3.11 Symmetry of trigonometric graphs · IB Maths AA

Key terms introduced here

Each is in the Maths AA glossary, with every other term in the subject.

Complementary angles: Two angles whose sum is a right angle, $\frac{\pi}{2}$ (or $90^\circ$ ); the angles $\theta$ and $\frac{\pi}{2} - \theta$ .
Point symmetry: A graph has point symmetry about a point $M$ if a rotation of $180^\circ$ about $M$ maps the graph onto itself; $M$ is the centre of symmetry.

Notes on 3.11 Symmetry of trigonometric graphs

20 min readLast reviewed 14 Oct 2026

Learning objectives

Derive the relationships between $\sin$ , $\cos$ and $\tan$ at $\theta$ , $-\theta$ , $\pi - \theta$ , $\pi + \theta$ and $\frac{\pi}{2} - \theta$ from the unit circle.
Describe each relationship as a symmetry of a graph: a line of symmetry, a centre of symmetry or a translation.
Determine whether a function built from circular functions is odd, even or neither.
Express a sine graph as a translation of a cosine graph, and the reverse.
Simplify trigonometric expressions and prove identities using the symmetry relationships.
Use symmetry to find every solution of a trigonometric equation from one known solution, including equations of the form $\sin A = \sin B$ and $\cos A = \cos B$ .

The graphs of sine, cosine and tangent are full of symmetry. The cosine curve is a mirror image of itself in the $y$ -axis, the sine curve is unchanged by a half-turn about the origin, and each curve lands on itself after a translation by one period. Every one of these geometric facts is also an algebraic identity, such as $\cos(-\theta) = \cos\theta$ or $\sin(\pi - \theta) = \sin\theta$ , true for every angle $\theta$ .

These identities turn hard-looking expressions into simple ones, connect the sine and cosine functions to each other, and explain why the solutions of a trigonometric equation come in symmetric families. They are also the reason a single calculator value is enough to find every solution of an equation.

Symmetry in the unit circle

On the unit circle, the point at angle $\theta$ is $P(\cos\theta, \sin\theta)$ . Reflecting or rotating $P$ in the axes and in the line $y = x$ gives four more points whose angles are simple expressions in $\theta$ . Comparing their coordinates with those of $P$ gives the symmetry relationships between the circular functions.

Unit circle with points at angles θ (cos θ, sin θ), π − θ (−cos θ, sin θ), π + θ (−cos θ, −sin θ), −θ (cos θ, −sin θ) at the corners of a dashed rectangle, and π/2 − θ (sin θ, cos θ), the reflection of θ in the dashed line y = x. — The four points at θ, π − θ, π + θ and −θ form a rectangle centred at O, so their coordinates differ only in sign. Reflecting in y = x swaps the two coordinates.

Transformation of $P$	New angle	New point	Relationship
Reflection in the $x$ -axis	$-\theta$	$(\cos\theta, -\sin\theta)$	$\cos(-\theta) = \cos\theta$ , $\sin(-\theta) = -\sin\theta$
Reflection in the $y$ -axis	$\pi - \theta$	$(-\cos\theta, \sin\theta)$	$\sin(\pi - \theta) = \sin\theta$ , $\cos(\pi - \theta) = -\cos\theta$
Rotation by $\pi$ about $O$	$\pi + \theta$	$(-\cos\theta, -\sin\theta)$	$\sin(\pi + \theta) = -\sin\theta$ , $\cos(\pi + \theta) = -\cos\theta$
Reflection in $y = x$	$\frac{\pi}{2} - \theta$	$(\sin\theta, \cos\theta)$	$\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta$ , $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$
Full turn	$\theta + 2\pi$	$(\cos\theta, \sin\theta)$	$\sin(\theta + 2\pi) = \sin\theta$ , $\cos(\theta + 2\pi) = \cos\theta$

The tangent relationships follow by dividing the sine by the cosine in each row:

\tan(-\theta) = -\tan\theta \qquad \tan(\pi - \theta) = -\tan\theta \qquad \tan(\pi + \theta) = \tan\theta \qquad \tan\left(\frac{\pi}{2} - \theta\right) = \frac{1}{\tan\theta}

Each holds wherever both sides are defined.

