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

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

Symmetric wrt Origin Suppose the function satisfies the equation $f(-x) = -f(x)$. Then it is symmetric with respect to the origin.

Figure 1.22: odd function

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 $2\pi$. When the function satisfies $f(x + L) = f(x)$, we say $f(x)$ has the period $L$. 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 1.23.

Graph of symmetric fct To draw the graph of symmetric function with respect to the origin, it is enough to check the values of $x \geq 0$.

Figure 1.23: SINE

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 $y \geq \frac{1}{2}$.

Figure 1.24: Example1-8

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.

Figure 1.25: Exercise1-8

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 figure1.27

symmetry wrt $y$ A function $f$ is symmetric with respect to the $y$-axis if and only if $f(-x) = -f(x)$.

Figure 1.26: even function

Figure 1.27: COSINE

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

Figure 1.28:

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 $x > \frac{1}{\sqrt{2}}$.

Figure 1.29: Example1-9

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.

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

Figure 1.30: Exercise1-9

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: $\lim_{x \to \frac{\pi}{2}-}\tan{x} = \infty$.

Left-hand limit We study $\lim_{x \to \frac{\pi}{2}-}$ in the next chapter. Just note that $x$ approaches $\frac{\pi}{2}$ taking smaller values of $\frac{\pi}{2}$.

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

By using the symmetry, we obtain the figure1.31.

Figure 1.31: TAN

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 $y < -1$.

Figure 1.32: Example1-10

From the figure, we have

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

Figure 1.33: Exercise1-10

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

(a) $30^{\circ}$ (b) $40^{\circ}$ (c) $72^{\circ}$

2.

Find the angle between the following line and $x$-axisprovided $0 \leq \theta < \pi$

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

3.

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

(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

(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 $\alpha,\beta$show the following identities hold

(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

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

#_#>18391#>4.enshu:1-2-4

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

5.

Derive the following formulas

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