Trigonometry Revision Notes — 9709 Paper 1

Sine, cosine and tangent as functions

Syllabus requirement · §1.5

“

sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using either degrees or radians).

”

The problem with SOH-CAH-TOA

At IGCSE, $\sin\theta$ meant opposite over hypotenuse in a right-angled triangle. That definition has a hard limit built into it: a right-angled triangle cannot contain an angle of $210^\circ$ , because the three angles must add to $180^\circ$ and one of them is already $90^\circ$ . So the old definition simply has nothing to say about $\sin 210^\circ$ .

And yet your calculator answers at once: $\sin 210^\circ = -0.5$ . Exam questions need angles like this all the time — a question gives you $\cos\theta = -\tfrac12$ and expects $\theta = 120^\circ$ and $240^\circ$ . For that to make sense, sine and cosine need a new definition that works for an angle of any size, positive or negative. Everything else in this note is built on it.

The unit circle definition

Draw a circle of radius $1$ centred at the origin. Start at the point $(1, 0)$ and rotate anticlockwise through an angle $\theta$ , arriving at a point $P$ . Then, by definition:

\cos\theta = \text{the } x\text{-coordinate of } P \qquad\qquad \sin\theta = \text{the } y\text{-coordinate of } P

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}

There is no triangle to run out of, so $\theta$ can be anything: $210^\circ$ , $-40^\circ$ , $1000^\circ$ . A negative angle means rotating clockwise instead. The tangent is undefined whenever $x = 0$ , because you cannot divide by zero.

Two quick checks.

It agrees with the old definition. For an acute $\theta$ , drop a perpendicular from $P$ to the $x$ -axis. You get a right-angled triangle with hypotenuse $1$ (the radius), opposite side $y$ and adjacent side $x$ . So $\dfrac{\text{opp}}{\text{hyp}} = \dfrac{y}{1} = y = \sin\theta$ . The new definition contains the old one.

It gives $\sin^2\theta + \cos^2\theta = 1$ straight away. $P$ lies on a circle of radius $1$ , so by Pythagoras $x^2 + y^2 = 1$ . Substituting $x = \cos\theta$ and $y = \sin\theta$ , $\cos^2\theta + \sin^2\theta = 1.$ You will use this identity a great deal in the section on the two identities. It is Pythagoras written in different letters.

cos θ and sin θ are the two coordinates of the point P you reach after rotating θ anticlockwise round a circle of radius 1. Their signs depend only on which quadrant P is in.

Signs, read off the picture

Because $\cos\theta$ and $\sin\theta$ are coordinates, their signs are just the signs of $x$ and $y$ — and everybody already knows those.

First quadrant ( $0^\circ$ to $90^\circ$ ): $x > 0$ and $y > 0$ , so $\sin$ , $\cos$ and $\tan$ are all positive.
Second quadrant ( $90^\circ$ to $180^\circ$ ): $x < 0$ , $y > 0$ . So $\sin$ is positive; $\cos$ is negative; and $\tan = \tfrac{y}{x}$ is negative.
Third quadrant ( $180^\circ$ to $270^\circ$ ): $x < 0$ , $y < 0$ . So $\sin$ and $\cos$ are both negative, and $\tan$ — negative over negative — is positive.
Fourth quadrant ( $270^\circ$ to $360^\circ$ ): $x > 0$ , $y < 0$ . So $\cos$ is positive, $\sin$ negative, $\tan$ negative.

Some students remember this as "CAST" (going anticlockwise from the fourth quadrant: Cos, All, Sin, Tan are the positive ones). You do not need a memory aid: picture which quadrant you are in, remember that $\cos$ is the across-coordinate and $\sin$ the up-coordinate, and the signs follow.

A clean demonstration: $\theta = 210^\circ$

$210^\circ$ is $180^\circ + 30^\circ$ , so rotating $210^\circ$ lands you $30^\circ$ past the negative $x$ -axis, in the third quadrant.

