Cosine and Sine

Section 4.2 Cosine and Sine

After exponential quantities the circular functions, sine and cosine, should be considered because they arise when imaginary quantities are involved in the exponential. — Leonhard Euler, Swiss mathematician, physicist, astronomer, logician and engineer, 1707 – 1783

Reminder: We say that the angle that is associated with the arc of the length 1 measures 1 radian. See Figure 4.26.

Figure 4.26. The angle that measures *1 radian*.

Reminder.

Figure 4.27. The unit circle, angles, and points.

\begin{equation*} \begin{array}{c|c|} x\amp \text{endpoint of the radius}\\ \text{in radians}\amp\ \text{corresponding to }x \\ \hline \frac{\pi}{6}\amp \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\\ \hline \frac{\pi}{4}\amp \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\\ \hline \frac{\pi}{3}\amp \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\\ \hline \frac{\pi}{2}\amp (0,1) \\ \hline \frac{2\pi}{3}\amp \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\\ \hline \frac{3\pi}{4}\amp \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\\ \hline \frac{5\pi}{6}\amp \left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\\ \hline \pi \amp (-1,0)\\ \hline \frac{3\pi}{2} \amp (0,-1)\\ \hline 2\pi \amp (1,0)\\ \end{array} \end{equation*}

Script 1:

\begin{equation*} \begin{array}{lll} \text{a real number } x \amp \to \amp \text{the unique corresponding point on the unit circle }X=(a,b)\\ \amp \to \amp x\mapsto a \text{ and } x\mapsto b\\ \amp \to \amp \text{ Call }a=\cos (x) \text{ and call }b=\sin (x)\\ \end{array} \end{equation*}

Script 2: See Figure 4.28.

Figure 4.28. From \(x\) to \(\cos x\) and \(\sin x\text{.}\)

Definition: The cosine of \(x\text{,}\) denoted \(\cos x\text{,}\) is the first coordinate of the endpoint of the radius of the unit circle corresponding to \(x\text{.}\)

Definition: The sine of \(x\text{,}\) denoted \(\sin x\text{,}\) is the second coordinate of the endpoint of the radius of the unit circle corresponding to \(x\text{.}\)

Example 4.2.1. Cosines and sines.

Complete the table:

Figure 4.29. The unit circle, angles, and points.

\begin{equation*} \begin{array}{c|c|c|c} x\amp \text{endpoint of the radius}\amp \cos x\amp \sin x\\ \text{in radians}\amp\ \text{corresponding to }x \amp \amp\\ \hline \frac{\pi}{6}\amp \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) \amp \amp\\ \hline \frac{\pi}{4}\amp \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\amp \amp \\ \hline \frac{\pi}{3}\amp \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\amp \amp \\ \hline \frac{\pi}{2}\amp (0,1) \amp \amp\\ \hline \frac{2\pi}{3}\amp \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\amp \amp\\ \hline \frac{3\pi}{4}\amp \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\amp \amp\\ \hline \frac{5\pi}{6}\amp \left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\amp \amp\\ \hline \pi \amp (-1,0)\amp \amp\\ \hline \frac{3\pi}{2} \amp (0,-1)\amp \amp\\ \hline 2\pi \amp (1,0)\amp \amp\\ \end{array} \end{equation*}

Solution

By definition: \(\cos x\) is the first coordinate of the corresponding point on the unit circle and \(\cos x\) is the first coordinate of the corresponding point on the unit circle.

\begin{equation*} \begin{array}{c|c|c|c} x\amp \text{endpoint of the radius}\amp \cos x\amp \sin x\\ \text{in radians}\amp\ \text{corresponding to }x \amp \amp\\ \hline \frac{\pi}{6}\amp \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) \amp \frac{\sqrt{3}}{2}\amp \frac{1}{2}\\ \hline \frac{\pi}{4}\amp \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\amp \frac{\sqrt{2}}{2} \amp \frac{\sqrt{2}}{2}\\ \hline \frac{\pi}{3}\amp \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\amp \frac{1}{2} \amp \frac{\sqrt{3}}{2} \\ \hline \frac{\pi}{2}\amp (0,1) \amp 0 \amp1 \\ \hline \frac{2\pi}{3}\amp \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\amp -\frac{1}{2} \amp \frac{\sqrt{3}}{2}\\ \hline \frac{3\pi}{4}\amp \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\amp -\frac{\sqrt{2}}{2} \amp \frac{\sqrt{2}}{2}\\ \hline \frac{5\pi}{6}\amp \left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\amp -\frac{\sqrt{3}}{2} \amp \frac{1}{2}\\ \hline \pi \amp (-1,0)\amp -1 \amp 0\\ \hline \frac{3\pi}{2} \amp (0,-1)\amp 0 \amp -1\\ \hline 2\pi \amp (1,0)\amp 1 \amp 0\\ \end{array} \end{equation*}

Example 4.2.2. More about cosines and sines.

Evaluate

\begin{equation*} \cos \left(-\frac{\pi}{4}\right) \text{ and } \cos \left(\frac{7\pi}{4}\right) \end{equation*}

Solution

See Figure 4.30 and Figure 4.31.

