Chapter 4.2 Translational Symmetry

In this chapter, we discuss a symmetry in which certain functions have. Specifically, for certain functions, a translation by the right amount will leave the function unchanged. We will discuss the periodic functions, discuss the symmetry the trigonometric functions posses, and define their inverses.

Periodicity

We first define period and fundamental period.

Period and Fundamental Period

Suppose that \(f\colon \mathbb R\to \mathbb R\).

A real number \(a\) is a period of \(f\) if for every \(x\) in \(\mathcal D(f)\), \(x+a\) is also in \(\mathcal D(f)\) and \[f(x) = f(x+a).\]

A positive period \(a\) of \(f\) is the fundamental period of \(f\) if there is no smaller positive period of \(f\).

Essentially, a function \(f\) is periodic if there is a vector \(\langle a,0\rangle\) so that \(V+f=f.\) That is, translating \(f\) by the right vector produces the same function \(f\).

The fundamental period is the smallest positive \(a\) so that it is still a period for \(f\).

Identify the period of a function using the graph with the next example.

Example 1

Below is a sketch of a function \(f\). What is the fundamental period of \(f\)? Give an example of at least one non-fundamental period.

Period is \(P=2\). A non-fundamental period is \(P=4\).

Not all functions have a fundamental period. See the next example.

Example 2

Is the function \(f\) given by \[f(x) = 2\] a periodic function? Does it have a fundamental period?

Yes, the function is periodic. For example \(f(1+x)=f(x)\) for all \(x\) in \(\mathbb{R}\). However, there is no fundamental period.

What kind of translations and scalings affect the fundamental period of a function? It turns out only one kind of scaling does.

Take \(m\) and \(b\) to be real numbers, where \(m\) is nonzero, and take \(g\) to be defined by \[g(x) = mx+b.\]

The composite \(f\circ g\) is periodic and, if \(a\) is a period of \(f\), then \(\frac{a}{m}\) is a period of \(f\circ g\).

Example 3

Take \(f\) to be a function with fundamental period equal to 2 and take \(g\) to be given by \[g(x) = 3x - 5.\] What is the fundamental period of \(f\circ g\)?

The function \(f\circ g\) would have fundamental period \(\frac{2}{3}\). The only thing that affects the period is \(x\)-scaling.

Only assymmetric scaling in \(x\) affects the fundamental period.

Sketching Trigonometric Functions

We have already seen examples of periodic functions. For example, the trigonometric functions are periodic functions.

• The fundamental period of the sine function is \(2\pi\).

• The fundamental period of the cosine function is \(2\pi\).

• The fundamental period of the tangent function is \(\pi\).

This is due to the fact that the functions are defined based on the unit circle and angles.

The graph of sine can be obtained by plotting points using standard angles.

First sketch the sine, cosine, and tangent functions.

Remember that the fundamental period of sine and cosine is \(2\pi\).

\(x\)

\(0\)

\(\frac{\pi}{6}\)

\(\frac{\pi}{4}\)

\(\frac{\pi}{3}\)

\(\frac{\pi}{2}\)

\(\frac{2\pi}{3}\)

\(\frac{3\pi}{4}\)

\(\frac{5\pi}{6}\)

\(\pi\)

\(\frac{7\pi}{6}\)

\(\frac{5\pi}{4}\)

\(\frac{4\pi}{3}\)

\(\frac{3\pi}{2}\)

\(\frac{5\pi}{3}\)

\(\frac{7\pi}{4}\)

\(\frac{11\pi}{6}\)

\(2\pi\)

\(y\)

\(0\)

\(\frac{1}{2}\)

\(\frac{\sqrt{2}}{2}\)

\(\frac{\sqrt{3}}{2}\)

\(1\)

\(\frac{\sqrt{3}}{2}\)

\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{2}\)

\(0\)

\(-\frac{1}{2}\)

\(-\frac{\sqrt{2}}{2}\)

\(-\frac{\sqrt{3}}{2}\)

\(-1\)

\(-\frac{\sqrt{3}}{2}\)

\(-\frac{\sqrt{2}}{2}\)

\(-\frac{1}{2}\)

\(0\)