The size of each value. The angle between $OP$ and the $x$ -axis is $210^\circ - 180^\circ = 30^\circ$ . That angle is called the related acute angle (some books say "reference angle"). The right-angled triangle it makes is exactly the $30^\circ$ triangle, so the two coordinates have sizes $\cos 30^\circ = \tfrac{\sqrt3}{2}$ and $\sin 30^\circ = \tfrac12$ .

The signs. Third quadrant: both coordinates negative. So

\cos 210^\circ = -\frac{\sqrt3}{2}, \qquad \sin 210^\circ = -\frac12, \qquad \tan 210^\circ = \frac{-\tfrac12}{-\tfrac{\sqrt3}{2}} = \frac{1}{\sqrt3}

Check the tangent against the sign rule: third quadrant, $\tan$ positive. ✓ And check $\sin 210^\circ = -0.5$ against your calculator. ✓

This is the whole method, and it is worth naming because you will use it constantly:

Find the related acute angle — how far $\theta$ is from the nearest part of the $x$ -axis ( $0^\circ$ , $180^\circ$ or $360^\circ$ ).
Work out the size of the ratio from that acute angle.
Put on the sign that the quadrant demands.

Related acute angle, by quadrant

0^\circ\text{–}90^\circ: \ \alpha = \theta

First quadrant — nothing to do

90^\circ\text{–}180^\circ: \ \alpha = 180^\circ - \theta

Second quadrant

180^\circ\text{–}270^\circ: \ \alpha = \theta - 180^\circ

Third quadrant

270^\circ\text{–}360^\circ: \ \alpha = 360^\circ - \theta

Fourth quadrant

From the circle to the three graphs

Now let $\theta$ increase steadily from $0^\circ$ and plot the height $y = \sin\theta$ against $\theta$ .

At $0^\circ$ the point is at $(1,0)$ , height $0$ .
Rising to $90^\circ$ , the point climbs to the top of the circle: height $1$ .
Falling back to $180^\circ$ , height returns to $0$ .
Down to $270^\circ$ : height $-1$ at the bottom.
Back to $360^\circ$ : height $0$ again, and the point is exactly where it started.

That traced-out shape is the sine wave, and the last line explains why it repeats every $360^\circ$ : going round again retraces the identical journey. Plot the across-coordinate $x = \cos\theta$ instead and you get the same wave started at its maximum, because at $\theta = 0$ the point is already at the far right.

The tangent graph is different in kind. $\tan\theta = \tfrac{y}{x}$ blows up whenever $x = 0$ , which happens at $90^\circ$ and $270^\circ$ — so the tangent curve has vertical asymptotes there rather than a smooth peak. And because the point $(x, y)$ and the point $(-x, -y)$ diametrically opposite it give the same quotient $\tfrac{y}{x}$ , the tangent graph repeats after only half a turn: its period is $180^\circ$ , not $360^\circ$ .

Sine starts at 0 and rises; cosine starts at 1 and falls; tangent repeats every 180° and shoots off to infinity at 90° and 270°.

	$y=\sin x$	$y=\cos x$	$y=\tan x$
Period	360° (2π)	360° (2π)	180° (π)
Range	$-1 \leqslant y \leqslant 1$	$-1 \leqslant y \leqslant 1$	all real values
Value at 0°	0	1	0
Maximum	$1$ , at $90^\circ$	$1$ , at $0^\circ$ and $360^\circ$	none
Minimum	$-1$ , at $270^\circ$	$-1$ , at $180^\circ$	none
Cuts the $x$ -axis at	$0^\circ, 180^\circ, 360^\circ$	$90^\circ, 270^\circ$	$0^\circ, 180^\circ, 360^\circ$
Symmetry	about $x = 90^\circ$	about $x = 0^\circ$	rotational about the origin
Asymptotes	none	none	$x = 90^\circ,\ 270^\circ,\ \ldots$

The tangent column is the one that catches people: half the period of the other two, and no maximum or minimum at all.

The four symmetry facts you will actually use

Finding all the solutions of an equation, later in this note, comes out of these facts, and each one is visible on the graphs above.

\sin(180^\circ - \theta) = \sin\theta \qquad \cos(360^\circ - \theta) = \cos\theta \qquad \tan(\theta + 180^\circ) = \tan\theta

and, running the curves backwards through $0^\circ$ ,

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

Read the first one off the sine graph: the curve is a mirror image in the vertical line $x = 90^\circ$ , so $30^\circ$ and $150^\circ$ sit at the same height. Read the second off the cosine graph: it is a mirror image in the vertical line $x = 0^\circ$ and, because it repeats every $360^\circ$ , in $x = 360^\circ$ too — so $50^\circ$ and $310^\circ$ sit at the same height. The third is just "the tangent graph repeats every $180^\circ$ ".

In radians, replace $180^\circ$ by $\pi$ and $360^\circ$ by $2\pi$ . Nothing else changes.

Using symmetry instead of a calculator

You are told that $\sin 40^\circ = 0.643$ , correct to 3 significant figures. Without using a calculator, write down

(a) another angle between $0^\circ$ and $360^\circ$ whose sine is $0.643$ ,
(b) the value of $\sin 140^\circ$ , $\sin 220^\circ$ and $\sin 320^\circ$ ,
(c) the value of $\cos 320^\circ$ , given that $\cos 40^\circ = 0.766$ .

Show full working

1
(a) Sine is positive in the first and second quadrants, so the partner angle is in the second quadrant. Use the mirror symmetry about $90^\circ$ : $180^\circ - 40^\circ = 140^\circ$
A horizontal line drawn across the sine curve between 0° and 180° cuts it twice, and because the curve is symmetric about 90°, the two crossings add up to 180°.
2
(b) $\sin 140^\circ$ is that partner, so $\sin 140^\circ = 0.643$ — the same value, positive, because $140^\circ$ is in the second quadrant where sine is positive.
3
$220^\circ = 180^\circ + 40^\circ$ , so its related acute angle is $220^\circ - 180^\circ = 40^\circ$ and it lies in the third quadrant, where sine is negative: $\sin 220^\circ = -0.643$
Size from the related acute angle; sign from the quadrant. Students often get the size right and the sign wrong.
4
$320^\circ = 360^\circ - 40^\circ$ , related acute angle $40^\circ$ , fourth quadrant, where sine is negative: $\sin 320^\circ = -0.643$
5
(c) $320^\circ$ is in the fourth quadrant, where cosine is positive. Its related acute angle is again $40^\circ$ , so $\cos 320^\circ = +0.766$
The same angle, the same related acute angle — but a different ratio, so a different sign. This is why you must ask about the sign separately for each of sin, cos and tan.

Answer

(a) $140^\circ$ (b) $\sin 140^\circ = 0.643$ , $\sin 220^\circ = -0.643$ , $\sin 320^\circ = -0.643$ (c) $\cos 320^\circ = 0.766$

Get into the habit of saying two separate sentences to yourself: "the size comes from the related acute angle" and "the sign comes from the quadrant". Almost every sign error in this topic is caused by trying to do both at once.

Common mistakes

$\sin 210^\circ = \sin 30^\circ = \tfrac12$
$\sin 210^\circ = -\tfrac12$
The related acute angle gives the size only. 210° is in the third quadrant, where the y-coordinate — and therefore the sine — is negative.
Giving $y = \tan x$ a maximum of $1$ and a minimum of $-1$
$\tan x$ takes every real value; it has no maximum and no minimum
Only sin and cos are coordinates on a unit circle and so trapped in [−1, 1]. tan is a quotient of them and runs off to infinity at 90° and 270°.
Drawing $y = \tan x$ with period $360^\circ$
Period $180^\circ$ — two complete branches between $0^\circ$ and $360^\circ$
Getting the tangent period wrong doubles or halves the number of solutions you find when solving equations, which loses the final accuracy mark.
Treating $\cos(-40^\circ)$ as $-\cos 40^\circ$
$\cos(-40^\circ) = \cos 40^\circ$
Cosine is the across-coordinate, and rotating 40° clockwise instead of anticlockwise does not change it. Sine and tangent do change sign; cosine does not.

Your turn

Do all of these without a calculator except where told otherwise — the point is the quadrant reasoning, not the arithmetic.