Figure 4.30. \(\cos \left(-\frac{\pi}{4}\right) =\frac{\sqrt{2}}{2}\text{.}\)

Figure 4.31. \(\cos \left(\frac{7\pi}{4}\right) =\frac{\sqrt{2}}{2}\text{.}\)

Five facts. Let \(x\) be a real number. Then:

\(-1\leq\cos x\leq 1\) and \(-1\leq\sin x\leq 1\text{.}\) See Figure 4.32.

Figure 4.32. \(-1\leq\cos x\leq 1\) and \(-1\leq\sin x\leq 1\text{.}\)
The function \(x\mapsto \cos x\) is even, i.e.

\begin{equation*} \cos(-x)=\cos x. \end{equation*}

See Figure 4.33.

Figure 4.33. Note that the points with the coordinates \((\cos x,\sin x)\) and \((\cos (-x),\sin (-x))\) are symmetric with respect to the \(x\)-axis. Thus \(\cos(-x)=\cos x\text{.}\)
The function \(x\mapsto \sin x\) is odd, i.e.

\begin{equation*} \sin(-x)=-\sin x. \end{equation*}

See Figure 4.34.

Figure 4.34. Note that the points with the coordinates \((\cos x,\sin x)\) and \((\cos (-x),\sin (-x))\) are symmetric with respect to the \(x\)-axis. Thus \(\sin(-x)=-\sin x\text{.}\)
\(\cos^2x+\sin^2x=1\text{.}\) See Figure 4.35.

Figure 4.35. Since the point \((\cos x, \sin x)\) lie on the unit circle, it follows that \(\cos^2x+\sin^2x=1\)>.
Functions \(x\mapsto \cos x\) and \(x\mapsto \sin x\) are \(periodic\text{,}\) i.e.

\begin{equation*} \cos (x+2\pi)=\cos x \text{ and } \sin (x+2\pi)=\sin x. \end{equation*}
Solution

Note that, for any \(x\in \mathbb{R}\text{,}\) the angles \(x\) and \(x+2\pi\) are associated with the same point on the unit circle. This implies

\begin{equation*} \cos (x+2\pi)=\cos x \text{ and } \sin (x+2\pi)=\sin x. \end{equation*}

More facts: Sign of \(\cos x\) and \(\sin x\) for \(0\leq x\lt 2\pi\text{:}\)

\begin{equation*} \begin{array}{c|c|c|c|c|c|c|c|c} \amp x=0 \amp 0\lt x\lt \frac{\pi}{2} \amp x=\frac{\pi}{2} \amp \frac{\pi}{2}\lt x\lt \pi \amp x=\pi \amp \pi\lt x\lt \frac{3\pi}{2} \amp x=\frac{3\pi}{2} \amp \frac{3\pi}{2}\lt x\lt 2\pi \\ \hline \cos x \amp 1\amp +\amp 0\amp -\amp -1\amp -\amp 0\amp +\\ \hline \sin x\amp 0\amp +\amp 1\amp +\amp 0\amp -\amp -1\amp -\\ \end{array} \end{equation*}

Example 4.2.3. Sign of cosine and sine.

Where is \(\cos \alpha \lt 0\text{?}\) Where is\(\sin \alpha \gt 0\text{?}\)

Solution

Suppose that \(\alpha\) is an angle with the vertex at the origin. Also, suppose that the positive ray of the \(x\)–axis is the initial ray of the angle \(\alpha\text{.}\) See Figure 4.30 and Figure 4.31.

Figure 4.36. If the terminal ray of the angle \(\alpha\) belongs to the \(2^{\text{nd}}\) or the \(3^{\text{rd}}\) quadrant then \(\cos \alpha\lt 0\text{.}\)

Figure 4.37. If the terminal ray of the angle \(\alpha\) belongs to the \(1^{\text{st}}\) or the \(2^{\text{nd}}\) quadrant then \(\sin \alpha \gt 0\) .

Two functions: See See Figure 4.38 and Figure 4.38.

Figure 4.38. The function \(f(x)=\cos x\text{.}\)

Figure 4.39. The function \(g(x)=\sin x\text{.}\)

All you need to know for now:

\begin{equation*} \begin{array}{c|c|c} \amp f(x)=\cos x \amp g(x)=\sin x\\ \hline \text{Domain} \amp \mathbb{R} \amp \mathbb{R}\\ \hline \text{Range} \amp [-1,1] \amp [-1,1]\\ \hline \text{Zeros} \amp \frac{\pi}{2}+k\pi,\ k\in \mathbb{Z} \amp k\pi, \ k\in \mathbb{Z}\\ \hline y=1 \amp x=2k\pi,\ k\in \mathbb{Z} \amp \frac{\pi}{2}+2k\pi, k\in \mathbb{Z}\\ \hline y=-1 \amp (2k+1)\pi, k\in \mathbb{Z} \amp -\frac{\pi}{2}+2k\pi,\ k\in \mathbb{Z}\\ \hline \text{Parity} \amp \text{Even} \amp \text{Odd}\\ \hline \text{Period} \amp T=2\pi \amp T=2\pi\\ \end{array} \end{equation*}