IB Analysis and Approaches HL · Functions · Section 14 of 15

Circular functions

3 worked examples · By Álvaro López

Key idea

Sine, cosine and tangent, seen as functions of a real number $x$ (an angle in radians), give curves that repeat with a fixed period. Knowing their shape, their period and their range lets you solve trigonometric equations in an interval and define their inverse functions, $arcsin$ , $arccos$ and $arctan$ . In the IB exam they appear in both papers, and at HL with compositions of inverse functions.

Sine and cosine

With the angle $x$ measured in radians (as is always done in the study of functions), $sin x$ and $cos x$ are the coordinates of the point on the unit circle that corresponds to the angle $x$ . As you go round the circle, the coordinate rises and falls smoothly, and its graph is a wave.

Graphs of y = sin x (orange) and y = cos x (blue) from minus 2 pi to 2 pi: they oscillate between -1 and 1 with period 2 pi

The graph of the cosine is that of the sine shifted $\frac{π}{2}$ to the left: $cos x = sin (x + \frac{π}{2})$ .

Tangent

Since $tan x = \frac{sin x}{cos x}$ , the tangent does not exist where the cosine is $0$ , that is, at $x = \frac{π}{2} + k π$ . At those values there are vertical asymptotes.

Graph of y = tan x from minus 2 pi to 2 pi: period pi, zero at the multiples of pi and dashed vertical asymptotes at x equal to pi over 2 plus k pi

Inverse functions

None of the three functions is one-to-one over its whole domain (they repeat values), so they have no inverse unless the domain is restricted to a piece where they are. The same piece is always chosen.

Definition · Inverse circular functions

Function	Restricted domain	Inverse	Domain of the inverse	Range of the inverse
$sin x$	$[- \frac{π}{2}, \frac{π}{2}]$	$arcsin x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$cos x$	$[0, π]$	$arccos x$	$[- 1, 1]$	$[0, π]$
$tan x$	$] - \frac{π}{2}, \frac{π}{2} [$	$arctan x$	$R$	$] - \frac{π}{2}, \frac{π}{2} [$

They are also written $sin^{- 1} x$ , $cos^{- 1} x$ and $tan^{- 1} x$ . They do not mean $\frac{1}{sin x}$ ; they mean the inverse function.

Graphs of arcsin x (orange, from -1 to 1), arccos x (blue, from -1 to 1) and arctan x (grey, with horizontal asymptotes y = pi/2 and y = -pi/2)

The graphs are the reflections in $y = x$ of the chosen pieces. Notice that $arctan x$ has two horizontal asymptotes, $y = \pm \frac{π}{2}$ , which correspond to the vertical asymptotes of the tangent.

The first line is always true (if $a$ is in the domain). The second is not: outside those intervals, the inverse returns the equivalent angle that does lie in them. For example, $arcsin (sin \frac{5 π}{6}) = \frac{π}{6}$ , not $\frac{5 π}{6}$ , because $\frac{5 π}{6}$ is outside $[- \frac{π}{2}, \frac{π}{2}]$ and $sin \frac{5 π}{6} = sin \frac{π}{6}$ .

Trigonometric equations in an interval

An equation such as $sin x = \frac{1}{2}$ has infinitely many solutions because of periodicity; the question fixes an interval so that you keep a finite number. The calculator or the inverse gives only one solution (the principal value); the others are obtained with the symmetries of the graph.

A good habit is to sketch the graph of the function and of the line $y = k$ over the interval: the solutions are the intersections, and the sketch shows how many there should be.

Worked examples

Example 1 Basic

Values of the inverse functions

Calculate the exact value, in radians, of:

a) $arcsin \frac{1}{2}$

b) $arccos (- \frac{2}{2})$

c) $arctan (- 1)$

d) $arcsin (sin \frac{5 π}{6})$

Solution

a) $arcsin \frac{1}{2}$ is the angle in $[- \frac{π}{2}, \frac{π}{2}]$ whose sine is $\frac{1}{2}$ : it is $\frac{π}{6}$ .

b) $arccos (- \frac{2}{2})$ is the angle in $[0, π]$ with cosine $- \frac{2}{2}$ : it is $π - \frac{π}{4} = \frac{3 π}{4}$ .

c) $arctan (- 1)$ is the angle in $] - \frac{π}{2}, \frac{π}{2} [$ whose tangent is $- 1$ : it is $- \frac{π}{4}$ .

d) Since $\frac{5 π}{6}$ is not in $[- \frac{π}{2}, \frac{π}{2}]$ , it cannot be simplified directly. Work out the sine: $sin \frac{5 π}{6} = \frac{1}{2}$ , and then $arcsin \frac{1}{2} = \frac{π}{6}$ .

Example 2 Medium

Trigonometric equations in an interval

a) Solve $2 sin x + 1 = 0$ for $0 \leq x \leq 2 π$ . Give the set of solutions.

b) Solve $tan x = 3$ for $- π \leq x \leq π$ . Give the set of solutions.

c) How many solutions does $cos x = \frac{1}{3}$ have for $- 2 π \leq x \leq 2 π$ ?

