7.1: Radian Measure.

Calculus with trigonometric functions is easier when angles are measured in radians. Radian measure of an angle and the trigonometric measures of that angle are all scaled by the length of the radius of a defining circle. For use in calculus you should put your calculator in RADIAN mode.

The circle in Figure \(\PageIndex{1}\) has radius 1. For the angle \(z^{\prime}\) (the angle \(\angle A O C\)) the radian measure, the sine, and the cosine are all dimensionless quantities:

\[\begin{aligned}
\text { Radian measure of } z^{\prime} &=\frac{\text { length of arc } \overbrace{A C}}{\text { length of the radius }}=\frac{z \quad \mathrm{~cm}, \mathrm{~m}, \text { in, }}{1 \quad \mathrm{~cm}, \mathrm{~m}, \text { in, }}=z \\
\text { sine of } z^{\prime} &=\frac{\text { length of the segment } \overline{A B}}{\text { length of the radius }}=\frac{y \quad \mathrm{~cm}, \mathrm{~m}, \text { in, } ?}{1 \mathrm{~cm}, \mathrm{~m}, \text { in, } ?}=y \\
\text { cosine of } z^{\prime}=& \frac{\text { length of the segment } \overline{O B}}{\text { length of the radius }}=\frac{x \quad \mathrm{~cm}, \mathrm{~m}, \text { in, } ?}{1 \mathrm{~cm}, \mathrm{~m}, \text { in, } ?}=x
\end{aligned}\]

Figure \(\PageIndex{1}\): A circle with radius 1. The radian measure of the angle \(z^{\prime}\) is \(z\), the length of the arc \(AC\) divided by the length of the radius, 1.

It is obvious¹ from the figure that for \(z^{\prime}\) in the first quadrant

\[\begin{aligned}
0< \text{ Length of } \overline{A B}< \text{ Length of } \overbrace{A C}\\
0< \sin z^{\prime} < z
\end{aligned}\]

The inequality, \(\sin z ^{\prime} < z\), is read, 'the sine of angle \(z ^{\prime}\) is less than \(z\), the radian measure of \(z ^{\prime}\).' We intentionally blur the distinction between the angle \(z ^{\prime}\) and \(z\), the radian measure of \(z ^{\prime}\), to the point that they are used interchangeably. The inequality

\[\sin z^{\prime}<z \quad \text { is usually replaced with } \quad \sin z<z \text {, } \label{7.1}\]

the sine of \(z\) is less than \(z\), where \(z\) is a positive number.

The definitions of \(\sin z\) and \(\cos z\) for angles that are not accute are extended by use of the unit circle, the circle with center at (0,0) and of radius 1. For \(z\) positive, consider the arc of length \(z\) counterclockwise along the unit circle from (1,0) to a point, \((x, y)\), in Figure \(\PageIndex{2}\)A. For \(z\) negative consider the arc of length \(|z|\) clockwise along the unit circle from (1,0) to a point, \((x, y)\), in Figure \(\PageIndex{2}\)B. In either case

\[\sin z=\frac{y}{1}=y \quad \cos z=\frac{x}{1}=x \label{7.2}\]

Figure \(\PageIndex{2}\): The unit circle with A. an arc of length \(z\) between 0 and \(2 \pi\) and B. an arc of length \(|z|\) for \(z\) a negative number.

From Figure \(\PageIndex{2}\)B it can be seen that if \(z\) is a negative number then

\[z<\sin z<0 \label{7.3}\]

A single statement that combines Equations \ref{7.1} and \ref{7.3} is written:

\[|\sin z|<|z| \quad \text { for all numbers } z \label{7.4}\]

We need this inequality for computing \([ \sin t] ^{\prime}\) in the next section, and we also need the inequality

\[|z|<|\tan z| \quad \text { for } \quad \frac{\pi}{2}<z<\frac{\pi}{2} \label{7.5}\]

To see this, for \(z > 0\), examine the circle with radius 1 in Figure \(\PageIndex{3}\),

\[z=\overbrace{\frac{A C}{1}}=\overbrace{A C}, \quad \text { and } \quad \tan z=\frac{\overline{C D}}{\overline{O C}}=\overline{C D} .\]

We need to show that \(\overbrace{A C}<\overline{C D}\) which appears reasonable from the figure, but we present a proof.

Proof. The area of the sector of the circle \(OAC\) is equal to the area of the whole circle times the ratio of the length of the arc \(\overbrace{A C}\) to the circumference of the whole circle. Thus

\[\text { Area of sector } O A C=\pi 1^{2} \times \frac{\overbrace{A C}}{2 \pi \times 1}=\frac{\overbrace{A C}}{2}\]

The area of the triangle \(\triangle O C D\) is

\[\text { Area } \triangle O C D=\frac{1}{2} \times 1 \times \overline{C D}=\frac{\overline{C D}}{2}\]

The sector \(OAC\) is contained in the triangle \(\triangle O C D\) so that the area of sector \(OAC\) is less than the area of triangle \(\triangle O C D\). Therefore

\[\frac{\overbrace{A C}}{2} < \frac{ \overline{C D}}{2}, \quad \overbrace{A C}<\overline{C D}, \quad \text { and } \quad z<\tan z .\]

Figure \(\PageIndex{3}\): The unit circle and an angle \(z\), \(0 < z < \pi /2\).

An argument for Equation \ref{7.5} for \(z < 0\) can be based on the reflection of Figure \(\PageIndex{3}\) about the interval \(OC\). End of Proof.