\(x\)	\(0\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	\(\frac{2\pi}{3}\)	\(\frac{3\pi}{4}\)	\(\frac{5\pi}{6}\)	\(\pi\)	\(\frac{7\pi}{6}\)	\(\frac{5\pi}{4}\)	\(\frac{4\pi}{3}\)	\(\frac{3\pi}{2}\)	\(\frac{5\pi}{3}\)	\(\frac{7\pi}{4}\)	\(\frac{11\pi}{6}\)	\(2\pi\)
\(y\)	\(1\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{1}{2}\)	\(0\)	\(-\frac{1}{2}\)	\(-\frac{\sqrt{2}}{2}\)	\(-\frac{\sqrt{3}}{2}\)	\(-1\)	\(-\frac{\sqrt{3}}{2}\)	\(-\frac{\sqrt{2}}{2}\)	\(-\frac{1}{2}\)	\(0\)	\(\frac{1}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(1\)

Remember that the fundamental period of the tangent function is \(\pi\).

Plot a few standard points and consider the behavior near \(\pm\frac{\pi}{2}\).

\(x\)	\(0\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	\(\frac{2\pi}{3}\)	\(\frac{3\pi}{4}\)	\(\frac{5\pi}{6}\)	\(\pi\)	\(\frac{7\pi}{6}\)	\(\frac{5\pi}{4}\)	\(\frac{4\pi}{3}\)	\(\frac{3\pi}{2}\)	\(\frac{5\pi}{3}\)	\(\frac{7\pi}{4}\)	\(\frac{11\pi}{6}\)	\(2\pi\)
\(y\)	\(0\)	\(\frac{\sqrt{3}}{3}\)	\(1\)	\(\sqrt{3}\)	undefined	\(-\sqrt{3}\)	\(-1\)	\(-\frac{\sqrt{3}}{3}\)	\(0\)	\(\frac{\sqrt{3}}{3}\)	\(1\)	\(\sqrt{3}\)	undefined	\(-\sqrt{3}\)	\(-1\)	\(-\frac{\sqrt{3}}{3}\)	\(0\)

The same kinds of transformations that act on other functions we have discussed also act on \(\sin\).

The most complicated transformed function will be of the form \[f(\theta) = A\sin(B(\theta-C)) + D.\]

In this case, \[f = T_D\circ S_A \circ \sin \circ S_B\circ T_{-C}.\]

Of course, the transformations act on any trigonometric function and act similarly on all of them.

The amplitude of a transformed \(\sin\) function, which is called a sinusoidal function, is half the difference between heights of the function’s crest and trough.

The function \(f\) that is given by \[f(\theta) = A\sin(B(\theta-C)) + D\] will therefore have an amplitude equal to \(|A|\) and fundamental period equal to \(\frac{2\pi}{|B|}\)

Example 4

Determine the amplitude of \(f\), where \[f(\theta) = -5\sin(\theta).\]

The amplitude of \(f\) is \(|-5|=5.\)

Practice identifying the fundamental period with this next example.

Example 5

Determine the fundamental period of \(f\), where

1. \(f(\theta) = -3\sin(\theta)+2\);
2. \(f(\theta) = \sin(3\theta)\);
3. \(f(\theta) = \cos(\pi \theta)\);
4. \(f(\theta) = \tan(7 \theta)\).

1. Sine has a period of \(2\pi\). The function \(f\) is a vertically scaled and vertically shifted version of the sine function so the period remains as \(2\pi\).

2. Sine has a period of \(2\pi\). The function \(f\) is a horizontally scaled version of the sine function so the period will be scaled by \(\tfrac{1}{3}\). The new period is \(\tfrac{2\pi}{3}\).

3. Cosine has a period of \(2\pi\). The function \(f\) is a horizontally scaled version of the cosine function so the period will be scaled by \(\tfrac{1}{\pi}\). The new period is \(\tfrac{2\pi}{\pi}=2\).

4. Tangent has a period of \(\pi\). The function \(f\) is a horizontally scaled version of the tangent function so the period will be scaled by \(\tfrac{1}{7}\). The new period is \(\tfrac{\pi}{7}\).

Another important feature of the trigonometric function is the phase shift.

