This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.11 Symmetry properties of trigonometric graphs
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Use sin(π − θ) = sin θ, cos(π − θ) = −cos θ, tan(π − θ) = −tan θ, and the partner results for π + θ, −θ and 2π − θ | HL only | "Given sin θ = k, write sin(π + θ) and tan(π − θ) in terms of k" (4 marks) |
| Prove these results from the unit circle or with compound angle identities | HL only | "Show that cos(3π/2 − θ) = −sin θ" (2 marks) |
| Use the complementary results sin(π/2 − θ) = cos θ, cos(π/2 − θ) = sin θ, tan(π/2 − θ) = cot θ | HL only | "Simplify sin(π/2 + x) − cos(π − x)" (2 marks) |
| Describe the lines of symmetry, centres of symmetry and periods of y = sin x, y = cos x and y = tan x | HL only | "Write down the equation of a line of symmetry of the graph" (1 to 2 marks) |
| Decide whether a function built from trigonometric functions is odd, even or neither | HL only | "Show that f is an odd function" (2 to 3 marks) |
| Relate one trigonometric graph to another by a translation or reflection | HL only | "Describe the transformation that maps y = sin x onto y = cos x" (2 marks) |
| Use symmetry and period to find every solution of an equation in an interval | HL only | "Solve sin x = 0.3 for −2π ≤ x ≤ 2π" (4 to 5 marks) |
Before you start
You need the unit circle definitions of sin θ and cos θ and the relationship between angles in different quadrants (SL 3.5), the graphs of sin, cos and tan with their periods (SL 3.7), and solving equations in a finite interval (SL 3.8). The proofs use the compound angle identities of 3.10. Odd and even functions are defined in AHL 2.14; section 5 below repeats what you need.
1The idea in one paragraph
A point on the unit circle at angle θ has coordinates (cos θ, sin θ). Reflect it in the y-axis and you reach the angle π − θ, with coordinates (−cos θ, sin θ): same sine, opposite cosine. Reflect it in the x-axis and you reach −θ; turn it half a turn and you reach π + θ. Every "related angle" result is one of those three moves, read off as a sign change. The same facts appear on the graphs as symmetries: sin(π − θ) = sin θ says the sine graph is its own mirror image in the line x = π/2; cos(−θ) = cos θ says the cosine graph is its own mirror image in the y-axis; tan(θ + π) = tan θ says the tangent graph repeats every π. Learn the picture, not the list, and use it to simplify expressions and to turn one solution of an equation into all of them.
2Four angles on one circle
Start with an acute angle θ and its point (cos θ, sin θ) on the unit circle. Figure 1 shows the three other points you can reach by reflecting it.
- π − θ is the reflection in the y-axis: x changes sign, y does not.
- −θ (the same point as 2π − θ) is the reflection in the x-axis: y changes sign, x does not.
- π + θ is a half-turn about the origin: both change sign.
Since cos is the x-coordinate, sin the y-coordinate, and tan = sin ÷ cos, the table writes itself.
| Angle | sin | cos | tan | Move on the circle |
|---|---|---|---|---|
| π − θ | sin θ | −cos θ | −tan θ | reflect in the y-axis |
| π + θ | −sin θ | −cos θ | tan θ | half-turn about O |
| −θ, or 2π − θ | −sin θ | cos θ | −tan θ | reflect in the x-axis |
| θ + 2π | sin θ | cos θ | tan θ | a full turn: nothing changes |
sin(π − θ) = sin θ, cos(π − θ) = −cos θ, tan(π − θ) = −tan θ. The picture in Figure 1 gives the other rows the same way.
The picture was drawn with θ acute, but the results hold for every θ, because the reflections do not care where θ starts. That is what makes them identities.
Proving them with compound angles. A "show that" question wants algebra, and 3.10 supplies it. Use sin π = 0 and cos π = −1.
For tan, dividing the sine by the cosine is safer than the tan(A − B) formula, which needs tan π = 0 and gets the same answer.
3Complementary angles and quarter turns
The angle π/2 − θ is the complement of θ: in a right-angled triangle the two non-right angles add up to π/2, so the side opposite one is adjacent to the other. That swaps sine and cosine.
The prefix "co" in cosine, cotangent and cosecant means "of the complement": cos θ is the sine of the complementary angle. The same moves combine. For example, cos(3π/2 − θ) = cos(3π/2)cos θ + sin(3π/2)sin θ = 0 − sin θ = −sin θ.