Solution

a) $2 sin x + 1 = 0$ gives $sin x = - \frac{1}{2}$ . The principal value is $arcsin (- \frac{1}{2}) = - \frac{π}{6}$ , which is not in $[0, 2 π]$ . The solutions are $x = - \frac{π}{6} + 2 π n$ and $x = π - (- \frac{π}{6}) + 2 π n = \frac{7 π}{6} + 2 π n$ . Taking $n$ so as to enter the interval gives

$x = \frac{7 π}{6}, x = \frac{11 π}{6}$

(the second is $- \frac{π}{6} + 2 π$ ). This is consistent with the graph: the sine is $- \frac{1}{2}$ in the third and fourth quadrants.

b) The principal value is $arctan 3 = \frac{π}{3}$ . The tangent has period $π$ , so $x = \frac{π}{3} + π n$ . In $[- π, π]$ there is room for $n = 0$ ( $x = \frac{π}{3}$ ) and $n = - 1$ ( $x = \frac{π}{3} - π = - \frac{2 π}{3}$ ). The set is ${- \frac{2 π}{3}, \frac{π}{3}}$ .

c) Between $0$ and $2 π$ , $cos x = \frac{1}{3}$ has two solutions ( $α$ and $2 π - α$ , with $α = arccos \frac{1}{3} \approx 1.23$ ). The interval $[- 2 π, 2 π]$ contains two full periods, and in each one there are two solutions: in total there are $4$ (seen by cutting the line $y = \frac{1}{3}$ with the graph of the cosine).

Example 3 Hard

Compositions with inverse functions

a) Calculate the exact value of $cos (arcsin \frac{5}{13})$ .

b) Calculate the exact value of $arctan \frac{1}{4} + arctan \frac{3}{5}$ .

c) Solve $arctan x + arctan 2 x = \frac{π}{4}$ . Give the exact solution.

Solution

a) Let $θ = arcsin \frac{5}{13}$ , an angle in $[- \frac{π}{2}, \frac{π}{2}]$ with $sin θ = \frac{5}{13} > 0$ . In that interval the cosine is non-negative, so

$cos θ = 1 - sin^{2} θ = 1 - \frac{25}{169} = \frac{144}{169} = \frac{12}{13}$

b) Let $S = arctan \frac{1}{4} + arctan \frac{3}{5}$ . The two terms are in $] 0, \frac{π}{2} [$ , so $S \in] 0, π [$ . Using the formula for the tangent of a sum:

$tan S = \frac{\frac{1}{4} + \frac{3}{5}}{1 - \frac{1}{4} \cdot \frac{3}{5}} = \frac{\frac{17}{20}}{\frac{17}{20}} = 1$

The only angle in $] 0, π [$ with tangent $1$ is $\frac{π}{4}$ . Therefore $S = \frac{π}{4}$ .

c) Take the tangent of both sides: $tan (arctan x + arctan 2 x) = \frac{x + 2 x}{1 - 2 x ^{2}} = tan \frac{π}{4} = 1$ . Then

$3 x = 1 - 2 x^{2} ⟹ 2 x^{2} + 3 x - 1 = 0 ⟹ x = \frac{- 3 \pm 17}{4}$

Taking the tangent may introduce false solutions, so they must be checked. For $x = \frac{- 3 - 17}{4} \approx - 1.78$ both arctangents are negative and their sum is negative, not $\frac{π}{4}$ : it is rejected. For $x = \frac{- 3 + 17}{4} \approx 0.281$ , the sum of the arctangents is positive and equals $0.273 + 0.512 = 0.785 \approx \frac{π}{4}$ . Solution:

$x = \frac{17 - 3}{4}$

In the exam: when you take "the tangent of both sides", you must check the solutions, because $tan β = 1$ is also satisfied by $β = - \frac{3 π}{4}$ , which is not $\frac{π}{4}$ . If the question asks for "the exact value", do not substitute decimals part-way through the problem.

Common mistakes

Summary

$sin$ and $cos$ : domain $R$ , range $[- 1, 1]$ , period $2 π$ , amplitude $1$ ; the sine is odd and the cosine is even.
$tan$ : domain $R$ without $\frac{π}{2} + k π$ , range $R$ , period $π$ , vertical asymptotes at $x = \frac{π}{2} + k π$ ; it is odd and has no amplitude.
$arcsin : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$ ; $arccos : [- 1, 1] \to [0, π]$ ; $arctan : R \to] - \frac{π}{2}, \frac{π}{2} [$ .
$sin (arcsin a) = a$ always (with $a$ in the domain); $arcsin (sin β) = β$ only in the principal interval.
Equations in an interval: principal value, symmetry (sine: $π - α$ ; cosine: $- α$ ; tangent: $+ π$ ), add periods and keep the solutions in the interval; in Paper 2, intersect graphs on the GDC in the appropriate angle mode.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp