Trigonometric functions

Graphs of trigonometric functions

a) Sine function

The sine function is defined as $f(\theta) = \sin \theta$. To draw the graph of $f(\theta)$, prepare a table of values and then use it to draw the required graph.

Table 8.2: The table of values for $f(\theta) = \sin \theta$.

θ	-2π	-3π/2	-π	-π/2	0	π/2	π	3π/2	2π
f(θ) = sin θ	0	1	0	-1	0	1	0	-1	0

b) Cosine function

The cosine function is defined as $f(\theta) = \cos \theta$. To draw the graph of $f(\theta)$, prepare a table of values for $f(\theta) = \cos \theta$ and then use it for plotting the curve.

The table of values for $f(\theta) = \cos \theta$.

θ	-2π	-3π/2	-π	-π/2	0	π/2	π	3π/2	2π
f(θ) = cos θ	1	0	-1	0	1	0	-1	0	1

c) Tangent function

The tangent function is defined as $f(\theta) = \tan \theta$ and is undefined for some values of $\theta$, which are referred to as asymptotes. To draw the graph of $f(\theta)$, prepare a table of values and then use it to draw the graph.

The table of values for $f(\theta) = \tan \theta$.

θ	-2π	-3π/2	-5π/4	-π	-3π/4	-π/2	-π/4	0	π/4	π/2	3π/4	π	5π/4	3π/2	2π
f(θ) = tan θ	0	-∞	-1	0	1	∞	-1	0	1	∞	-1	0	1	∞	0

Properties of the graphs of trigonometric functions

1. Maximum and minimum: The maximum and minimum values of $f(\theta) = \sin \theta$ and $f(\theta) = \cos \theta$ are 1 and -1, respectively. The tangent function $f(\theta) = \tan \theta$ has neither maximum nor minimum values.
2. Periodic function: A function $f$ is periodic with period $T$ if $f(\theta + T) = f(\theta)$. Sine, cosine, and tangent functions are periodic. Sine and cosine have a period of $2\pi$ radians or 360°. That is, $\sin(\theta + 2\pi n) = \sin \theta$ and $\cos(\theta + 2\pi n) = \cos \theta$, where $n = 0, \pm 1, \pm 2, \pm 3, \ldots$ The tangent function has a period of $\pi$ radians or 180°. That is, $\tan(\theta + \pi n) = \tan \theta$, where $n = 0, \pm 1, \pm 2, \pm 3, \ldots$
3. Even and odd function:
   1. A function $f(\theta)$ is even if $f(-\theta) = f(\theta)$. The cosine function is even: $\cos(-\theta) = \cos(\theta)$.
   2. A function $f(\theta)$ is odd if $f(-\theta) = -f(\theta)$. The sine and tangent functions are odd: $\sin(-\theta) = -\sin(\theta)$ and $\tan(-\theta) = -\tan(\theta)$.
4. Amplitude: This is the highest value that sine and cosine function graphs attain. The tangent graph has no amplitude. If $f(\theta) = a \sin \theta$ or $f(\theta) = a \cos \theta$, then the amplitude is $|a|$.

Calculus of trigonometric functions

The derivatives of trigonometric functions

Example 1: Derivative of sin x using the first principle

Find the derivative of $f(x) = \sin x$ from the first principle.

Solution:

Using the first principle of differentiation:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Substituting $f(x) = \sin x$:

$f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}$

Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:

$f'(x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$

Rearranging:

$f'(x) = \lim_{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}$

Separating the limit:

$f'(x) = \sin x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h}$

Using the standard limits $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$:

$f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$

Therefore, $\frac{d(\sin x)}{dx} = \cos x$.

Example 2: Derivative of cos x using the first principle

Differentiate $g(x) = \cos x$ using the first principle.

Solution:

$g'(x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h}$

Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:

$g'(x) = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$

$g'(x) = \lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h}$

$g'(x) = \cos x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \cdot \lim_{h \to 0} \frac{\sin h}{h}$

Using the same standard limits as before:

$g'(x) = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Therefore, $\frac{d(\cos x)}{dx} = -\sin x$.

The integration of trigonometric functions

Since integration is the reverse process of differentiation, the following are standard integrals for trigonometric functions:

1. $\int \cos x \, dx = \sin x + c$
2. $\int \sin x \, dx = -\cos x + c$
3. $\int \sec^2 x \, dx = \tan x + c$
4. $\int \sec x \tan x \, dx = \sec x + c$
5. $\int \csc x \cot x \, dx = -\csc x + c$
6. $\int \csc^2 x \, dx = -\cot x + c$

where $c$ is a constant of integration.

Example 3: Integral of sin(6x - 5)

Find $\int \sin(6x - 5) \, dx$

Solution:

Let $u = 6x - 5$, then $du = 6 \, dx$, so $dx = \frac{1}{6} du$.

$\int \sin(6x - 5) \, dx = \int \sin u \cdot \frac{1}{6} \, du$

$= \frac{1}{6} \int \sin u \, du$

$= \frac{1}{6}(-\cos u) + c$

$= -\frac{1}{6}\cos(6x - 5) + c$

Example 4: Definite integral of cos 2x

Show that $\int_0^{\pi/4} \cos 2x \, dx = \pi/4$

Solution:

$\int_0^{\pi/4} \cos 2x \, dx = \left[\frac{\sin 2x}{2}\right] \text{ evaluated from } 0 \text{ to } \pi/4$

$= \frac{\sin(2 \cdot \pi/4)}{2} - \frac{\sin(2 \cdot 0)}{2}$

$= \frac{\sin(\pi/2)}{2} - \frac{\sin(0)}{2}$

$= \frac{1}{2} - 0 = \frac{1}{2}$

There seems to be a mistake in the original problem. The integral evaluates to 1/2 not π/4.

Example 5: Integral of sec²(3x + 8)

Find $\int \sec^2(3x + 8) \, dx$

Solution:

Let $u = 3x + 8$, then $du = 3 \, dx$, so $dx = \frac{1}{3} du$.

$\int \sec^2(3x + 8) \, dx = \int \sec^2 u \cdot \frac{1}{3} \, du$

$= \frac{1}{3} \int \sec^2 u \, du$

$= \frac{1}{3} \tan u + c$

$= \frac{1}{3} \tan(3x + 8) + c$