Phase Shift

Take the functions \(f_1\), \(f_2\), and \(f_3\) to be given by \[f_1(\theta) = \sin(B(\theta -C)), \quad f_2(\theta) = \cos(B(\theta -C)),\quad {\rm and}\quad f_3(\theta) = \tan(B(\theta -C)).\]

Take the functions \(g_1\), \(g_2\), and \(g_3\) to be given by \[g_1(\theta) = \sin(B\theta), \quad g_2(\theta) = \cos(B\theta),\quad {\rm and}\quad g_3(\theta) = \tan(B\theta).\]

The functions \(f_1\), \(f_2\), and \(f_3\) are, respectively, the functions \(g_1\), \(g_2\), and \(g_3\) shifted to the right by \(C\).

The quantity \(K\) that is given by the equality \[C = \frac{2\pi K}{B}\] is the phase shift.

In terms of transformations, any vertical translation is called a vertical shift.

Vertical Shift

If \(f\) is a function, and \(D\) is a real number, then the function \(g\) that is given by \[g(\theta) = f(\theta) + D\] is the function \(f\) with a vertical shift of \(D\).

The transformation principle can be used to plot the function given below \[A\cdot g(B(x-C))+D = T_D\circ S_A \circ g \circ S_B\circ T_{-C}\] or \[Ag(Bx-BC))+D.\]

1. Start with the graph of sine, cosine, or tangent. You may wish to start with some basic points in the original graphs.
2. Do the translation \(T_{-BC}\) (\(x\)-values)
3. Do the scaling \(S_{B}\) (\(x\)-values)
4. Do scaling \(S_{A}\) (\(y\)-values)
5. Do scaling \(S_{A}\) (\(y\)-values)
6. Do the translation \(T_{D}\) (\(y\)-values)

Example 6

Sketch the function that is given by the equation \(y = 2\sin\!\big(3\big(x-\frac{1}{2}\big)\big) + 1\).

For transformation purposes, let’s distribute the \(3:\)

\[2\sin(3x-\tfrac{3}{2})+1\]

Start with the graph of \(\sin(x)\). Make note of the period.

Points:

• \((0,0)\)
• \((\tfrac{\pi}{2},1)\)
• \((\pi,0)\)
• \((\tfrac{3\pi}{2},-1)\)
• \((2\pi,0)\)
• Period is \(2\pi\)

Translate in \(x\)-direction by \(T_{-3/2}\):

Points:

• \((\tfrac{3}{2},0)\)
• \((\tfrac{\pi}{2}+\tfrac{3}{2},1)\)
• \((\pi+\tfrac{3}{2},0)\)
• \((\tfrac{3\pi}{2}+\tfrac{3}{2},-1)\)
• \((2\pi+\tfrac{3}{2},0)\)
• Period is \(2\pi\)

Scale in \(x\)-direction by \(S_{3}\).

Points:

• \((\tfrac{1}{2},0)\)
• \((\tfrac{\pi}{6}+\tfrac{1}{2},1)\)
• \((\tfrac{\pi}{3}+\tfrac{1}{2},0)\)
• \((\tfrac{\pi}{2}+\tfrac{1}{2},-1)\)
• \((\tfrac{2\pi}{3}+\tfrac{1}{2},0)\)
• Period is \(\tfrac{2\pi}{3}\)

Scale in \(y\)-direction by \(S_{2}\).

Points:

• \((\tfrac{1}{2},0)\)
• \((\tfrac{\pi}{6}+\tfrac{1}{2},2)\)
• \((\tfrac{\pi}{3}+\tfrac{1}{2},0)\)
• \((\tfrac{\pi}{2}+\tfrac{1}{2},-2)\)
• \((\tfrac{2\pi}{3}+\tfrac{1}{2},0)\)
• Period is \(\tfrac{2\pi}{3}\)

Translate in \(y\)-direction by \(T_{1}\).

Points:

• \((\tfrac{1}{2},1)\)
• \((\tfrac{\pi}{6}+\tfrac{1}{2},3)\)
• \((\tfrac{\pi}{3}+\tfrac{1}{2},1)\)
• \((\tfrac{\pi}{2}+\tfrac{1}{2},-1)\)
• \((\tfrac{2\pi}{3}+\tfrac{1}{2},1)\)
• Period is \(\tfrac{2\pi}{3}\)

Identify the following information by analyzing its formula.