1
State the quadrant each angle lies in, its related acute angle, and the sign of its sine, cosine and tangent.
(a) $160^\circ$ (b) $250^\circ$ (c) $295^\circ$ (d) $-70^\circ$
Stuck? Show hint
For (d), a negative angle means rotating clockwise from $(1, 0)$ . Where do you end up after turning $70^\circ$ clockwise?
Show solution
1. 1
  (a) $160^\circ$ is between $90^\circ$ and $180^\circ$ , so it is in the second quadrant. Related acute angle $= 180^\circ - 160^\circ = 20^\circ$ .
2. 2
  Second quadrant: $x < 0$ , $y > 0$ . So $\sin$ is positive, $\cos$ is negative, $\tan$ is negative.
3. 3
  (b) $250^\circ$ is between $180^\circ$ and $270^\circ$ : third quadrant. Related acute angle $= 250^\circ - 180^\circ = 70^\circ$ .
4. 4
  Third quadrant: $x < 0$ , $y < 0$ . So $\sin$ negative, $\cos$ negative, $\tan$ positive.
  The third quadrant is the one people get wrong: two negatives divide to give a positive tangent.
5. 5
  (c) $295^\circ$ is between $270^\circ$ and $360^\circ$ : fourth quadrant. Related acute angle $= 360^\circ - 295^\circ = 65^\circ$ .
6. 6
  Fourth quadrant: $x > 0$ , $y < 0$ . So $\sin$ negative, $\cos$ positive, $\tan$ negative.
7. 7
  (d) Turning $70^\circ$ clockwise from $(1,0)$ lands you below the positive $x$ -axis, which is the fourth quadrant (it is the same place as $360^\circ - 70^\circ = 290^\circ$ ). Related acute angle $70^\circ$ ; $\sin$ negative, $\cos$ positive, $\tan$ negative.
  Adding 360° to a negative angle to bring it into 0°–360° is always safe — it is a whole extra turn, so it lands on exactly the same point.
Answer
(a) 2nd, $20^\circ$ ; $+,-,-$ (b) 3rd, $70^\circ$ ; $-,-,+$ (c) 4th, $65^\circ$ ; $-,+,-$ (d) 4th, $70^\circ$ ; $-,+,-$
2
Given that $\cos 25^\circ = 0.906$ and $\tan 25^\circ = 0.466$ , write down the value of
(a) $\cos 155^\circ$ (b) $\cos 335^\circ$ (c) $\tan 205^\circ$ (d) $\tan 155^\circ$
Stuck? Show hint
Write each angle as $180^\circ \pm 25^\circ$ or $360^\circ - 25^\circ$ first, so the related acute angle is obvious.
Show solution
1. 1
  (a) $155^\circ = 180^\circ - 25^\circ$ , so the related acute angle is $25^\circ$ and the quadrant is the second.
2. 2
  In the second quadrant cosine is negative, so $\cos 155^\circ = -0.906$
3. 3
  (b) $335^\circ = 360^\circ - 25^\circ$ : related acute angle $25^\circ$ , fourth quadrant, where cosine is positive. $\cos 335^\circ = 0.906$
4. 4
  (c) $205^\circ = 180^\circ + 25^\circ$ : related acute angle $25^\circ$ , third quadrant, where tangent is positive. $\tan 205^\circ = 0.466$
  This is the direct statement of tan(θ + 180°) = tan θ — adding half a turn leaves the tangent completely unchanged.
5. 5
  (d) $155^\circ = 180^\circ - 25^\circ$ : second quadrant, where tangent is negative. $\tan 155^\circ = -0.466$
Answer
(a) $-0.906$ (b) $0.906$ (c) $0.466$ (d) $-0.466$
3
(a) Sketch $y = \cos x$ for $-360^\circ \leqslant x \leqslant 360^\circ$ , marking the coordinates of every maximum, minimum and $x$ -axis crossing.
(b) Hence state how many solutions the equation $\cos x = 0.3$ has in that interval.
(c) How many solutions does $\tan x = 5$ have in $0^\circ \leqslant x \leqslant 720^\circ$ ?
Show solution
1. 1
  (a) Cosine starts at its maximum. Over $-360^\circ$ to $360^\circ$ you fit two complete cycles.
2. 2
  Maxima (where $y = 1$ ): $(-360^\circ, 1)$ , $(0^\circ, 1)$ , $(360^\circ, 1)$ . Minima (where $y = -1$ ): $(-180^\circ, -1)$ and $(180^\circ, -1)$ .
3. 3
  Crossings of the $x$ -axis are the odd multiples of $90^\circ$ : $-270^\circ$ , $-90^\circ$ , $90^\circ$ , $270^\circ$ .
  Mark schemes for sketches check that the curve is in the right place, not just the right shape, so mark the key points.
4. 4
  (b) Draw the horizontal line $y = 0.3$ . It lies between $-1$ and $1$ , so it cuts the curve. Each complete cycle of the cosine curve is cut twice by any horizontal line strictly between $-1$ and $1$ .
5. 5
  There are two complete cycles in $-360^\circ \leqslant x \leqslant 360^\circ$ , so there are $2 \times 2 = \mathbf{4}$ solutions.
  Counting cycles then doubling is more reliable than trying to list the angles. It is also the check to run at the end of every trigonometric equation you solve.
6. 6
  (c) The tangent graph has period $180^\circ$ , so $0^\circ$ to $720^\circ$ contains $720 \div 180 = 4$ complete branches.
7. 7
  A horizontal line cuts each branch of the tangent graph exactly once (each branch sweeps from $-\infty$ to $+\infty$ ), so there are $\mathbf{4}$ solutions.
Answer
(b) $4$ solutions (c) $4$ solutions