A quick sign check that always works. For any expression like tan(3π − x), treat x as small and positive, find the quadrant of the whole angle, and read the sign of the ratio there. 3π − x is just below 3π, which is the same direction as π − x, in the second quadrant, where tan is negative. So tan(3π − x) = −tan x. Then decide whether the ratio stays the same (multiples of π added or subtracted) or swaps to its co-ratio (odd multiples of π/2).
Worked example 1 (Paper 1). Simplify sin(π − x) + cos(π/2 + x) + tan(π + x) cos x.
Worked example 2 (Paper 1). Given that sin θ = k and 0 < θ < π/2, write in terms of k: sin(π + θ), cos(π/2 + θ), cos(θ − π) and tan(π − θ).
The step cos(θ − π) = cos(π − θ) uses the evenness of cosine, which is the next idea.
4The symmetries of the three graphs
Each identity in the table is a symmetry of a graph: a reflection, a half-turn or a slide that maps the graph onto itself.
The sine graph and π − θ. The inputs θ and π − θ are the same distance either side of π/2, so they are mirror images in the vertical line x = π/2. sin(π − θ) = sin θ says the heights there are equal: the line x = π/2 is a line of symmetry of y = sin x. Figure 2(a) shows it. By the period, so are x = 3π/2, x = −π/2 and every x = π/2 + kπ.
The cosine graph and π − θ. For cosine the heights at θ and π − θ are opposite, cos(π − θ) = −cos θ. A point and its opposite either side of (π/2, 0) are swapped by a half-turn about that point, so (π/2, 0) is a centre of symmetry (rotational symmetry of order 2) of y = cos x. Figure 2(b) shows it.
The tangent graph. tan(π − θ) = −tan θ makes (π/2, 0) a centre of symmetry of y = tan x, even though tan is undefined at π/2 itself: the half-turn swaps the two branches either side of the asymptote. Every point (kπ/2, 0) is such a centre, both the zeros and the points under the asymptotes, and tan(π + θ) = tan θ says the whole graph repeats every π. Tan has no lines of symmetry. Figure 3 marks all of this.
| Graph | Lines of symmetry | Centres of symmetry | Period |
|---|---|---|---|
| y = sin x | x = π/2 + kπ | (kπ, 0) | 2π |
| y = cos x | x = kπ | (π/2 + kπ, 0) | 2π |
| y = tan x | none | (kπ/2, 0) | π |
Here k is any integer. Notice the pattern: the lines of symmetry pass through the maxima and minima, and the centres of symmetry sit on the axis crossings.
5Odd and even: symmetry about the origin and the y-axis
The row for −θ in the table has a name. A function is even if f(−x) = f(x) for every x in its domain; its graph is symmetric in the y-axis. A function is odd if f(−x) = −f(x) for every x; its graph has rotational symmetry of order 2 about the origin. Figure 4 shows the two standard examples.
- cos x is even: cos(−x) = cos x. So is sec x = 1/cos x.
- sin x and tan x are odd: sin(−x) = −sin x, tan(−x) = −tan x. So are cosec x and cot x.
These combine like signs in multiplication: odd × odd is even, odd × even is odd, even × even is even. To prove a claim, substitute −x and simplify; to disprove one, a single counterexample is enough.
Worked example 3 (Paper 1). Decide whether each function is odd, even or neither: f(x) = x³ cos x, g(x) = sin²x + cos x, h(x) = x + cos x.
So f is odd, g is even, and h is neither. The counterexample must be checked against both possibilities; −π − 1 is neither π − 1 nor 1 − π.
6One graph from another
Symmetries also relate different functions. Since cos x = sin(x + π/2), the cosine graph is the sine graph translated π/2 to the left, as Figure 5 shows. Equally, sin x = cos(x − π/2), a translation π/2 to the right. Combining a reflection with a translation gives the same result: cos x = sin(π/2 − x) says cosine is sine reflected in the y-axis and then translated.
The same idea links tan and cot: cot x = tan(π/2 − x), so y = cot x is y = tan x reflected in the y-axis and translated π/2 to the right. Questions in this area use the language of transformations from SL 2.11: say what kind of transformation it is, which way, and by how much.
7Every solution from one
This is where symmetry earns most marks. A calculator (or arcsin, arccos, arctan) gives one solution. The symmetries give the rest, and the period repeats them.
