Inverse Functions and Logarithms

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Inverse Trig Functions

Sine and Tangent

The functions sine, tangent and secant are not one-to-one, since they repeat (every $2\pi$ for sine and secant, and every $\pi$ for tangent). To get inverse functions, we must restrict their domains. We could do this in many ways, but the convention is:

SINE: We restrict the domain to $[-\pi/2, \pi/2]$ to ensure our function is one-to-one. By definition, $\sin^{-1}(x)$ is the angle in $[-\pi/2,\pi/2]$ whose sine is $x$. This only makes sense if $-1 \le x \le 1$.

$$\theta = \sin^{-1}(x) \quad \Longleftrightarrow \quad \sin(\theta)=x, \quad \hbox{ for }-\pi/2 \le \theta \le \pi/2.$$

Examples: $\sin^{-1}(1/2)$ is the angle $\theta$ (in the restricted domain) for which $\sin(\theta)=1/2$. Thus $\theta=\pi/6$; i.e. $\sin^{-1}(1/2)=\pi/6$.
DO: Find $\sin^{-1}\left(\frac{\sqrt 3}{2}\right)$

TANGENT: We restrict the domain to $(-\pi/2,\pi/2)$. By definition, $\tan^{-1}(x)$ is the angle in $(-\pi/2,\pi/2)$ whose tangent value is $x$. Here, $x$ can be any real number; can you see why?

$$\theta=\tan^{-1}(x) \quad \Longleftrightarrow \quad \tan(\theta)=x, \quad \hbox{ for } -\pi/2 < \theta < \pi/2.$$

DO: Find $\tan^{-1}(\sqrt 3)$.

Some Facts

In practice, $\sin^{-1}$ and $\tan^{-1}$ come up a lot, $\sec^{-1}$ comes up occasionally, while $\cos^{-1}$, $\cot^{-1}$ and $\csc^{-1}$ almost never come up.

Another name for $\sin^{-1}(x)$ is $\arcsin(x)$. Likewise, $\arctan(x)$ and $\arccos(x)$ mean $\tan^{-1}(x)$ and $\cos^{-1}(x)$.

The "exponent" $-1$ has more than one meaning, which you must infer from the context. The notation $\sin^{-1}(x)$ should not be confused with expressions like $\sin^2(x)$. $\sin^2(x)$ is shorthand for $(\sin(x))^2$, but $\sin^{-1}(x)$ is not $\sin(x)^{-1}$. It is the inverse sine of $x$. The same applies to $\cos^{-1}(x)$ and $\tan^{-1}(x)$. Similarly, $f^{-1}(x)$ is the inverse of $f$, not $1/f(x)$. These are inverse functions, not reciprocals.

Advanced play for those who are interested

The function $\cos^{-1}$ is closely related to $\sin^{-1}$. Specifically, $\cos^{-1}(x) = \frac{\pi}{2}-\sin^{-1}(x)$. This is because $\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$.

Similarly, since $\cot(\theta)=1/\tan(\theta)$, $\cot^{-1}(x)=\tan^{-1}(1/x)$.

Likewise, $\sec^{-1}(x)=\cos^{-1}(1/x)$

DO: Carefully play with these statements and see if you can see why they are true.

The Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

$f$ vs. $f'$

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Inverse Trig Functions

Sine and Tangent

Some Facts

Advanced play for those who are interested