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1.4 Trig Functions and Their Inverses

Two common ways to measure an angle: degrees, radians.

A whole circle is \( 360^\circ \) or \( 2\pi \) radians.

What is a radian?

Figure: Coordinate plane showing a circle of radius r with an arc s and angle theta.

Circle radius \( r \)

Circular segment w/ included angle \( \theta \)

\( s \): arc length

\[ s = r \theta \]

The angle \( \theta \) such that \( s = r \) (arc length = radius) is called a radian \( \approx 57^\circ \).

Conversion:

\[ 360^\circ = 2\pi \text{ rad} \]

In calculus, ALWAYS use radians!

So, \( 1 \text{ rad} = \frac{360^\circ}{2\pi} \rightarrow 1 \text{ rad} = \frac{180}{\pi} \text{ deg} \)

\( 1^\circ = \frac{2\pi}{360} \text{ rad} \rightarrow 1 \text{ deg} = \frac{\pi}{180} \text{ rad} \)

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Basic Trig Functions

Sine, cosine, tangent, cosecant, secant, cotangent

Figure: Right triangle labeled with hypotenuse, opposite, adjacent, and angle theta.

\[ \begin{aligned} \sin \theta &= \frac{\text{opp}}{\text{hyp}} \\ \cos \theta &= \frac{\text{adj}}{\text{hyp}} \\ \tan \theta &= \frac{\text{opp}}{\text{adj}} = \frac{\sin \theta}{\cos \theta} \end{aligned} \]\[ \begin{aligned} \csc \theta &= \frac{1}{\sin \theta} = \frac{\text{hyp}}{\text{opp}} \\ \sec \theta &= \frac{1}{\cos \theta} = \frac{\text{hyp}}{\text{adj}} \\ \cot \theta &= \frac{1}{\tan \theta} = \frac{\text{adj}}{\text{opp}} = \frac{\cos \theta}{\sin \theta} \end{aligned} \]

Periodic Functions

These are periodic functions: \( f(x+p) = f(x) \) where \( p \) is the period.

For example, \( f(x) = \sin(x) \) has a period of \( 2\pi \), so it repeats every \( 2\pi \) of \( x \).

Figure: Sine wave graph on x-y axes with labels for period 2 pi and amplitude.

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Transformations and Inverses of Trigonometric Functions

We can transform them, too (vertical shift, horizontal stretch, etc.)

\[ f(x) = a \sin(bx + c) + d \]

a: vertical stretch/compress (changes amplitude)
b: frequency (horizontal stretch/compress)

c: horizontal shift (phase shift = \(\frac{c}{b}\))
d: vertical shift

Period = \(\frac{2\pi}{b}\)

Inverses of \(\sin(x), \cos(x), \tan(x)\)

These functions are not one-to-one, so we must restrict the domain.

\(\sin(x)\)

\(\sin(x)\): restrict domain to \([-\frac{\pi}{2}, \frac{\pi}{2}]\)

range: \([-1, 1]\)

\(\sin^{-1}(x)\)

Figure: Graph of arcsin(x) on x-y axes, passing through (-1, -pi/2), (0,0), and (1, pi/2).

\(\sin^{-1}(x)\) is NOT \((\sin(x))^{-1}\)

\(\sin^{-1}(x) \neq \frac{1}{\sin x}\)

\(\sin^{-1}(x)\): domain \([-1, 1]\)

range \([-\frac{\pi}{2}, \frac{\pi}{2}]\)

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\(\cos(x)\)

\(\cos(x)\): domain \([0, \pi]\)

range \([-1, 1]\)

\(\cos^{-1}(x)\)

Figure: Graph of arccos(x) on x-y axes, showing a decreasing curve from (-1, pi) to (1, 0).

\(\cos^{-1}(x)\): domain \([-1, 1]\)

range \([0, \pi]\)

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Trigonometric and Inverse Trigonometric Functions

Tangent Function: \(\tan(x)\)

The graph of the tangent function shows vertical asymptotes at odd multiples of \(\frac{\pi}{2}\).

