Graph of Trig Functions

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

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

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 $x$ from 0 to $\pi$ and corresponding values of $y$.

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

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

Figure 1.1:

Example 1..8   Solve the following inequality.

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

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 $(x,y)$ such that $y \geq \frac{1}{2}$. Thus,

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

Exercise 1..8   Solve the following inequality.

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

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 $(x,y)$ 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 $(x,y)$ such that $y \leq \frac{\sqrt{2}}{2}$. See the figure. Putting these toghether, we have

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

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 $y$-axis. Furthermore, $f$ satisfies $f(x + 2\pi) = \cos(x+2\pi) = \cos{x} = f(x)$. Thus, $f(x)$ has the period $2\pi$.

Thus to draw the graph of $y = \cos{x}$, it is enough to check the values of $x$ from 0 to $\pi$. From these information, we can draw the graph of $y = \cos{x}$ by checking the values of $x$ from 0 to $\pi$ and corresponding values of $y$.

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

Now using symmetry of the function, we obtain the figure1.2

Figure 1.2:

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

Example 1..9   Solve the following inequality.

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

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 $(x,y)$ so that the value of $x > \frac{1}{\sqrt{2}}$. From the figure, we have

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

Exercise 1..9   Solve the following inequality.

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

SOLUTION The region satisfying $0 \leq x \leq 2\pi$ and $\cos{x} > \frac{\sqrt{3}}{2}$ is the set of points $(x,y)$ such that the value of $x > \frac{\sqrt{3}}{2}$.

From this figure, we have

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

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:

$$f(x + \pi) = \tan{(x + \pi)} = \frac{\sin{(x + \pi)}}{\cos{(x + \pi)}} = \frac{-\sin{x}}{-\cos{x}} = \tan{x}$$

Thus, $f(x)$ 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 $x$ such that $0 < x < \\ frac{\pi}{2}$ and corresponding $y$. We note that the function $f$ is not defined at $x = \frac{\pi}{2}$. To overcome this problem, we use the limit: $\lim_{x \to \frac{\pi}{2}-}\tan{x} = \infty$.

$$\begin{array}{c\vert c\vert c\vert c\vert c\vert c} x & 0 & \... ...qrt{3}}{3} & \frac{\sqrt{2}}{2} & \sqrt{3} & \infty \end{array}$$

By using the symmetry, we obtain the figure1.3.

Figure 1.3:

Example 1..10   Solve the following inequality.

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

.

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

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

Exercise 1..10   Solve the following inequality.

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

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

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

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 $x$-axis?Dprovided $0 \leq \theta < \pi$?D

(a) $${y = \sqrt{3}x}$$ (b) $${y = \frac{1}{\sqrt{3}}x}$$

3.

Using the picture , Solve the following inequalities, provide $0 \leq \theta \leq 2\pi$?D

(a) $${\cos{\theta} > \frac{\sqrt{3}}{2}}$$ (b) $${\sqrt{3}\tan{\theta} > 1}$$ (c) $${1 < 2\sin{\theta} \leq \sqrt{2}}$$

4.

Evaluate the followings?D

(a) $${\tan^{-1}{0}}$$ (b) $${\sin^{-1}{\frac{1}{2}}}$$ (c) $${\sin{\left(\cos^{-1}{\frac{1}{2}}\right)}}$$

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) $${\cos^{2}{\frac{\alpha}{2}} = \frac{1 + \cos{\alpha}}{2}}$$

(d) $${\sin{\alpha}\cos{\beta} = \frac{1}{2}\left(\sin{(\alpha + \beta)} + \sin{(\alpha - \beta)}\right)}$$

(e) $${\sin{\alpha} + \sin{\beta} = 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}}}$$

2.

Find the value of the followings:

(a) $${\cos{\frac{5\pi}{4}}}$$ (b) $${\sin{\frac{7\pi}{12}}}$$ (c) $${\cos{\frac{\pi}{8}}}$$

3.

Find the value of the followings?D

(a) $${\sin^{-1}{(\frac{-1}{2})}}$$ (b) $${\cos^{-1}{(-1)}}$$ (c) $${\tan^{-1}{(-1)}}$$ (d) $${\tan^{-1}{\sqrt{3}}}$$

4.

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

5.

Derive the following formulas?D

(a) $${\sin^{-1}{(-x)} = - \sin^{-1}{x}}$$ (b) $${\cos^{-1}{(-x)} = \pi - \cos^{-1}{x}}$$