- For sin x = c: if α is one solution, so is π − α (mirror line x = π/2). Then add or subtract 2π.
- For cos x = c: if α is one solution, so is −α, or equivalently 2π − α (cos is even). Then add or subtract 2π.
- For tan x = c: if α is one solution, so is α + π. Keep adding or subtracting π.
Worked example 4 (Paper 2). Solve sin x = 0.3 for −2π ≤ x ≤ 2π.
Figure 6 shows the four solutions on the graph. Adding 2π to α gives 6.588…, beyond 2π, so there are no more.
Worked example 5 (Paper 2). Solve cos x = −0.6 and tan x = −2, each for 0 ≤ x ≤ 2π.
The guide's TOK question asks how an equation can have infinitely many discrete solutions. This is the answer: one solution, a symmetry and a period generate all of them, spaced out along the line. The general solution is not required, but the pattern behind it is exactly what you are using.
8Where marks are lost
Mixing up which ratio keeps its sign. For π − θ, only sine keeps its sign; for π + θ, only tangent does; for −θ, only cosine does. Draw Figure 1 in the margin instead of guessing.
Forgetting that π/2 swaps the ratio. sin(π/2 − θ) is cos θ, not sin θ. Adding or subtracting a multiple of π keeps the ratio; an odd multiple of π/2 swaps it for its co-ratio.
Giving only the calculator's solution. sin x = 0.3 has two solutions in every interval of length 2π. The second is π − α, not −α and not α + π.
Using sine's partner for cosine. For cos x = c the partner is 2π − α (or −α), not π − α.
Proving "odd" with one example. f(1) = −f(−1) does not show f is odd; you must show f(−x) = −f(x) for all x. One counterexample is enough only to show a function is not odd or even.
Calling tan's asymptote a line of symmetry. The graph of tan x has no lines of symmetry; the points (kπ/2, 0) are centres of symmetry.
Leaving solutions outside the interval, or in degrees. Check each value against the interval, and work in radians unless told otherwise.
9Work it right
- For any related angle, sketch the unit circle with θ acute, mark the new point, and read off the sign of sin, cos or tan.
- In a "show that", expand with the compound angle identity and write the values sin π = 0, cos π = −1 (or those at π/2, 3π/2) on the line.
- For "in terms of k", find the missing ratio of θ first with Pythagoras, choosing the sign from the quadrant of θ.
- For odd or even, substitute −x, simplify with sin(−x) = −sin x and cos(−x) = cos x, and compare with f(x) and −f(x) in words.
- For an equation, write α from the calculator, then the symmetry partner (π − α, 2π − α or α + π), then add or subtract the period until you leave the interval.
- For a graph, name each transformation fully: translation by how much and which way, or reflection in which line.
10Try it
Marks in brackets. Q1 to Q4 and Q6 are Paper 1 style (no calculator). Q5 is Paper 2 style.
Q1. Given that cos θ = k and 0 < θ < π/2, write each of the following in terms of k.
(a) cos(π − θ) 1 mark
(b) sin(θ + π/2) 1 mark
(c) tan(π + θ) 2 marks
Q2. Show that cos(3π/2 − θ) + sin(π + θ) = −2 sin θ. 3 marks
Q3. Determine whether each function is odd, even or neither, giving a reason.
(a) f(x) = x² sin x 2 marks
(b) g(x) = x sin x + cos 2x 2 marks
(c) h(x) = sin x + cos x 2 marks
Q4. Solve sin x = −√3/2 for −π ≤ x ≤ 2π. 5 marks
Q5. Paper 2. One solution of cos x = 0.4 is x = 1.159, to 3 decimal places.
(a) Using the symmetry of the graph of y = cos x, write down all the solutions of cos x = 0.4 for −2π ≤ x ≤ 2π. 3 marks
(b) Hence solve cos x = −0.4 for 0 ≤ x ≤ 2π. 3 marks
Q6. The function f is defined by f(x) = sin x + sin 3x.