The figure is drawn with $\theta$ in the first quadrant, but the relationships hold for every angle. A reflection or rotation of the circle moves every point in the same way, whichever quadrant the point starts in, so the coordinates change in the same way too.

Negative angles: odd and even

Reflection in the $x$ -axis changes the sign of the angle. The $x$ -coordinate stays the same and the $y$ -coordinate changes sign, so

\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta \qquad \tan(-\theta) = -\tan\theta

In the language of 2.14, cosine is an even function and sine and tangent are odd functions.

Line symmetry of the cosine graph

Because $\cos(-x) = \cos x$ , the points $(x, \cos x)$ and $(-x, \cos x)$ are both on the graph of $y = \cos x$ . They are mirror images in the $y$ -axis, so the whole graph is symmetrical about the $y$ -axis.

y = cos x

The cosine graph is its own reflection in the y-axis: the points at θ and −θ are level with each other.

Point symmetry of the sine graph

Because $\sin(-x) = -\sin x$ , the points $(x, \sin x)$ and $(-x, -\sin x)$ are both on the graph of $y = \sin x$ . The second is the image of the first after a rotation of $180^\circ$ about the origin.

The graph of every odd function has point symmetry about the origin, and the graph of every even function has line symmetry in the $y$ -axis. So the graphs of $y = \sin x$ and $y = \tan x$ have point symmetry about $O$ , and the graph of $y = \cos x$ has line symmetry in the $y$ -axis.

y = sin x

A half-turn about the origin maps the sine graph onto itself: the points at θ and −θ are the same distance from O on opposite sides.

Determine whether each function is odd, even or neither.

(a) $f(x) = \sin x\cos x$ (b) $g(x) = x^2\cos 3x$ (c) $h(x) = \sin x + \cos x$

Show the solution

Replace $x$ by $-x$ and use $\sin(-x) = -\sin x$ , $\cos(-x) = \cos x$ .

(a) $f(-x) = \sin(-x)\cos(-x) = (-\sin x)(\cos x) = -f(x)$ , so $f$ is odd.

(b) $g(-x) = (-x)^2\cos(-3x) = x^2\cos 3x = g(x)$ , so $g$ is even.

(c) $h(-x) = -\sin x + \cos x$ . A single value shows that this is neither $h(x)$ nor $-h(x)$ in general: $h\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$ , but $h\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0$ , which is neither $\sqrt{2}$ nor $-\sqrt{2}$ . So $h$ is neither odd nor even.

Supplementary angles: π − θ

Reflection in the $y$ -axis turns the angle $\theta$ into $\pi - \theta$ (in degrees, $180^\circ - \theta$ ). The height of the point does not change, but its horizontal coordinate changes sign:

\sin(\pi - \theta) = \sin\theta \qquad \cos(\pi - \theta) = -\cos\theta \qquad \tan(\pi - \theta) = -\tan\theta

The first of these is behind the rule "the second solution of $\sin x = k$ is $\pi - \alpha$ " used in 3.8.

Lines of symmetry of the sine graph

Replacing $\theta$ by $\frac{\pi}{2} + t$ in $\sin(\pi - \theta) = \sin\theta$ gives

\sin\left(\frac{\pi}{2} - t\right) = \sin\left(\frac{\pi}{2} + t\right)

The values of $\sin x$ at equal distances $t$ either side of $x = \frac{\pi}{2}$ are the same, so the sine graph is symmetrical about the vertical line $x = \frac{\pi}{2}$ , which passes through its maximum. Because the graph repeats every $2\pi$ and is also symmetrical about the line through its minimum, it has a line of symmetry through every turning point: $x = \frac{\pi}{2} + k\pi$ for every integer $k$ .

y = sin x

The sine graph is symmetrical about the vertical line through each maximum and minimum, and has point symmetry about each zero, such as (π, 0).

The same reasoning applied to the other graphs gives a full list of their symmetries.