Example 7

Take \(f\) to be the function given by \(y = 3\sin(5(\theta - \tfrac{\pi}{4})) + 1.\) What is the amplitude, period, phase shift, and vertical shift?

The amplitude is \(A=3\), the period is \(P=\tfrac{2\pi}{5}\), the vertical shift is \(D=1\). The phase shift is the number \(K\) such that \(C=\frac{2\pi K}{B}\), which in this example is the number \(K\) that satisfies \(\tfrac{\pi}{4}=\frac{2\pi K}{5}\) or \(K=\frac{5}{8}\).

The sine and cosine function are related. Try the following example.

Example 8

Write \(\cos\) as a sinusoidal function.

Hence \(\sin(x+\tfrac{\pi}{2})=\cos(x)\)

The recipocal of sine, cosine and tangent are important, so we name them.

Secant Cosecant Cotangent

Define the secant function by \[\sec(\theta) = \frac{1}{\cos(\theta)}.\]

Define the cosecant function by \[\csc(\theta) = \frac{1}{\sin(\theta)}.\]

Define the cotangent function by \[\cot(\theta) = \frac{1}{\tan(\theta)}.\]

Understanding how to evaluate these functions with this example.

Example 9

Evaluate the three reciprocal trigonometric functions at \(\frac{4\pi}{3}\).

• \(\sin(\tfrac{4\pi}{3})=-\frac{\sqrt{3}}{2}\) so \(\csc(\tfrac{4\pi}{3})=-\frac{2}{\sqrt{3}}\)
• \(\cos(\tfrac{4\pi}{3})=-\frac{1}{2}\) so \(\sec(\tfrac{4\pi}{3})=-2\)
• \(\tan(\tfrac{4\pi}{3})=-\sqrt{3}\) so \(\cot(\tfrac{4\pi}{3})=-\frac{1}{\sqrt{3}}\)

Sketching these functions can be achieved using \(y\)-axis inversion.

Example 10

Sketch the secant function.

Here is \(\cos(x)\)

There are zeros at \(x=\tfrac{\pi}{2}\) and \(x=\tfrac{3\pi}{2}\) and any odd multiple of \(\tfrac{\pi}{2}\). So when we do inversion, they become vertical asymptotes.

Try cosecant next.

Example 11

Sketch the cosecant function.

Here is \(\sin(x)\)

There are zeros at \(x=0\) and \(x=\pi\) and any multiple of \(\pi\). So when we do inversion, they become vertical asymptotes.

Next, try cotangent.

Example 12

Sketch the cotangent function.

Here is \(\tan(x)\)

There are zeros at \(x=\tfrac{\pi}{2}\) and \(x=\tfrac{3\pi}{2}\) and any odd multiple of \(\tfrac{\pi}{2}\). So when we do inversion, they become vertical asymptotes. And anywhere there are VA, we wil get zeros: so zeros at multiples at odd multiples of \(\pi/2\)

Inverse Trigonometric Functions

The functions \(\cos\), \(\sin\), and \(\tan\) do not pass the horizontal line test.

No periodic functions can!

So, they are not invertible. However, we can restrict them to domains where they are invertible.

Example 13

Restrict the cosine function to a domain where it is invertible

Restrict to \([0,\pi]\)

Now the function is invertible.

We now focus on sine.

Example 14

Restrict the sine function to a domain where it is invertible.

Restrict to \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\)

Now the function is invertible.

Next, we restrict the tangent function.

Example 15

Restrict the tangent function to a domain where it is invertible.

Restrict to \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\)

Now the function is invertible.

In general and as expected, it is difficult to evaluate the inverse of a trigonometric function.

Input a length (for sine or cosine) or slope (for tangent) and output a corresponding angle.

Take \({\rm Cos}\), \({\rm Sin}\), and \({\rm Tan}\) to respectively be the principle cosine, sine, and tangent functions, the respective restrictions of \(\cos\), \(\sin\), and \(\tan\) to \(\left[0, \pi\right]\), \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Notation: \(\cos^{-1}\), \(\sin^{-1}\), and \(\tan^{-1}\) are the inverse relations, where \(\arccos\), \(\arcsin\), and \(\arctan\) are the inverses of the respective principle trigonometric functions.