(a) Show that f(π − x) = f(x). 3 marks
(b) Hence write down the equation of a line of symmetry of the graph of y = f(x). 1 mark
(c) Show that f is an odd function. 2 marks
11In one breath
Put θ on the unit circle at (cos θ, sin θ): reflecting in the y-axis gives π − θ, where only sine keeps its sign; a half-turn gives π + θ, where only tangent does; reflecting in the x-axis gives −θ, where only cosine does. Adding π keeps the ratio, an odd multiple of π/2 swaps it for its co-ratio, and a compound angle expansion proves any of them. On the graphs these are symmetries: sine has mirror lines x = π/2 + kπ and centres (kπ, 0), cosine has mirror lines x = kπ and centres (π/2 + kπ, 0), tangent has centres (kπ/2, 0), no mirror lines and period π. Cosine is even, sine and tangent odd, and cos x = sin(x + π/2) makes cosine a translated sine. To solve an equation, take one solution α, add its partner (π − α for sine, 2π − α for cosine, α + π for tangent), and step by the period until you leave the interval.
Answers
Q1. (a) cos(π − θ) = −cos θ = −k. (b) sin(θ + π/2) = cos θ = k. (c) tan(π + θ) = tan θ = sin θ ÷ cos θ, and sin θ = √(1 − k²) because θ is acute, so tan(π + θ) = √(1 − k²)/k. (a) A1. (b) A1. (c) M1 for using tan(π + θ) = tan θ with sin θ = √(1 − k²), A1 for √(1 − k²)/k. A negative answer in (c) scores M1 A0.
Q2. cos(3π/2 − θ) = cos(3π/2)cos θ + sin(3π/2)sin θ = 0 − sin θ = −sin θ, and sin(π + θ) = sin π cos θ + cos π sin θ = −sin θ. So the sum is −sin θ − sin θ = −2 sin θ. M1 for expanding either term correctly with a compound angle identity or a justified unit circle argument, A1 for −sin θ for each term, A1 for the conclusion.
Q3. (a) f(−x) = (−x)² sin(−x) = −x² sin x = −f(x), so f is odd. (b) g(−x) = (−x)sin(−x) + cos(−2x) = x sin x + cos 2x = g(x), so g is even. (c) h(π/4) = √2/2 + √2/2 = √2 and h(−π/4) = −√2/2 + √2/2 = 0; since 0 is neither √2 nor −√2, h is neither. (a) M1 for substituting −x, A1 for "odd" with −f(x) shown. (b) M1, A1 likewise. (c) M1 for a counterexample or a general argument, A1 for "neither" with both comparisons made.
Q4. The reference angle is π/3 (sin(π/3) = √3/2). Sine is negative in the third and fourth quadrants, so in 0 ≤ x ≤ 2π, x = π + π/3 = 4π/3 and x = 2π − π/3 = 5π/3. Subtracting 2π gives −2π/3 and −π/3, both in −π ≤ x ≤ 2π. x = −2π/3, −π/3, 4π/3, 5π/3. A1 for the reference angle π/3, M1 for using symmetry to place the solutions where sin x < 0, A1 for 4π/3 and 5π/3, M1 for subtracting 2π, A1 for −2π/3 and −π/3. Extra values outside the interval lose the final A1.
Q5. (a) cos is even, so −1.159 is also a solution; the period 2π then gives 2π − 1.159 = 5.124 and −5.124. x = −5.124, −1.159, 1.159, 5.124. (b) cos(π − x) = −cos x, so if cos x = 0.4 then cos(π − x) = −0.4 and cos(π + x) = −0.4. From x = 1.159: π − 1.159 = 1.982 and π + 1.159 = 4.301. x = 1.98, 4.30 (3 s.f.). (a) A1 for −1.159, M1 for using the period, A1 for ±5.124. (b) M1 for using cos(π − x) = −cos x (or the graph), A1 for 1.98, A1 for 4.30. Solving (b) from scratch with arccos(−0.4) ignores "hence" and loses the M1.
Q6. (a) f(π − x) = sin(π − x) + sin(3π − 3x). sin(π − x) = sin x, and sin(3π − 3x) = sin(π − 3x + 2π) = sin(π − 3x) = sin 3x. So f(π − x) = sin x + sin 3x = f(x). (b) x and π − x are mirror images in x = π/2, and they give equal values of f, so x = π/2 is a line of symmetry. (c) f(−x) = sin(−x) + sin(−3x) = −sin x − sin 3x = −f(x), so f is odd. (a) M1 for sin(π − x) = sin x, M1 for simplifying sin(3π − 3x) using the period, A1 for the conclusion. (b) A1 for x = π/2. (c) M1 for substituting −x, A1 for −f(x) with the conclusion.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section AHL 3.11 Relationships between trigonometric functions and the symmetry properties of their graphs. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.