Graph of Trig Functions

We have been using $\theta$ to express the angle. But from now on, we use as independent variable and as dependent variable. In other words, we write

$\displaystyle y = \sin{x}, \ y = \cos{x}, \ y = \tan{x}$

We study the graphs of these functions. First one is the graph of $f(x) = \sin{x}$ . Since $f(-x) = \sin(-x) = -\sin{x} = -f(x)$ , $f(x) = \sin{x}$ is an odd function and symmetric with respect the origin. Also, the function satisfies $f(x + 2\pi) = \sin(x + 2\pi) = \sin{x}$ . Thus $f(x) = \sin{x}$ has the period $2\pi$ . ^1.4 From these information, we can draw the graph of $y = \sin{x}$ by checking the values of from 0 to $\pi$ and corresponding values of .

$\begin{displaymath}\begin{array}{c\vert c\vert c\vert c\vert c\vert c\vert c\ver... ...\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \end{array}\end{displaymath}$

Using the fact that it is symmetric with respect to the origin, we obtain the figure 1.1.

**Figure 1.1:**
$\includegraphics[width=7cm]{SOFTFIG-1/Fig1-2-0-1_gr2.eps}$

Example 1..8 Solve the following inequality.

$\displaystyle 2\sin{x} \geq 1,\ 0 \leq x \leq 2\pi$

SOLUTION Since $2\sin{x} \geq 1$ , we have $\sin{x} \geq \frac{1}{2}$ . The region satisfying $0 \leq x \leq 2\pi$ and $\sin{x} \geq \frac{1}{2}$ is the set of points such that $y \geq \frac{1}{2}$ . Thus,

$\displaystyle \frac{\pi}{6} \leq x \leq \frac{5\pi}{6}\ensuremath{\ \blacksquare}$

Exercise 1..8 Solve the following inequality.

$\displaystyle 1 < 2\sin{x} \leq \sqrt{2},\ 0 \leq x \leq 2\pi$

SOLUTION We separate $1 < 2\sin{x} \leq \sqrt{2}$ into two inequalities such as $1 < 2\sin{x}$ and $\sin{x} > \frac{1}{2}$ . The region satisfying $0 \leq x \leq 2\pi$ and $\sin{x} > \frac{1}{2}$ is the set of points such that $y \geq \frac{1}{2}$ . Now $2\sin{x} \leq \sqrt{2}$ implies that $\sin{x} \leq \frac{\sqrt{2}}{2}$ . Thus the region satisfying $0 \leq x \leq 2\pi$ and $\sin{x} \leq \frac{\sqrt{2}}{2}$ is the set of all points such that $y \leq \frac{\sqrt{2}}{2}$ . See the figure. Putting these toghether, we have

$\displaystyle \frac{\pi}{6} < x \leq \frac{\pi}{4}, \frac{3\pi}{4} \leq x < \frac{5\pi}{6}\ \ensuremath{\ \blacksquare}$

Next we probe the graph of the function $f(x) = \cos{x}$ . Since $f(-x) = \cos(-x) = \cos{x} = f(x)$ , $f(x) = \cos{x}$ is an even function and symmetric with respect to the -axis. Furthermore, satisfies $f(x + 2\pi) = \cos(x+2\pi) = \cos{x} = f(x)$ . Thus, has the period $2\pi$ .
Thus to draw the graph of $y = \cos{x}$ , it is enough to check the values of from 0 to $\pi$ . From these information, we can draw the graph of $y = \cos{x}$ by checking the values of from 0 to $\pi$ and corresponding values of .

$\begin{displaymath}\begin{array}{c\vert c\vert c\vert c\vert c\vert c\vert c\ver... ...2} & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{3}}{2} & -1 \end{array}\end{displaymath}$
Now using symmetry of the function, we obtain the figure1.2

Figure 1.2:

$\includegraphics[width=7cm]{SOFTFIG-1/Fig1-2-0-1_gr3.eps}$

Now note that the functions $y = \sin{x}$ and $y = \cos{x}$ satisfy the relation $\sin(x+\frac{\pi}{2}) = \cos{x}$ .

$\includegraphics[width=7cm]{SOFTFIG-1/Fig1-2-0-1_gr1.eps}$

Example 1..9 Solve the following inequality.

$\displaystyle \sqrt{2}\cos{x} > 1,\ 0 \leq x \leq 2\pi$

SOLUTION Since $\sqrt{2}\cos{x} > 1$ , we have $\cos{x} > \frac{1}{\sqrt{2}}$ . The region satisfying $0 \leq x \leq 2\pi$ and $\cos{x} > \frac{1}{\sqrt{2}}$ is the set of points so that the value of $x > \frac{1}{\sqrt{2}}$ . From the figure, we have

$\displaystyle 0 \leq x < \frac{\pi}{4}, \frac{7\pi}{4} < x \leq 2\pi\ \ensuremath{\ \blacksquare}$

Exercise 1..9 Solve the following inequality.