Graph	Lines of symmetry	Centres of symmetry
$y = \sin x$	$x = \frac{\pi}{2} + k\pi$ (through every maximum and minimum)	$(k\pi, 0)$ (every zero)
$y = \cos x$	$x = k\pi$ (through every maximum and minimum)	$\left(\frac{\pi}{2} + k\pi, 0\right)$ (every zero)
$y = \tan x$	none	$\left(\frac{k\pi}{2}, 0\right)$ (every zero and every point where an asymptote meets the $x$ -axis)

The centres of symmetry of the sine graph come from $\sin(\pi + t) = -\sin t$ and $\sin(\pi - t) = \sin t$ : together they give $\sin(\pi + t) = -\sin(\pi - t)$ , so the values at equal distances either side of $x = \pi$ are equal in size and opposite in sign.

Half-turns and periods: π + θ

Rotating $P$ through $\pi$ about the origin turns $\theta$ into $\pi + \theta$ and changes the sign of both coordinates:

\sin(\pi + \theta) = -\sin\theta \qquad \cos(\pi + \theta) = -\cos\theta \qquad \tan(\pi + \theta) = \frac{-\sin\theta}{-\cos\theta} = \tan\theta

The last identity says that $\tan$ has period $\pi$ . The first two say that translating the graph of $y = \sin x$ or $y = \cos x$ by $\pi$ to the left gives the same result as reflecting it in the $x$ -axis.

Combining the relationships gives others. For example, $\sin(\theta - \pi)$ can be simplified by writing it as $\sin(-(\pi - \theta))$ :

\sin(\theta - \pi) = -\sin(\pi - \theta) = -\sin\theta

using first that sine is odd and then the supplementary angle relationship. Alternatively, $\theta - \pi = (\theta + \pi) - 2\pi$ , so $\sin(\theta - \pi) = \sin(\theta + \pi) = -\sin\theta$ .

Complementary angles: π/2 − θ

In a right-angled triangle, the two acute angles are complementary. The side opposite one of them is adjacent to the other, so the sine of each angle is the cosine of the other. The unit circle extends this to every angle: reflecting $P$ in the line $y = x$ swaps its coordinates and turns the angle $\theta$ into $\frac{\pi}{2} - \theta$ , so

\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta \qquad \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta \qquad \tan\left(\frac{\pi}{2} - \theta\right) = \frac{1}{\tan\theta}

The name cosine records this: the cosine of an angle is the sine of its complement. (The reciprocal $\frac{1}{\tan\theta}$ is $\cot\theta$ , defined in 3.9.)

Prove that $\sin\left(\frac{\pi}{2} + \theta\right) = \cos\theta$ and $\cos\left(\frac{\pi}{2} + \theta\right) = -\sin\theta$ for all $\theta$ .

Show the solution

Write $\frac{\pi}{2} + \theta$ as $\frac{\pi}{2} - (-\theta)$ and use the complementary angle relationships, then the odd and even properties:

\sin\left(\frac{\pi}{2} + \theta\right) = \sin\left(\frac{\pi}{2} - (-\theta)\right) = \cos(-\theta) = \cos\theta

\cos\left(\frac{\pi}{2} + \theta\right) = \cos\left(\frac{\pi}{2} - (-\theta)\right) = \sin(-\theta) = -\sin\theta

Each step uses an identity that holds for every angle, so both results hold for all $\theta$ .

Translating between sine and cosine graphs

Reading $\sin\left(x + \frac{\pi}{2}\right) = \cos x$ as a statement about graphs: translating the sine curve $\frac{\pi}{2}$ to the left gives the cosine curve. In the same way $\cos\left(x - \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2} - x\right) = \sin x$ (cosine is even), so translating the cosine curve $\frac{\pi}{2}$ to the right gives the sine curve. Any sinusoidal function can therefore be written using either sine or cosine.

y = sin x
y = cos x = sin(x + π/2)

Each feature of the sine curve, such as the maximum at (π/2, 1), appears on the cosine curve π/2 further left, at (0, 1).

