Trigonometrically Functions

• $f(x) = \sin x$: Domain: $\{x: x \in \mathbb{R}\}$; Range: $\{f(x): -1 \le f(x) \le 1\}$
• $f(x) = \cos x$: Domain: $\{x: x \in \mathbb{R}\}$; Range: $\{f(x): -1 \le f(x) \le 1\}$
• $f(x) = \tan x = \frac{\sin x}{\cos x}$: Domain: $\{x: x \in \mathbb{R}, x \ne \frac{(2n+1)\pi}{2}, n \in \mathbb{Z}\}$; Range: $\{f(x): f(x) \in \mathbb{R}\}$

The sine, cosine, and tangent functions are periodic. This means their graphs repeat a pattern over a fixed interval. This interval is called the period.

Both sine and cosine functions have a period of $2\pi$ radians ($360^\circ$). They oscillate between -1 and 1.

• For $y = \sin \theta$, the first period for $\theta \ge 0$ is between 0 and $2\pi$. The $n$th period is between $2(n-1)\pi$ and $2n\pi$.
• For $y = \cos \theta$, the $n$th period is between $2(n-1)\pi$ and $2n\pi$.

Example 1: Graph of $f(\theta) = \cos \theta$ for $-2\pi \le \theta \le 2\pi$

Table of values:

Properties of the cosine function:

1. Minimum and maximum values are -1 and 1, respectively.
2. Period is $2\pi$.
3. Cosine is an even function: $f(-x) = f(x)$.
4. Amplitude is 1.

The tangent function, $f(\theta) = \tan \theta$, is also periodic, but with a period of $\pi$ radians ($180^\circ$). It is an odd function, meaning $f(-\theta) = -f(\theta)$, and its graph is symmetrical about the origin.

The tangent function is undefined where $\cos \theta = 0$, which occurs at $\theta = \frac{(2n+1)\pi}{2}$, where $n$ is any integer. These values create vertical asymptotes where $f(\theta)$ approaches $\pm\infty$.

Example 2: Graph of $f(\theta) = \tan \theta$

Table of values (note the asymptotes):

Example 3

Give the domain and range of $f(x) = 2\sin x$.

Solution:

Domain: Since the sine function is defined for all real numbers, the domain of $f(x) = 2\sin x$ is $\{x: x \in \mathbb{R}\}$.

Range: Because $-1 \le \sin x \le 1$, multiplying by 2 gives $-2 \le 2\sin x \le 2$. Thus, the range is $\{f(x): -2 \le f(x) \le 2\}$.

Example 4

Determine the domain and range of $g(x) = \frac{2}{3}\cos x$, for $-\frac{2\pi}{3} \le x \le \frac{2\pi}{3}$.

Solution:

Domain: Given as $-\frac{2\pi}{3} \le x \le \frac{2\pi}{3}$.

Range: Since $-1 \le \cos x \le 1$, multiplying by $\frac{2}{3}$ gives $-\frac{2}{3} \le \frac{2}{3}\cos x \le \frac{2}{3}$. Thus, the range is $\{g(x): -\frac{2}{3} \le g(x) \le \frac{2}{3}\}$.

Inverse trigonometric functions are the inverses of the basic trigonometric functions. They are used to find angles from trigonometric ratios.

1. $y = \sin^{-1} x$ (or $y = \arcsin x$) if and only if $x = \sin y$ and $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.
2. $y = \cos^{-1} x$ (or $y = \arccos x$) if and only if $x = \cos y$ and $0 \le y \le \pi$.
3. $y = \tan^{-1} x$ (or $y = \arctan x$) if and only if $x = \tan y$ and $-\frac{\pi}{2} < y < \frac{\pi}{2}$.

Inverses of other trigonometric functions can be defined in terms of these:

1. $\sec^{-1} y = \cos^{-1}\left(\frac{1}{y}\right)$, for $|y| \ge 1$
2. $\csc^{-1} y = \sin^{-1}\left(\frac{1}{y}\right)$, for $|y| \ge 1$
3. $\cot^{-1} y = \tan^{-1}\left(\frac{1}{y}\right)$, for $y > 0$ or $\cot^{-1} y = \pi + \tan^{-1}\left(\frac{1}{y}\right)$, for $y < 0$

Example 5

Show that $\sin^{-1}x + 2\sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right) = \pi/2$

Solution:

Let $\sin^{-1} x = \theta$, then $x = \sin \theta$.

$\sin^{-1} x + 2\sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right) = \theta + 2\sin^{-1}\left(\sqrt{\frac{1-\sin\theta}{2}}\right)$

$= \theta + 2\sin^{-1}\left(\sqrt{\frac{1-\cos(\frac{\pi}{2} - \theta)}{2}}\right)$

$= \theta + 2\sin^{-1}\left(\sqrt{\sin^2(\frac{\pi}{4} - \frac{\theta}{2})}\right)$

$= \theta + 2(\frac{\pi}{4} - \frac{\theta}{2}) = \theta + \frac{\pi}{2} - \theta = \frac{\pi}{2}$

Example 6

If $\tan^{-1}\left(\frac{x-4}{16}\right) + \tan^{-1}\left(\frac{x+12}{13}\right) = \frac{\pi}{4}$, find the positive value of x.