Warning: This notation is not completely standard.

Example 16

Calculate

1. \(\arccos(-1)\)

2. \(\arccos\!\left(-\frac{\sqrt{3}}{2}\right)\)

3. \(\arccos\!\left(-\frac{1}{\sqrt{2}}\right)\)

4. \(\arccos\!\left(-\frac{1}{2}\right)\)

5. \(\arccos(0)\)

6. \(\arcsin\!\left(\frac{1}{2}\right)\)

7. \(\arcsin\!\left(\frac{1}{\sqrt{2}}\right)\)

8. \(\arctan\!\left(\sqrt{3}\right)\)

9. \(\arctan(1)\)

1. \(\arccos(-1)=y\) is the same as \(-1=\cos(y)\) where \(y\in[0,\pi]\). Based on the unit circle we get \(y=\pi\), so \(\arccos(-1)=\pi\).

2. \(\arccos\!\left(-\frac{\sqrt{3}}{2}\right)=y\) is the same as \(-\frac{\sqrt{3}}{2}=\cos(y)\) where \(y\in[0,\pi]\). Based on the unit circle we get \(y=\tfrac{5\pi}{6}\), so \(\arccos\!\left(-\frac{\sqrt{3}}{2}\right)=\tfrac{5\pi}{6}\).

3. \(\arccos\!\left(-\frac{1}{\sqrt{2}}\right)=y\) is the same as \(-\frac{1}{\sqrt{2}}=\cos(y)\) where \(y\in[0,\pi]\). Based on the unit circle we get \(y=\tfrac{3\pi}{4}\), so \(\arccos\!\left(-\frac{1}{\sqrt{2}}\right)=\tfrac{3\pi}{4}\).

4. \(\arccos\!\left(-\frac{1}{\sqrt{2}}\right)=y\) is the same as \(-\frac{1}{2}=\cos(y)\) where \(y\in[0,\pi]\). Based on the unit circle we get \(y=\tfrac{2\pi}{3}\), so \(\arccos\!\left(-\frac{1}{2}\right)=\tfrac{2\pi}{3}\).

5. \(\arccos\!\left(0\right)=y\) is the same as \(0=\cos(y)\) where \(y\in[0,\pi]\). Based on the unit circle we get \(y=\tfrac{\pi}{2}\), so \(\arccos\!\left(0\right)=\tfrac{\pi}{2}\).

6. \(\arcsin\!\left(\frac{1}{2}\right)=y\) is the same as \(\frac{1}{2}=\sin(y)\) where \(y\in[-\tfrac{\pi}{2},\frac{\pi}{2}]\). Based on the unit circle we get \(y=\tfrac{\pi}{6}\), so \(\arcsin\!\left(\frac{1}{2}\right)=\tfrac{\pi}{6}\).

7. \(\arcsin\!\left(\frac{1}{\sqrt{2}}\right)=y\) is the same as \(\frac{1}{\sqrt{2}}=\sin(y)\) where \(y\in[-\tfrac{\pi}{2},\frac{\pi}{2}]\). Based on the unit circle we get \(y=\tfrac{\pi}{4}\), so \(\arcsin\!\left(\frac{1}{\sqrt{2}}\right)=\tfrac{\pi}{4}\).

8. \(\arctan\!\left(\sqrt{3}\right)=y\) is the same as \(\sqrt{3}=\tan(y)\) where \(y\in(-\tfrac{\pi}{2},\frac{\pi}{2})\). Based on the unit circle we get \(y=\tfrac{\pi}{3}\), so \(\arctan\!\left(\sqrt{3}\right)=\tfrac{\pi}{3}\).

9. \(\arctan\!\left(1\right)=y\) is the same as \(1=\tan(y)\) where \(y\in(-\tfrac{\pi}{2},\frac{\pi}{2})\). Based on the unit circle we get \(y=\tfrac{\pi}{4}\), so \(\arctan\!\left(1\right)=\tfrac{\pi}{4}\).