Relationship	Graph statement
$\cos x = \sin\left(x + \frac{\pi}{2}\right)$	$y = \cos x$ is $y = \sin x$ translated $\frac{\pi}{2}$ to the left
$\sin x = \cos\left(x - \frac{\pi}{2}\right)$	$y = \sin x$ is $y = \cos x$ translated $\frac{\pi}{2}$ to the right
$-\sin x = \cos\left(x + \frac{\pi}{2}\right)$	$y = -\sin x$ is $y = \cos x$ translated $\frac{\pi}{2}$ to the left
$-\cos x = \sin\left(x - \frac{\pi}{2}\right)$	$y = -\cos x$ is $y = \sin x$ translated $\frac{\pi}{2}$ to the right

These translations are not the only ones that work. Because the graphs are periodic, adding any multiple of $2\pi$ to the translation gives the same result; for example, $y = \cos x$ is also $y = \sin x$ translated $\frac{3\pi}{2}$ to the right.

(a) Find the smallest positive value of $p$ for which the graph of $y = \cos 2x$ is the graph of $y = \sin 2x$ translated $p$ units to the left.

(b) Hence write $f(x) = 4\cos\left(2x\right) - 1$ in the form $a\sin(b(x + c)) + d$ , where $a > 0$ , $b > 0$ and $0 < c < \pi$ .

Show the solution

(a) Translating $y = \sin 2x$ by $p$ to the left gives $y = \sin(2(x + p)) = \sin(2x + 2p)$ . This equals $\cos 2x = \sin\left(2x + \frac{\pi}{2}\right)$ for all $x$ when $2p = \frac{\pi}{2}$ , that is $p = \frac{\pi}{4}$ .

The translation is $\frac{\pi}{4}$ , not $\frac{\pi}{2}$ : the horizontal stretch halves every horizontal distance, including the gap between the sine and cosine curves.

(b) From part (a), $\cos 2x = \sin\left(2\left(x + \frac{\pi}{4}\right)\right)$ , so

f(x) = 4\sin\left(2\left(x + \frac{\pi}{4}\right)\right) - 1

with $a = 4$ , $b = 2$ , $c = \frac{\pi}{4}$ and $d = -1$ , and $0 < \frac{\pi}{4} < \pi$ .

Using symmetry to simplify expressions

The relationships reduce the circular function of any angle of the form $\pm\theta + \frac{k\pi}{2}$ to $\pm\sin\theta$ , $\pm\cos\theta$ or $\pm\tan\theta$ (or its reciprocal). Two checks decide the result:

Which function. Adding or subtracting $\pi$ (or $2\pi$ ) keeps the same function; adding or subtracting $\frac{\pi}{2}$ (or $\frac{3\pi}{2}$ ) swaps sine and cosine.
Which sign. Imagine $\theta$ as a small positive acute angle, find the quadrant of the new angle, and use the sign of the original function in that quadrant.

For example, for $\cos\left(\frac{3\pi}{2} + \theta\right)$ : the $\frac{3\pi}{2}$ swaps cosine to sine; with $\theta$ small, $\frac{3\pi}{2} + \theta$ is in the fourth quadrant, where cosine is positive. So $\cos\left(\frac{3\pi}{2} + \theta\right) = \sin\theta$ .

Given that $\sin\alpha = 0.6$ and $0 < \alpha < \frac{\pi}{2}$ , find the value of

(a) $\sin(\pi - \alpha)$ (b) $\cos(\pi + \alpha)$ (c) $\tan(-\alpha)$ (d) $\cos\left(\frac{\pi}{2} - \alpha\right)$ (e) $\sin\left(\alpha + \frac{3\pi}{2}\right)$

Show the solution

First find $\cos\alpha$ and $\tan\alpha$ . Since $\alpha$ is acute, $\cos\alpha > 0$ :

\cos\alpha = \sqrt{1 - 0.6^2} = 0.8 \qquad \tan\alpha = \frac{0.6}{0.8} = 0.75

(a) $\sin(\pi - \alpha) = \sin\alpha = 0.6$ .