Solution:

Let $A = \tan^{-1}\left(\frac{x-4}{16}\right)$ and $B = \tan^{-1}\left(\frac{x+12}{13}\right)$. Then $\tan A = \frac{x-4}{16}$ and $\tan B = \frac{x+12}{13}$.

Given $A + B = \frac{\pi}{4}$, taking the tangent of both sides:

$\tan(A + B) = \tan\frac{\pi}{4}$

$\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$

$\frac{\frac{x-4}{16} + \frac{x+12}{13}}{1 - \frac{(x-4)(x+12)}{16 \cdot 13}} = 1$

$\frac{13(x-4) + 16(x+12)}{208 - (x^2 + 8x - 48)} = 1$

$13x - 52 + 16x + 192 = 208 - x^2 - 8x + 48$

$29x + 140 = 256 - x^2 - 8x$

$x^2 + 37x - 116 = 0$

Using the quadratic formula, $x = \frac{-37 \pm \sqrt{37^2 - 4(1)(-116)}}{2}$

$x = \frac{-37 \pm \sqrt{1369 + 464}}{2} = \frac{-37 \pm \sqrt{1833}}{2}$

$x = \frac{-37 \pm 42.81}{2}$

The positive value is $x \approx \frac{5.81}{2} \approx 2.905$.

Example 7

Express $\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)$ in its simplest form.

Solution:

Using the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan\left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{\tan\frac{x}{2} + \tan\frac{\pi}{4}}{1 - \tan\frac{x}{2} \tan\frac{\pi}{4}}$

Since $\tan\frac{\pi}{4} = 1$:

$= \frac{\tan\frac{x}{2} + 1}{1 - \tan\frac{x}{2}}$

If you wanted to express this in terms of $\sin x$ and $\cos x$, you could use the half-angle formulas:

$\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} = \frac{1 - \cos x}{\sin x}$

Substituting the first half-angle formula into the simplified expression:

$\frac{\frac{\sin x}{1 + \cos x} + 1}{1 - \frac{\sin x}{1 + \cos x}} = \frac{\sin x + 1 + \cos x}{1 + \cos x - \sin x}$

Example 8

If $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{\pi}{2}$, prove that $x^2 + y^2 + z^2 + 2xyz = 1$.

Solution:

Let $A = \sin^{-1}x$, $B = \sin^{-1}y$, and $C = \sin^{-1}z$. Then $x = \sin A$, $y = \sin B$, and $z = \sin C$.

Given $A + B + C = \frac{\pi}{2}$, we can write $A + B = \frac{\pi}{2} - C$.

Taking the cosine of both sides:

$\cos(A + B) = \cos\left(\frac{\pi}{2} - C\right)$

$\cos A \cos B - \sin A \sin B = \sin C$

Since $\sin A = x$, $\sin B = y$, and $\sin C = z$, we have $\cos A = \sqrt{1 - x^2}$ and $\cos B = \sqrt{1 - y^2}$.

$\sqrt{1 - x^2}\sqrt{1 - y^2} - xy = z$

$\sqrt{(1 - x^2)(1 - y^2)} = z + xy$

Squaring both sides:

$(1 - x^2)(1 - y^2) = (z + xy)^2$

$1 - x^2 - y^2 + x^2y^2 = z^2 + 2xyz + x^2y^2$

$1 - x^2 - y^2 = z^2 + 2xyz$

$x^2 + y^2 + z^2 + 2xyz = 1$

Example 9

Solve the equation $\cos^{-1} x + \cos^{-1} 3x = \frac{\pi}{2}$.

Solution:

Let $A = \cos^{-1} x$ and $B = \cos^{-1} 3x$. Then $\cos A = x$ and $\cos B = 3x$.

Given $A + B = \frac{\pi}{2}$, so $B = \frac{\pi}{2} - A$.

Taking the cosine of both sides:

$\cos B = \cos\left(\frac{\pi}{2} - A\right)$

$3x = \sin A$

Since $\cos A = x$, we know that $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - x^2}$.

$3x = \sqrt{1 - x^2}$

Squaring both sides:

$9x^2 = 1 - x^2$

$10x^2 = 1$

$x^2 = \frac{1}{10}$

$x = \pm\frac{1}{\sqrt{10}} = \pm\frac{\sqrt{10}}{10}$

Since the range of $\cos^{-1} x$ is $[0, \pi]$, and given the original equation, we must have $x$ positive. Therefore, $x = \frac{\sqrt{10}}{10}$.

The graphs of inverse trigonometric functions are reflections of the corresponding trigonometric functions across the line $y = x$. Because trigonometric functions are not one-to-one over their entire domain, their inverses have restricted domains.

The domain of $y = \sin^{-1} x$ is $[-1, 1]$, and its range (also called the principal values) is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Example 10: Graph of $\theta = \tan^{-1} x$

Table of values:

From the graph:

• Domain: $\{x: x \in \mathbb{R}\}$
• Range: $\left\{-\frac{\pi}{2} < \theta < \frac{\pi}{2}\right\}$

Example 11: Graph of $\theta = \cos^{-1} x$

Table of values:

From the graph:

• Domain: $\{x: -1 \le x \le 1\}$
• Range: $\{0 \le \theta \le \pi\}$