Domain: \((-\pi/2, \pi/2)\)
Range: \((-\infty, \infty)\)

Figure: Graph of tan(x) with vertical asymptotes at x = -pi/2 and x = pi/2, passing through the origin.

Inverse Tangent Function: \(\tan^{-1}(x)\)

The inverse tangent function, or arctangent, is the reflection of the tangent function over the line \(y=x\), restricted to its principal branch.

Domain: \((-\infty, \infty)\)
Range: \((-\pi/2, \pi/2)\)

Figure: Graph of arctan(x) with horizontal asymptotes at y = pi/2 and y = -pi/2, passing through the origin.

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Inverse Trigonometric Ranges and Quadrants

\(\sin^{-1}(x)\) has range of \([-\pi/2, \pi/2]\) so \(\sin^{-1}(x)\) can only give back an angle in Quadrant IV or Quadrant I.

For example, \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\) means to find an angle whose sine is \(\frac{\sqrt{2}}{2}\) and it is in \([-\pi/2, \pi/2]\) (Quadrants IV, I).

What angles have sine equal to \(\frac{\sqrt{2}}{2}\)?

\[ \frac{\pi}{4}, \quad \text{\sout{\(\frac{3\pi}{4}\)}} \]

But since \(\sin^{-1}(x)\) can only give a QI or QIV angle, we must discard \(\frac{3\pi}{4}\) since it is in QII.

So, \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\) only.

Cosine Inverse Properties

Similarly, \(\cos^{-1}(x)\) has range \([0, \pi]\) (QI, QII).

For example, \(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\) and NOT \(-\frac{\pi}{3}\) since \(-\frac{\pi}{3}\) is NOT in QI or QII.

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Example: Inverse Trigonometric Functions

\[ \cos^{-1}\left(\cos\left(\frac{4\pi}{3}\right)\right) = ? \]

Call the inner term \( x \):

First, find \( x \): \[ x = \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \]

Work on the unit circle!!!!

Now, evaluate the outer function: \[ \cos^{-1}\left(-\frac{1}{2}\right) \rightarrow \text{find an angle whose cosine is } -\frac{1}{2} \]

That is in the interval \( [0, \pi] \) (QI, QII).

= \( \frac{2\pi}{3} \)

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Example: Inverse Sine Function

\[ \sin^{-1}\left(\sin\left(\frac{\pi}{4}\right)\right) \]

Call the inner term \( x \): \[ x = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

Now, evaluate the outer function: \[ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \rightarrow \text{find angles whose sine is } \frac{\sqrt{2}}{2} \]

and in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) (QI, QIV).

= \( \frac{\pi}{4} \)

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Example: Rewriting Trigonometric Expressions

Rewrite \(\cos(\tan^{-1}(x))\) without trigonometric functions.

Step 1: Work Inside-Out

Start with the inner function and assign it a variable:

\[ \cos(\underbrace{\tan^{-1}(x)}_{\text{call this } \theta}) \]

Let \(\theta = an^{-1}(x)\). This implies:

\[ \tan(\theta) = x = \frac{x}{1} = \frac{\text{opp}}{\text{adj}} \]

Step 2: Build a Triangle

Using the relationship \(\tan(\theta) = \frac{x}{1}\), we can define the sides of a right triangle:

\(\text{opp} = x\)
\(\text{adj} = 1\)

By the Pythagorean theorem, the hypotenuse is \(\sqrt{x^2 + 1}\).

Right triangle with angle theta, opposite side x, adjacent side 1, and hypotenuse \sqrt{x^2 + 1}. — Figure: Right triangle with angle theta, opposite side x, adjacent side 1, and hypotenuse \(\sqrt{x^2 + 1}\).

Step 3: Evaluate the Outer Function

Now, find the cosine of the angle \(\theta\) using the triangle:

\[ \cos(\theta) = \frac{\text{adj}}{\text{hyp}} = \frac{1}{\sqrt{x^2 + 1}} \]

Final Result:

\[ \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}} \]