1The idea in one paragraph

In a right-angled triangle sine and cosine are ratios of sides, which only makes sense for angles between 0 and 90°. The unit circle frees them. Draw a circle of radius 1 about the origin, turn an arm through the angle θ from the positive x-axis, and call the point where the arm meets the circle P. Then cos θ is the x-coordinate of P and sin θ is its y-coordinate, for any angle at all: obtuse, reflex, negative, or more than a full turn. For acute angles this agrees with SOH CAH TOA; for the rest it is the definition. The signs come from the quadrant, large angles borrow the values of the acute angle the arm makes with the x-axis, tan θ is the gradient of the arm, and because sin θ = sin(π − θ), the sine rule can be satisfied by two angles.

2From the triangle to the circle

The unit circle is the circle with centre O(0, 0) and radius 1. Angles are measured from the positive x-axis: anticlockwise is positive and clockwise is negative. The arm at angle θ meets the circle at P, and

P = (cos θ, sin θ). Cosine is the x-coordinate, sine is the y-coordinate, for every angle θ.

Figure 1(a) shows why this is not a new idea. For an acute θ, drop a perpendicular from P to the x-axis. The triangle has hypotenuse OP = 1, so SOH CAH TOA gives adjacent = cos θ and opposite = sin θ: exactly the coordinates of P. Figure 1(b) keeps the same definition when θ is obtuse: P is now to the left of the y-axis, so cos θ is negative, while sin θ is still positive.

Three facts come straight out of the picture.

• Both values lie between −1 and 1. P is on a circle of radius 1, so neither coordinate can be bigger than 1 in size: −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. An equation such as sin x = 1.3 has no solution, and 3.8 will use that.
• A full turn changes nothing. θ and θ + 2π put the arm in the same place, so sin(θ + 2π) = sin θ and cos(θ + 2π) = cos θ. This is why 3.7 calls these functions periodic.
• The four axis angles can be read off. At θ = 0, P = (1, 0); at π/2, (0, 1); at π, (−1, 0); at 3π/2, (0, −1). So cos π = −1, sin(3π/2) = −1, and so on, with no triangle needed.

A circle of radius r scales everything by r: the point at distance r from O along the arm at angle θ is (r cos θ, r sin θ), which is how 3.7 places a seat on a wheel.

Example. The arm at 150° makes 30° with the negative x-axis, above the axis and to the left, so P = (−cos 30°, sin 30°) = (−√3/2, 1/2): cos 150° = −√3/2 and sin 150° = 1/2.

3Signs in the four quadrants

Because cos θ is an x-coordinate and sin θ a y-coordinate, their signs are the signs of x and y in whichever quadrant P lies. Figure 2 collects them.

Learn the reason, not the table: the quadrant tells you the signs of x and y. The word CAST, read anticlockwise from the fourth quadrant (Cos, All, Sin, Tan), is a quick check.

4Tangent, and the line through the origin

The booklet defines tan θ = sin θ ÷ cos θ, and the unit circle says what that means. OP runs from (0, 0) to (cos θ, sin θ), so its gradient is sin θ ÷ cos θ. tan θ is the gradient of the arm. Figure 3 also shows the name: extended, OP meets the tangent line x = 1 at height tan θ.

tan θ is undefined when cos θ = 0, that is at θ = π/2, 3π/2 and every odd multiple of π/2. The arm is vertical there, and a vertical line has no gradient (2.1). Everywhere else tan θ can take any real value, positive or negative, with no upper limit.

The line through the origin at angle θ. A line through O with gradient tan θ has equation y = x tan θ, where θ is the angle from the positive x-axis to the line. The line through O at 2π/3 to the positive x-axis is y = x tan(2π/3) = −√3 x. Going the other way, the line y = −x has gradient −1, so tan θ = −1 and θ = 3π/4 (135°), the angle between 0 and π whose tangent is −1.

5Related angles: one acute angle does all the work

Figure 4 is the most useful picture on the page. Take an acute angle θ and its point P = (cos θ, sin θ). Three other angles give the same coordinates up to sign.