Remember that sine and arcsine are inverses, cosine and arccosine are inverses, and tangent and arctangent are inverses. However, the values must make sense within the respective domain.

Example 17

1. Calculate \(\arcsin\left(\sin\left(\frac{\pi}{11}\right)\right)\)
2. Calculate \(\arctan\left(\tan\left(-\frac{\pi}{11}\right)\right)\)
3. Calculate \(\arccos\left(\cos\left(\frac{8\pi}{9}\right)\right)\)
4. Calculate \(\arcsin\left(\sin\left(\frac{4\pi}{5}\right)\right)\).
5. Calculate \(\arctan\left(\tan\left(\frac{8\pi}{7}\right)\right)\).
6. Calculate \(\arcsin\left(\cos\left(\frac{8\pi}{7}\right)\right)\)
7. Calculate \(\cos\left(\arcsin\left(2a-1\right)\right)\). For what values of \(a\) will this makes sense?

1. Since sine and arcsine are inverses, and \(\tfrac{\pi}{11}\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), then \(\arcsin\left(\sin\left(\frac{\pi}{11}\right)\right)=\frac{\pi}{11}\).

2. Since tangent and arctangent are inverses, and \(-\tfrac{\pi}{11}\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\), then \(\arctan\left(\tan\left(-\frac{\pi}{11}\right)\right)=-\frac{\pi}{11}\)

3. Since cosine and arccosine are inverses, and \(\tfrac{8\pi}{9}\in[0,\pi)\), then \(\arccos\left(\cos\left(\frac{8\pi}{9}\right)\right)=\frac{8\pi}{9}\)

4. The function sine and arcsine are inverses, but \(\tfrac{4\pi}{5}\not\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\). So we use symmetry: \(\sin(\tfrac{4\pi}{5})=\sin(\tfrac{\pi}{5})\) and \(\tfrac{\pi}{5}\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) so we have \(\arcsin\left(\sin\left(\frac{4\pi}{5}\right)\right)=\frac{\pi}{5}\).

5. The function tangent and arctangent are inverses, but \(\tfrac{8\pi}{7}\not\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\). So we use symmetry: \(\tan(\tfrac{8\pi}{7})=\tan(\tfrac{8\pi}{7})\) and \(\tfrac{\pi}{5}\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\) so we have \(\arctan\left(\tan\left(\frac{8\pi}{7}\right)\right)=\frac{\pi}{7}\).

6. Notice that \(\cos(\tfrac{8\pi}{7})=\sin(\tfrac{8\pi}{7}+\frac{\pi}{2})=\sin(\tfrac{23\pi}{14})\). So \(\arcsin\left(\cos\left(\frac{8\pi}{7}\right)\right)=\arcsin\left(\sin\left(\frac{23\pi}{14}\right)\right)\). However, \(\tfrac{23\pi}{14}\not\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), so we need to use symmetry: \(\sin(\tfrac{23\pi}{14})=\sin(-\tfrac{5\pi}{14})\). Thus \(\arcsin\left(\cos\left(\frac{8\pi}{7}\right)\right)=\arcsin\left(\sin(-\tfrac{5\pi}{14})\right)=-\tfrac{5\pi}{14}\)

7. The function \(\arcsin(\theta)\) has a domain equal to the range of \(\sin(\theta))\), which is \([-1,1]\). So \(\cos\left(\arcsin\left(2a-1\right)\right)\) only makes sense as long as \(2a-1\) is in \([-1,1]\), meaning \(-1\leq 2a-1\leq 1\) or \(0\leq a\leq 1.\) Since \(\cos^2(\theta)+\sin^2(\theta)=1\), we have that \(\cos(\theta)=\pm \sqrt{1-\sin^2(\theta)}.\) Thus \[\cos\left(\arcsin\left(2a-1\right)\right)=\pm\sqrt{1-\sin^2(\arcsin(2a-1))}=\pm\sqrt{1-2a+1}=\pm\sqrt{-2a+2}\]

When involving two trigonometric functions that are not inverses of each other, the key to the calculation is using the inverse relationship and the standard angles on the unit circle.