In addition to the basic trigonometric identities, (\(\sin ^{2} t+\cos ^{2} t=1, \tan t=\sin t / \cos t\), etc.) the double angle trigonometric formulas are critical to this chapter:

\[\sin (A+B)=\sin A \cos B+\cos A \sin B \quad \cos (A+B)=\cos A \cos B-\sin A \sin B \label{7.6}\]

From these you are asked to prove in Exercises 7.1.3

\[\begin{aligned}
\sin x-\sin y &=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) \\
\cos x-\cos y &=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)
\end{aligned}\]

 

Exercises for Section 7.1 Radian Measure.

Exercise 7.1.1 For small positive values of \(z\), \(\sin z < z\) (Equation \ref{7.4}), but 'just barely so'.

1. Compute \[z-\sin z \quad \text { for } \quad z=0.1 \quad \text { for } \quad z=0.01 \quad \text { and for } \quad z=0.001\]
2. Compute \[\frac{\sin z}{z} \quad \text { for } \quad z=0.1 \quad \text { for } \quad z=0.01 \quad \text { and for } \quad z=0.001 \text {. }\]
3. Note that the slope of the tangent to \(y = \sin t\) at (0,0) is \[[\sin t]_{t=0}^{\prime}=\lim _{h \rightarrow 0} \frac{\sin (0+h)-\sin 0}{h}=\lim _{h \rightarrow 0} \frac{\sin h}{h}.\] What is your best estimate of \[[\sin t]_{t=0}^{\prime} ?\]

Exercise 7.1.2 For small positive values of \(z\), \(z < \tan z\) (Equation \ref{7.5}), but 'just barely so'.

1. Compute \[\tan z-z \quad \text { for } \quad z=0.01 \quad \text { for } \quad z=0.001 \quad \text { and for } \quad z=0.0001\]
2. Compute \[\frac{\tan z}{z} \text { for } z=0.01 \quad \text { for } \quad z=0.001 \quad \text { and for } \quad z=0.0001\]
3. Note that the slope of the tangent to \(y = \tan t\) at (0,0) is \[[\tan t]_{t=0}^{\prime}=\lim _{h \rightarrow 0} \frac{\tan (0+h)-\tan 0}{h}=\lim _{h \rightarrow 0} \frac{\tan h}{h}\] What is your best estimate of \[[\tan t]_{t=0}^{\prime} ?\]

Exercise 7.1.3 We need the identities

\[\begin{aligned}
\sin x-\sin y &=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \\
\cos x-\cos y &=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}
\end{aligned}\]

in the next two exercises and in the next section. It is unlikely that you remember them from a trigonometry class. We hope you do remember, however, the double angle formulas \ref{7.6},

\[\sin (A+B)=\sin A \cos B+\cos A \sin B \quad \cos (A+B)=\cos A \cos B-\sin A \sin B\]

1. Use \(\sin (A+B)=\sin A \cos B+\cos A \sin B\) and the identities, \(\sin (-A)=-\sin A\) and \(\cos (-A)=\cos A\), to show that \[\sin (A-B)=\sin A \cos B-\cos A \sin B\]
2. Use the equations \[\begin{aligned}
   &\sin (A+B)=\sin A \cos B+\cos A \sin B \\
   &\sin (A-B)=\sin A \cos B-\cos A \sin B
   \end{aligned}\] to show that \[\sin (A+B)-\sin (A-B)=2 \cos A \sin B \label{7.7}\]
3. Solve for \(A\) and \(B\) in \[\begin{aligned}
   &A+B=x \\
   &A-B=y
   \end{aligned}\]
4. Substitute the values for \(A + B, A − B, A\), and \(B\) into Equation \ref{7.7} to obtain \[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\].
5. Use \(\cos (A+B)=\cos A \cos B-\sin A \sin B\) to show that \[\cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\] The argument will be similar to the previous steps.

Exercise 7.1.4 Use steps (i) − (iv) below to show that at all numbers \(z\),

\[\lim _{h \rightarrow 0} \sin (z+h)=\sin z \label{7.8}\]

and therefore conclude that the sine function is continuous.

1. Write the trigonometric identity, \[\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\] with \(x = z + h\) and \(y = z\).
2. Justify the inequality in the following statement. \[|\sin (z+h)-\sin z|=2\left|\cos \left(\frac{2 z+h}{2}\right)\right|\left|\sin \left(\frac{h}{2}\right)\right|<2 \times 1 \times \frac{|h|}{2}=|h|\]
3. Suppose \(\epsilon\) is a positive number. Find a positive number \(\delta\) so that \[\text { if } \quad|(z+h)-z|=|h|<\delta \quad \text { then } \quad|\sin (z+h)-\sin z|<\epsilon\]
4. Is the previous step useful?

Exercise 7.1.5 Use the Inequality \ref{7.4} \(|\sin z|<|z|\) and the trigonometric identity,

\[\cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\]

to argue that

\[\lim _{h \rightarrow 0} \cos (z+h)=\cos z\]

and therefore conclude that the cosine function is continuous.

Hint: Look at the steps (i) - (iv) of Exercise 7.1.4.

 

 

¹ Should this not be obvious then reflect the figure about the horizontal line through \(O, B\), and \(C\) and let \(A ^{\prime}\) be the image of \(A\) under the reflection. The length of the chord \(A ^{\prime} BA\) is less than the length of the arc, \(A ^{\prime} CA\) (the straight line path is the shortest path between two points).