) \label{eq:21}\]
Similarly, we have \[ \cos A \cos B = \dfrac{1}{2} ( \cos (A+B) + \cos (A-B))\label{eq:22} \] and \[ \sin A \sin B = \dfrac{1}{2} (\cos (A-B) - \cos (A+B)) \label{eq:23}\]
These boxed equations are the most common trigonometric identities. They appear often and should just roll off of your tongue.
We will also need to understand the behaviors of trigonometric functions. In particular, we know that the sine and cosine functions are periodic. They are not the only periodic functions, as we shall see. [Just visualize the teeth on a carpenter’s saw.] However, they are the most common periodic functions.
A periodic function \(f (x)\) satisfies the relation \[ f(x+p) = f(x), \quad \text{for all }x \nonumber\] for some constant \(p\). If \(p\) is the smallest such number, then \(p\) is called the period. Both the sine and cosine functions have period \(2π\). This means that the graph repeats its form every \(2π\) units. Similarly, \(\sin bx\) and \(\cos bx\) have the common period \(p = \frac{2π}{b} \). We will make use of this fact in later chapters.
Related to these are the inverse trigonometric functions. For example, \(f (x) = \sin^{−1} x\), or \(f (x) = \arcsin x\). Inverse functions give back angles, so you should think \[ \theta = \sin^{-1} x \quad \Leftrightarrow \quad x=\sin \theta\label{eq:24} \] Also, you should recall that \(y = \sin^{−1} x = \arcsin x\) is only a function if \(− \frac{π}{2} ≤x ≤ \frac{π}{2}\). Similar relations exist for \(y = \cos^{−1} x = \arccos x\) and \(\tan^{−1} x = \arctan x\).
In Feynman’s Surely You’re Joking Mr. Feynman!, Richard Feynman (1918-1988) talks about his invention of his own notation for both trigonometric and inverse trigonometric functions as the standard notation did not make sense to him.
Once you think about these functions as providing angles, then you can make sense out of more complicated looking expressions, like \(\tan \left(\sin ^{-1} x\right)\). Such expressions often pop up in evaluations of integrals. We can untangle this in order to produce a simpler form by referring to expression \(\eqref{eq:24}\). \(\theta=\sin ^{-1} x\) is simple an angle whose sine is \(x\). Knowing the sine is the opposite side of a right triangle divided by its hypotenuse, then one just draws a triangle in this proportion as shown in Figure \(\PageIndex{1}\). Namely, the side opposite the angle has length \(x\) and the hypotenuse has length \(1\). Using the Pythagorean Theorem, the missing side (adjacent to the angle) is simply \(\sqrt{1-x^{2}}\). Having obtained the lengths for all three sides, we can now produce the tangent of the angle as \[\tan \left(\sin ^{-1} x\right)=\frac{x}{\sqrt{1-x^{2}}} .\nonumber \]
