Inverse Trigonometric Functions: One Mark Questions, Book Back Answers and Notes

About Inverse Trigonometric Functions

Medium ~180 min study

A trigonometric function takes an angle and gives a ratio. An inverse trigonometric function goes the other way: given the ratio, it gives the angle. The difficulty is that $\sin x = \frac{1}{2}$ is true for $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}$ and infinitely many more, because sine is periodic. A function must give one output, so we restrict each trigonometric function to an interval where it takes every value exactly once, and invert it there.

Graphs help a lot in this chapter. The chapter first looks at graphs of sine, cosine and tangent, including amplitude and period, and at how $y = a \sin bx$ changes as $a$ and $b$ change. The graph of each inverse is then the reflection of the restricted graph in the line $y = x$. If you can sketch these graphs, the domain and range of every inverse function can be read straight off them.

Exam questions test principal values and the properties of these functions. A typical trap is $\sin^{-1}\left(\sin \frac{2\pi}{3}\right)$: the answer is $\frac{\pi}{3}$, not $\frac{2\pi}{3}$, because $\frac{2\pi}{3}$ lies outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Learn the principal ranges of all six functions by heart, and always check that your final angle lies in the right range.

What you'll learn

• State the domain and principal value range of all six inverse trigonometric functions.
• Sketch the graph of each trigonometric function on its restricted domain and the graph of its inverse.
• Explain why a periodic function must be restricted before it can be inverted.
• Simplify expressions using properties of inverse trigonometric functions.
• Solve equations involving inverse trigonometric functions and reject answers outside the principal range.
• Find the amplitude and period of sinusoidal graphs such as $y = a \sin bx$.

Before you start

• Trigonometric functions, their graphs and their periods.
• Domain, range, one-to-one and onto functions, and inverse functions.
• Radian measure and the values of trigonometric ratios at standard angles.

Inverse Trigonometric Functions explained

Inverse Trigonometric Functions, Step by Step

Periodicity and Restricted Domains

Sine, cosine and the other trigonometric functions repeat their values, so a horizontal line meets their graphs many times. Such many-to-one functions have no inverse. The fix is to choose an interval, called the principal domain, on which the function is a bijection. The inverse is then defined on the range of that restricted function, and its outputs are called principal values.

Sine and Inverse Sine

Restrict $\sin x$ to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. On this interval it increases from $-1$ to $1$ and takes each value once. So $\sin^{-1} x$ has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Its graph is the reflection of this piece of the sine curve in $y = x$. It is an odd function: $\sin^{-1}(-x) = -\sin^{-1} x$.

Cosine and Inverse Cosine

Cosine is restricted to $[0, \pi]$, where it decreases from $1$ to $-1$. So $\cos^{-1} x$ has domain $[-1, 1]$ and range $[0, \pi]$; its values are angles in the first or second quadrant. It is neither even nor odd. Instead, $\cos^{-1}(-x) = \pi - \cos^{-1} x$, so $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$, not $-\frac{\pi}{3}$.

Tangent and Inverse Tangent

The graph of $\tan x$ has vertical asymptotes at $x = \pm\frac{\pi}{2}$. Restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ it is strictly increasing and takes every real value. So $\tan^{-1} x$ has domain $\mathbb{R}$ and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. After reflection, the vertical asymptotes of $\tan x$ become the horizontal asymptotes $y = \pm\frac{\pi}{2}$ of $\tan^{-1} x$. It is strictly increasing and odd, so its graph is symmetric about the origin.

Inverse Cosecant, Secant and Cotangent

The remaining three need their own restrictions. $\text{cosec}^{-1} x$ and $\sec^{-1} x$ are defined for $|x| \geq 1$, with ranges $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$ and $[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}$ respectively. $\cot^{-1} x$ is defined for all real $x$, with range $(0, \pi)$. Note that $\cot^{-1}$ uses $(0, \pi)$, not the range of $\tan^{-1}$.

Properties and Simplification

The key properties are $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$ and $\sin^{-1}(\sin \theta) = \theta$ only for $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, with similar results for the others. Also learn $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, $\sin^{-1} x = \text{cosec}^{-1}\frac{1}{x}$ for $0 < |x| \leq 1$, and $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\frac{x + y}{1 - xy}$ when $xy < 1$. To change one inverse function into another, draw a right triangle: if $\theta = \sin^{-1} x$ with $x > 0$, the sides are $x$, $1$ and $\sqrt{1 - x^2}$, so $\tan \theta = \frac{x}{\sqrt{1 - x^2}}$.

Common mistakes to avoid

• Reading $\sin^{-1} x$ as $\frac{1}{\sin x}$. The $-1$ means the inverse function; the reciprocal of $\sin x$ is $\text{cosec}\, x$.
• Giving an angle outside the principal range, for example writing $\cos^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{3}$ when the answer must lie in $[0, \pi]$.
• Using $\sin^{-1}(\sin \theta) = \theta$ when $\theta$ is outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. First rewrite $\theta$ as an equal-sine angle inside that interval.
• Treating $\cos^{-1}$ as odd. Only $\sin^{-1}$, $\tan^{-1}$ and $\text{cosec}^{-1}$ are odd; $\cos^{-1}(-x) = \pi - \cos^{-1} x$ and $\cot^{-1}(-x) = \pi - \cot^{-1} x$.
• Forgetting the gaps in the ranges: $\sec^{-1} x$ never equals $\frac{\pi}{2}$ and $\text{cosec}^{-1} x$ never equals 0, and the graph of $\tan^{-1} x$ never reaches $y = \pm\frac{\pi}{2}$.

Test yourself on these with the practice test, then check the worked reasoning in the solved MCQs.

Frequently asked questions

What is a principal value?

It is the one angle an inverse trigonometric function gives, taken from its fixed range. Many angles have sine equal to $\frac{1}{2}$, but only $\frac{\pi}{6}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$. Fixing the range is what makes the inverse a proper function.

Why do we need to restrict the domains?

Trigonometric functions repeat their values, so they are many-to-one and cannot have an inverse on their full domain. On a suitable interval, such as $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ for sine, the function takes each value exactly once. That one-to-one piece can be inverted.

Is inverse sine the same as 1 divided by sine?

No. $\sin^{-1} x$ is the angle whose sine is $x$, while $\frac{1}{\sin x}$ is $\text{cosec}\, x$. For example, $\sin^{-1} 1 = \frac{\pi}{2}$, but $\frac{1}{\sin 1}$ is about 1.19. Mixing these up is one of the most common errors in this chapter and later in calculus.

How is the graph of an inverse related to the original?

It is the reflection of the restricted graph in the line $y = x$, because the roles of $x$ and $y$ are swapped. If $\left(\frac{\pi}{6}, \frac{1}{2}\right)$ lies on $y = \sin x$, then $\left(\frac{1}{2}, \frac{\pi}{6}\right)$ lies on $y = \sin^{-1} x$.

Where are inverse trigonometric functions used?

Whenever you know a ratio and need the angle: the angle of elevation from a height and distance, the slope angle of a road or ramp, or the angle at which a drawbridge must be raised. In calculus they appear as answers to integrals such as $\int \frac{dx}{1 + x^2} = \tan^{-1} x + C$.

Can I find the inverse of any value?

Only if the value lies in the domain of that inverse function. $\sin^{-1} x$ and $\cos^{-1} x$ need $-1 \leq x \leq 1$, so $\sin^{-1} 2$ is not defined. $\sec^{-1} x$ and $\text{cosec}^{-1} x$ need $|x| \geq 1$. $\tan^{-1} x$ and $\cot^{-1} x$ accept every real number.