Example 18

1. Calculate \(\tan\left(\arcsin\left(\frac{3}{11}\right)\right)\)
2. Calculate \(\sin\left(\arctan\left(\frac{5}{7}\right)\right)\)

1. The expression \(\arcsin\left(\frac{3}{11}\right)\) is equal to some angle \(\theta\) in quadrant I so that its corresponding point on the unit circle is \[(\cos(\theta),\sin(\theta))=(\cos(\theta),\tfrac{3}{11}).\] Use the fact that \[\cos^2(\theta)+\left(\frac{3}{11}\right)^2=1\] and that \(\theta\) is in quadrant I to get that \[\cos(\theta)=\frac{\sqrt{112}}{11}.\] Therefore, \[\begin{align*} \tan\left(\arcsin\left(\frac{3}{11}\right)\right)&=\tan(\theta)\\ &=\frac{\sin(\theta)}{\cos(\theta)}\\ &=\frac{\frac{3}{11}}{\frac{\sqrt{118}}{11}}\\ &=\frac{3}{\sqrt{112}}. \end{align*}\]

2. The expression \(\arctan\left(\frac{5}{7}\right)\) is equal to some angle \(\theta\) in quadrant I so that its corresponding point on the unit circle is \[\tan(\theta)=\frac{5}{7}\] Use the fact that \[\begin{align*}\tan^2(\theta)+1&=\sec^2(\theta)\\ \left(\frac{5}{7}\right)^2+1&=\sec^2(\theta)\\ \end{align*}\] and that \(\theta\) is in quadrant I to get that \[\sec(\theta)=\frac{\sqrt{74}}{7},\] which implies that \[\cos(\theta)=\frac{7}{\sqrt{74}}.\] Now use \[\begin{align*}\sin^2(\theta)+\cos^2(\theta)&=1\\ \sin^2(\theta)+\left(\frac{7}{\sqrt{74}}\right)^2&=1\\ \end{align*}\] to obtain that \[\sin(\theta)=\frac{5}{\sqrt{74}}.\] Therefore, \[\begin{align*} \sin\left(\arctan\left(\frac{5}{7}\right)\right)&=\sin(\theta)\\ &=\frac{5}{\sqrt{74}}. \end{align*}\]

Equations Involving Trigonometric Functions

Very few equations involving trigonometric functions can be exactly solved.

Most require approximation methods.

Sometimes, such equations can be solved exactly and such problems can arise naturally.

Example 19

Find all values of \(\theta\) that satisfy the equality \[\tan^2(\theta) = 1.\]

Taking square-root means \[\tan(\theta)=\pm 1\], which happens when \(x=\tfrac{\pi}{4}+k\pi\), \(x=\tfrac{3\pi}{4}+k\pi\) where \(k\) is an integer.

In this example, pay attention to how we rewrite the trigonometric equation into a polynomial equation.

Example 20

Find all values of \(\theta\) that satisfy the equality \[3\sin^2(5\theta) - 4\sin(5\theta) + 1 = 0.\]

Notice that if \(u=\sin(\theta)\), then \(3u^2+4u+1=0\). So we solve the quadratic equation to get \(u=-1\) or \(u=-\frac{1}{3}\). Which means we solve \(\sin(5\theta)=-1\) or \(\sin(5\theta)=-\tfrac{1}{3}\). The solution set is therefore, \(x=\frac{3\pi}{10}+\frac{2\pi}{5} k\) and \(x=\frac{\arcsin(-\tfrac{1}{3})}{5}+\frac{2\pi}{5} k\) where \(k\) is an integer.

Practice this idea of rewriting with the next example. Also, pay attention to the range of functions we are studying!

Example 21

Find all values of \(\theta\) that satisfy the equality \[\sin^2(\theta) - 5\sin(\theta) + 6 = 0.\]

Notice that if \(u=\sin(\theta)\), then \(u^2+5u+6=0\). So we solve the quadratic equation to get \(u=-3\) or \(u=-2\). Which means we solve \(\sin(\theta)=-3\) or \(\sin(\theta)=-2\). There is no solution set for this problem.

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© Copyright 2024 by the POC Writing Team: Bryan Carrillo, Yat Sun Poon, and David Weisbart. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the POC Writing Team.