$\displaystyle \cos{x} > \frac{\sqrt{3}}{2},\ 0 \leq x \leq 2\pi$

SOLUTION The region satisfying $0 \leq x \leq 2\pi$ and $\cos{x} > \frac{\sqrt{3}}{2}$ is the set of points such that the value of $x > \frac{\sqrt{3}}{2}$ .
From this figure, we have

$\displaystyle 0 \leq x < \frac{\pi}{6}, \frac{7\pi}{6} < x \leq 2\pi\ensuremath{\ \blacksquare}$

At the end we investigate the graph of function $f(x) = \tan{x}$ . Since $f(-x) = \tan(-x) = \frac{\sin{(-x)}}{\cos{(-x)}} = \frac{-\sin{x}}{\cos{x}} = -\tan{x}$ , $f(x) = \tan{x}$ must be symmetric with respect to the origin. Furthermore, the function satisfies the following:

$\displaystyle f(x + \pi) = \tan{(x + \pi)} = \frac{\sin{(x + \pi)}}{\cos{(x + \pi)}} = \frac{-\sin{x}}{-\cos{x}} = \tan{x}$
Thus, is a periodic function with the period $\pi$ . This means that to draw the graph of the function $y = \tan{x}$ , it is enough to check the values of such that $0 < x < \\ frac{\pi}{2}$ and corresponding . We note that the function is not defined at $x = \frac{\pi}{2}$ . To overcome this problem, we use the limit: $\lim_{x \to \frac{\pi}{2}-}\tan{x} = \infty$ .

$\begin{displaymath}\begin{array}{c\vert c\vert c\vert c\vert c\vert c} x & 0 & \... ...qrt{3}}{3} & \frac{\sqrt{2}}{2} & \sqrt{3} & \infty \end{array}\end{displaymath}$
By using the symmetry, we obtain the figure1.3.

Figure 1.3:

$\includegraphics[width=7cm]{SOFTFIG-1/Fig1-2-0-1_gr4.eps}$

Example 1..10 Solve the following inequality.

$\displaystyle \tan{x} > 1,\ 0 \leq x \leq 2\pi$
.

SOLUTION The region satisfying $\tan{x} > 1$ is the set of points such that either and or and . From the figure, we have

$\displaystyle \frac{\pi}{4} < x < \frac{\pi}{2}, \frac{5\pi}{4} < x < \frac{3\pi}{2}\ensuremath{\ \blacksquare}$

Exercise 1..10 Solve the following inequality.

$\displaystyle \sqrt{3}\tan{x} > 1,\ 0 \leq x \leq 2\pi$

SOLUTION The region satisfying $\tan{x} > \frac{1}{\sqrt{3}}$ is the set of points such that either and $y > \sqrt{3}/3$ or and $y < -\sqrt{3}/3$ . Thus, we have

$\displaystyle \frac{\pi}{6} < x < \frac{\pi}{2}, \frac{7\pi}{6} < x < \frac{3\pi}{2}\ensuremath{\ \blacksquare}$

Exercise A

1.

Express the following angle by the radian?D
(a) $30^{\circ}$ (b) $40^{\circ}$ (c) $72^{\circ}$

2.

Find the angle between the following line and -axis?Dprovided $0 \leq \theta < \pi$ ?D
(a) $\displaystyle{y = \sqrt{3}x}$ (b) $\displaystyle{y = \frac{1}{\sqrt{3}}x}$

3.

Using the picture , Solve the following inequalities, provide $0 \leq \theta \leq 2\pi$ ?D
(a) $\displaystyle{\cos{\theta} > \frac{\sqrt{3}}{2}}$ (b) $\displaystyle{\sqrt{3}\tan{\theta} > 1}$ (c) $\displaystyle{1 < 2\sin{\theta} \leq \sqrt{2}}$

4.

Evaluate the followings?D
(a) $\displaystyle{\tan^{-1}{0}}$ (b) $\displaystyle{\sin^{-1}{\frac{1}{2}}}$ (c) $\displaystyle{\sin{\left(\cos^{-1}{\frac{1}{2}}\right)}}$

Exercise B

1.

For all angle $\alpha,\beta$ ?Cshow the following identities hold?D
(a) $\sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$
(b) $\cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$
(c) $\displaystyle{\cos^{2}{\frac{\alpha}{2}} = \frac{1 + \cos{\alpha}}{2}}$
(d) $\displaystyle{\sin{\alpha}\cos{\beta} = \frac{1}{2}\left(\sin{(\alpha + \beta)} + \sin{(\alpha - \beta)}\right)}$
(e) $\displaystyle{\sin{\alpha} + \sin{\beta} = 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}}}$

2.

Find the value of the followings:
(a) $\displaystyle{\cos{\frac{5\pi}{4}}}$ (b) $\displaystyle{\sin{\frac{7\pi}{12}}}$ (c) $\displaystyle{\cos{\frac{\pi}{8}}}$

3.

Find the value of the followings?D
(a) $\displaystyle{\sin^{-1}{(\frac{-1}{2})}}$ (b) $\displaystyle{\cos^{-1}{(-1)}}$ (c) $\displaystyle{\tan^{-1}{(-1)}}$ (d) $\displaystyle{\tan^{-1}{\sqrt{3}}}$

4.

Show $\displaystyle{\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}}$ is true for all ?D

5.

Derive the following formulas?D
(a) $\displaystyle{\sin^{-1}{(-x)} = - \sin^{-1}{x}}$ (b) $\displaystyle{\cos^{-1}{(-x)} = \pi - \cos^{-1}{x}}$

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