(b) $\cos(\pi + \alpha) = -\cos\alpha = -0.8$ .

(c) $\tan(-\alpha) = -\tan\alpha = -0.75$ .

(d) $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha = 0.6$ .

(e) Adding $\frac{3\pi}{2}$ swaps sine to cosine. The angle $\alpha + \frac{3\pi}{2}$ is in the fourth quadrant, where sine is negative, so

\sin\left(\alpha + \frac{3\pi}{2}\right) = -\cos\alpha = -0.8

Check with the identities: $\sin\left(\alpha + \frac{3\pi}{2}\right) = \sin\left(\left(\alpha + \frac{\pi}{2}\right) + \pi\right) = -\sin\left(\frac{\pi}{2} + \alpha\right) = -\cos\alpha$ .

Expressions with several terms are simplified term by term, then collected.

Simplify fully

(a) $\sin(\pi - x) + \cos\left(\frac{\pi}{2} + x\right)$

(b) $\dfrac{\sin(\pi + \theta)\cos(-\theta)}{\cos\left(\frac{\pi}{2} - \theta\right)}$ , where $\sin\theta \neq 0$ .

Show the solution

(a) $\sin(\pi - x) = \sin x$ and $\cos\left(\frac{\pi}{2} + x\right) = -\sin x$ , so

\sin(\pi - x) + \cos\left(\frac{\pi}{2} + x\right) = \sin x - \sin x = 0

(b) Replace each factor:

\frac{(-\sin\theta)(\cos\theta)}{\sin\theta} = -\cos\theta

Cancelling $\sin\theta$ is allowed because $\sin\theta \neq 0$ is given.

Pairs of terms that are complementary or supplementary often cancel or combine, which is useful for sums with many terms.

Find the exact value of $\cos^2\left(\frac{\pi}{12}\right) + \cos^2\left(\frac{5\pi}{12}\right)$ .

Show the solution

The angles $\frac{\pi}{12}$ and $\frac{5\pi}{12}$ add up to $\frac{6\pi}{12} = \frac{\pi}{2}$ , so they are complementary:

\cos\left(\frac{5\pi}{12}\right) = \cos\left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \sin\left(\frac{\pi}{12}\right)

Therefore, by the Pythagorean identity,

\cos^2\left(\frac{\pi}{12}\right) + \cos^2\left(\frac{5\pi}{12}\right) = \cos^2\left(\frac{\pi}{12}\right) + \sin^2\left(\frac{\pi}{12}\right) = 1

Neither value had to be found.

Using symmetry to find all solutions

Every solution of a trigonometric equation is related to every other by a symmetry of the graph. Once one solution is known, the lines of symmetry give a second, and the period gives the rest.

Symmetry of a transformed graph

For a transformed function such as $f(x) = a\sin(bx) + d$ , the lines of symmetry move with the graph. They pass through the maxima and minima, which for $\sin(bx)$ are at $bx = \frac{\pi}{2} + k\pi$ , that is $x = \frac{\pi}{2b} + \frac{k\pi}{b}$ . If $x_1$ is a solution of $f(x) = k$ , its reflection in the nearest line of symmetry $x = m$ is $2m - x_1$ , which is also a solution.

Given that $x = 0.2$ is a solution of $3\sin 2x + 1 = k$ , find, without finding $k$ , the other solution of this equation in the interval $0 \le x \le \pi$ .

Show the solution

The graph of $y = 3\sin 2x + 1$ has period $\pi$ and its first maximum where $2x = \frac{\pi}{2}$ , at $x = \frac{\pi}{4}$ . The line $x = \frac{\pi}{4}$ is a line of symmetry, and $x = 0.2$ lies on the rising part just before it.

Reflecting $0.2$ in $x = \frac{\pi}{4}$ :

x = 2 \times \frac{\pi}{4} - 0.2 = \frac{\pi}{2} - 0.2 = 1.37 \ (3 \text{ s.f.})

Adding the period $\pi$ to either solution goes beyond $\pi$ , so the other solution is $x = 1.37$ .