Practise graph questionsReal past-paper questions · Graphs of sine, cosine and tangent functions

Checking your access…

Can you do all of these?

Sketch $y = \sin x$ , $y = \cos x$ and $y = \tan x$ for $0^\circ \leqslant x \leqslant 360^\circ$ from memory
State the amplitude, period and range of $y = 1 - 3\cos 2x$
Write down $\cos 150^\circ$ and $\sin\tfrac{3\pi}{4}$ exactly, with the right signs
Explain why $\sin^{-1}(0.5)$ is a single number but $\sin\theta = 0.5$ has many solutions
Prove an identity by converting everything to $\sin$ and $\cos$ and working one side only
Turn an equation containing $\tan\theta$ and $\sin\theta$ into a quadratic in one function
Reject an impossible value such as $\cos\theta = 1.4$ , and say why
Find all four solutions of a quadratic in $\cos\theta$ over $0^\circ \leqslant \theta \leqslant 360^\circ$
Solve $\sin 2\theta = 0.5$ over $0^\circ \leqslant \theta \leqslant 360^\circ$ and get all four angles
Derive $\sin^2\theta + \cos^2\theta = 1$ from Pythagoras on a triangle of hypotenuse $1$
Write both rearrangements, $\sin^2\theta = 1-\cos^2\theta$ and $\cos^2\theta = 1-\sin^2\theta$ , without pausing
Expand $(\sin\theta + \cos\theta)^2$ correctly — all three terms
Factorise a numerator so a whole bracket cancels, rather than cancelling out of a sum
Rewrite $\dfrac{1}{\tan\theta}$ as $\dfrac{\cos\theta}{\sin\theta}$ , and combine fractions over a common denominator
Substitute $u = \sin^2\theta$ to turn a quartic into a quadratic, and restore the $\pm$ when you square-root
Transform the interval for $2\theta$ or $\theta + 30^\circ$ , then transform every answer back
Work in radians on an interval that does not start at zero, discarding a principal value that falls outside it
Find the asymptotes and axis crossings of $y = \tan\left(x + \tfrac14\pi\right)$
Given $\sin\beta = a$ with $\beta$ obtuse, write $\cos\beta = -\sqrt{1 - a^2}$ with the right sign
Solve $5\tan^3\theta = 2\tan\theta$ by factorising, keeping $\tan\theta = 0$ and both signs of the square root