(Check: $3\sin 0.4 + 1 = 2.168\ldots$ and $3\sin(2 \times 1.3708) + 1 = 3\sin 2.7416 + 1 = 2.168\ldots$ )

Equations of the form sin A = sin B and cos A = cos B

The symmetry relationships also say exactly when two angles have the same sine or the same cosine. The angles $B$ and $\pi - B$ have the same sine, and so does any angle a whole number of turns away from either. Similarly, $B$ and $-B$ have the same cosine, and the tangent repeats every $\pi$ :

\sin A = \sin B \iff A = B + 2k\pi \ \text{ or } \ A = \pi - B + 2k\pi

\cos A = \cos B \iff A = B + 2k\pi \ \text{ or } \ A = -B + 2k\pi

\tan A = \tan B \iff A = B + k\pi

where $k \in \mathbb{Z}$ . These give every solution of an equation in which the same function is applied to two different expressions.

Solve $\sin 3x = \sin x$ for $0 \le x \le \pi$ .

Show the solution

From $\sin A = \sin B$ with $A = 3x$ and $B = x$ :

Case 1: $3x = x + 2k\pi$ , so $2x = 2k\pi$ and $x = k\pi$ . In the interval: $x = 0$ or $x = \pi$ .

Case 2: $3x = \pi - x + 2k\pi$ , so $4x = \pi + 2k\pi$ and $x = \frac{\pi}{4} + \frac{k\pi}{2}$ . In the interval: $x = \frac{\pi}{4}$ or $x = \frac{3\pi}{4}$ .

x = 0, \ \frac{\pi}{4}, \ \frac{3\pi}{4}, \ \pi

Check $x = \frac{3\pi}{4}$ : $\sin\frac{9\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ .

When the two sides use different functions, a complementary angle relationship makes them the same function first.

Solve $\cos 2x = \sin x$ for $0 \le x \le 2\pi$ .

Show the solution

Write $\sin x = \cos\left(\frac{\pi}{2} - x\right)$ , so the equation is $\cos 2x = \cos\left(\frac{\pi}{2} - x\right)$ .

Case 1: $2x = \frac{\pi}{2} - x + 2k\pi$ , so $3x = \frac{\pi}{2} + 2k\pi$ and $x = \frac{\pi}{6} + \frac{2k\pi}{3}$ . In the interval: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ (the next, $\frac{13\pi}{6}$ , is too large).

Case 2: $2x = -\left(\frac{\pi}{2} - x\right) + 2k\pi$ , so $x = -\frac{\pi}{2} + 2k\pi$ . In the interval: $x = \frac{3\pi}{2}$ , already found.

x = \frac{\pi}{6}, \ \frac{5\pi}{6}, \ \frac{3\pi}{2}

The same answers come from writing $\cos 2x = 1 - 2\sin^2 x$ and solving a quadratic in $\sin x$ (3.8). The two methods are equivalent; the symmetry method gives the general solution directly.

Exam-style questions on symmetry of trigonometric graphs

No calculator.

(a) Show that $\cos\left(\frac{\pi}{2} + \theta\right) + \sin(\pi + \theta) = -2\sin\theta$ .

(b) Hence solve $\cos\left(\frac{\pi}{2} + \theta\right) + \sin(\pi + \theta) = 1$ for $0 \le \theta \le 2\pi$ .

(c) The graph of $y = \cos\left(x + \frac{\pi}{2}\right)$ can be obtained from the graph of $y = \sin x$ by a single transformation. Describe two different transformations that do this.

Show the solution

(a) $\cos\left(\frac{\pi}{2} + \theta\right) = \cos\left(\frac{\pi}{2} - (-\theta)\right) = \sin(-\theta) = -\sin\theta$ , and $\sin(\pi + \theta) = -\sin\theta$ (rotation by $\pi$ on the unit circle). So

\cos\left(\frac{\pi}{2} + \theta\right) + \sin(\pi + \theta) = -\sin\theta - \sin\theta = -2\sin\theta

Each identity used is named, because the result is given.

(b) Using part (a), $-2\sin\theta = 1$ , so $\sin\theta = -\frac{1}{2}$ . Sine is negative in the third and fourth quadrants and the related angle is $\frac{\pi}{6}$ :

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{or} \quad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

(c) Since $\cos\left(x + \frac{\pi}{2}\right) = -\sin x$ , the graph is $y = -\sin x$ . It is obtained from $y = \sin x$ by a reflection in the $x$ -axis, or by a translation of $\pi$ units horizontally (left or right), because $\sin(x \pm \pi) = -\sin x$ . (A rotation of $180^\circ$ about the origin does not work: since sine is odd, it maps the sine curve onto itself.)

The second question combines symmetry with a known exact value.

No calculator. It is given that $\cos\frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}$ .

(a) Find the exact values of $\cos\frac{4\pi}{5}$ , $\cos\frac{6\pi}{5}$ and $\sin\frac{3\pi}{10}$ .

(b) Show that $\sin(2\pi - \theta) = -\sin\theta$ for all $\theta$ .

(c) Hence find the value of $\displaystyle\sum_{k=1}^{9} \sin\left(\frac{k\pi}{5}\right)$ .

Show the solution

(a) $\frac{4\pi}{5} = \pi - \frac{\pi}{5}$ , so $\cos\frac{4\pi}{5} = -\cos\frac{\pi}{5} = -\frac{1 + \sqrt{5}}{4}$ .

$\frac{6\pi}{5} = \pi + \frac{\pi}{5}$ , so $\cos\frac{6\pi}{5} = -\cos\frac{\pi}{5} = -\frac{1 + \sqrt{5}}{4}$ .

$\frac{3\pi}{10} = \frac{\pi}{2} - \frac{\pi}{5}$ , so $\sin\frac{3\pi}{10} = \cos\frac{\pi}{5} = \frac{1 + \sqrt{5}}{4}$ .

(b) Sine has period $2\pi$ and is odd, so $\sin(2\pi - \theta) = \sin(-\theta) = -\sin\theta$ .

(c) Pair the term with $k$ with the term with $10 - k$ . Since $\frac{(10 - k)\pi}{5} = 2\pi - \frac{k\pi}{5}$ , part (b) gives

\sin\left(\frac{(10 - k)\pi}{5}\right) = -\sin\left(\frac{k\pi}{5}\right)

so the pairs $k = 1$ and $9$ , $2$ and $8$ , $3$ and $7$ , $4$ and $6$ each add up to $0$ . The remaining term is $k = 5$ : $\sin\pi = 0$ . So

\sum_{k=1}^{9} \sin\left(\frac{k\pi}{5}\right) = 0

This reflects the point symmetry of the sine graph about $(\pi, 0)$ .

Self-check

To do:
I can derive the relationships for $-\theta$ , $\pi - \theta$ , $\pi + \theta$ and $\frac{\pi}{2} - \theta$ from reflections and rotations of the unit circle.
To do:
I can explain that cosine is even and sine and tangent are odd, and what this means for their graphs.
To do:
I can state the lines of symmetry and centres of symmetry of the sine, cosine and tangent graphs.
To do:
I can write a sine graph as a translation of a cosine graph, and the reverse, including for $\sin(bx)$ and $\cos(bx)$ .
To do:
I can simplify expressions such as $\cos\left(\frac{3\pi}{2} + \theta\right)$ by deciding the function and the sign.
To do:
I can use complementary and supplementary angles to evaluate sums and exact values.
To do:
I can find a second solution of an equation by reflecting a known solution in a line of symmetry.
To do:
I can solve equations of the form $\sin A = \sin B$ and $\cos A = \cos B$ by considering both cases.

Questions on 3.11

15 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on 20 October 2026

Question 1 · AA-001877 marks

It is given that $\sin\dfrac{\pi}{9} = p$ , where $0 < p < 1$ .

(a)(i)
Write down, in terms of $p$ , the value of $\sin\dfrac{8\pi}{9}$ .
1 mark
Markscheme (a)(i)
$\sin\dfrac{8\pi}{9} = \sin\left(\pi - \dfrac{\pi}{9}\right) = p$ A1
(a)(ii)
Write down, in terms of $p$ , the value of $\sin\dfrac{10\pi}{9}$ .
1 mark
Markscheme (a)(ii)
$\sin\dfrac{10\pi}{9} = \sin\left(\pi + \dfrac{\pi}{9}\right) = -p$ A1
(b)
Find, in terms of $p$ , the value of $\cos\dfrac{11\pi}{18}$ .
2 marks
Markscheme (b)
$\cos\dfrac{11\pi}{18} = \cos\left(\dfrac{\pi}{2} + \dfrac{\pi}{9}\right)$ (M1)
$= -\sin\dfrac{\pi}{9} = -p$ A1
Note: Accept use of the compound angle identity for the (M1).
(c)
Find, in terms of $p$ , the value of $\tan\dfrac{17\pi}{9}$ .
3 marks
Markscheme (c)
$\cos\dfrac{\pi}{9} = \sqrt{1 - p^2}$ (positive since $\dfrac{\pi}{9}$ is acute) (A1)
$\tan\dfrac{17\pi}{9} = \tan\left(2\pi - \dfrac{\pi}{9}\right) = -\tan\dfrac{\pi}{9}$ (M1)
$= -\dfrac{p}{\sqrt{1 - p^2}}$ A1
Note: Award (A1)(M1)A0 for $\dfrac{p}{\sqrt{1-p^2}}$ .

Question 2 · AA-001887 marks

Let $\alpha$ be the solution of the equation $\cos x = 0.35$ for which $0 < \alpha < \dfrac{\pi}{2}$ .

(a)
Find the value of $\alpha$ .
1 mark
(b)
Write down, in terms of $\alpha$ , the four solutions of $\cos x = 0.35$ for $-2\pi \le x \le 2\pi$ .
2 marks
(c)
Hence solve the equation $\cos x = -0.35$ for $0 \le x \le 2\pi$ .
2 marks
(d)
Hence solve the equation $\cos 2x = -0.35$ for $0 \le x \le \pi$ .
2 marks

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Question 3 · AA-001898 marks

A large observation wheel rotates at a constant speed. The height, $h$ metres, of a capsule above the ground $t$ minutes after it passes the lowest point of the wheel is modelled by

h(t) = 20 - 18\cos\left(\frac{\pi t}{15}\right), \quad t \ge 0.

(a)
Write down the time taken for the wheel to complete one rotation.
1 mark
(b)
The model can also be written in the form $h(t) = 18\sin\left(\dfrac{\pi}{15}(t - q)\right) + 20$ , where $q > 0$ . Find the smallest possible value of $q$ .
3 marks
(c)
Find the first time at which the capsule is 30 metres above the ground.
2 marks
(d)
By using the symmetry of the graph of $h$ , find the length of time, during each rotation, for which the capsule is more than 30 metres above the ground.
2 marks

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Try a quick-fire question

One of 34 short multiple-choice questions on 3.11. Pick an answer to see why it's right or wrong.

The diagram shows $y = \sin x$ . Which line is a line of symmetry of the curve?

Graph of y = sin x for −π ≤ x ≤ 2π, with x-axis marks at −π, −π/2, π/2, π, 3π/2 and 2π and y-axis marks at −1 and 1.

Key terms introduced here

Notes on 3.11 Symmetry of trigonometric graphs

Symmetry in the unit circle

Negative angles: odd and even

Line symmetry of the cosine graph

Point symmetry of the sine graph

Supplementary angles: π − θ

Lines of symmetry of the sine graph

Half-turns and periods: π + θ

Complementary angles: π/2 − θ

Translating between sine and cosine graphs

Using symmetry to simplify expressions

Using symmetry to find all solutions

Symmetry of a transformed graph

Equations of the form sin A = sin B and cos A = cos B

Exam-style questions on symmetry of trigonometric graphs

Self-check

Questions on 3.11

Markscheme (a)(i)

Markscheme (a)(ii)

Markscheme (b)

Markscheme (c)

Try a quick-fire question

Carry on with 3.11 in Augment