• π − θ is P reflected in the y-axis: (−cos θ, sin θ).
• π + θ is P turned half a turn about O: (−cos θ, −sin θ).
• −θ, which is the same arm as 2π − θ, is P reflected in the x-axis: (cos θ, −sin θ).

Reading off coordinates, and dividing y by x for tangent, gives the relationships the guide lists.

Two of these deserve a name. cos(−θ) = cos θ, so cosine is unchanged when the angle changes sign (in 3.7 this makes its graph symmetric in the y-axis). sin(−θ) = −sin θ, so sine changes sign with the angle. And one more, from reflecting in the line y = x, which swaps the coordinates: sin(π/2 − θ) = cos θ and cos(π/2 − θ) = sin θ. That is the old fact that the sine of an angle in a right-angled triangle is the cosine of the other acute angle.

Although Figure 4 draws θ acute, the rules hold for every θ, because they come from reflections and turns of the whole plane. That is what lets you simplify expressions.

The method for any angle. To find a ratio of an angle that is not acute:

1. Find the quadrant P lies in, after removing whole turns of 2π.
2. Find the related acute angle: the acute angle between the arm and the x-axis (not the y-axis).
3. Take the ratio of that acute angle, and give it the sign of the quadrant.

For 4π/3: it is π + π/3, so P is in the third quadrant and the related acute angle is π/3. Cosine is negative there, so cos(4π/3) = −cos(π/3) = −1/2.

6Exact values

Most angles have sines and cosines that can only be written as decimals. A handful have exact values, surds and fractions that are the precise answer, and Paper 1 is built on them. They all come from two triangles, shown in Figure 5.

Half a square. Cut a square of side 1 along a diagonal. The angles are π/4 (45°), π/4 and π/2, and Pythagoras gives the diagonal √(1² + 1²) = √2. So sin(π/4) = cos(π/4) = 1/√2 = √2/2, and tan(π/4) = 1.

Half an equilateral triangle. Cut an equilateral triangle of side 2 down the middle. The half has angles π/3 (60°), π/6 (30°) and π/2, sides 1 and 2, and height √(2² − 1²) = √3. Opposite π/6 is the side of 1, opposite π/3 is the side √3.

A pattern makes the sine row easy to carry: √0/2, √1/2, √2/2, √3/2, √4/2. The cosine row is the same list backwards. Note that 1/√3 and √3/3 are the same exact value, one with the root rationalised; so are 1/√2 and √2/2. Either form earns the mark unless a question asks for a particular one.

Multiples. Every multiple of π/6 or π/4 is a special angle at a related acute angle of π/6, π/4 or π/3 (or an axis angle), so its ratios are the table's values with the quadrant's sign. Figure 6 puts all sixteen on the circle.

Six worked in a row, each by the three-step method of section 5.

7The ambiguous case of the sine rule

In 3.2 the sine rule was used only when it gave one answer. Here is the case it can give two. The source is a fact from section 5:

sin(180° − θ) = sin θ. So sin B = k, with 0 < k < 1, is satisfied by an acute B and by the obtuse angle 180° − B. The sine rule cannot tell them apart.

Your calculator's sin⁻¹ gives only the acute one. Whether the obtuse one also makes a triangle is your job to check: it does if the obtuse angle plus the given angle is less than 180°.

When it happens. You are given two sides and an angle that is not between them, and the given angle is opposite the shorter of the two sides. Figure 7 shows the reason. Fix angle A and side AC = b. The side a = BC is hinged at C and swings like a pendulum; the base line through A is crossed by the circle of radius a about C in two places, B₁ and B₂. Both triangles AB₁C and AB₂C contain angle A, side b and side a. The triangle CB₁B₂ is isosceles, so its base angles are equal, and that makes the angle at B₁ in one triangle and the angle at B₂ in the other add up to 180°.

Figure 8 shows all four possibilities, with h = b sin A the height of C above the base. If a < h the swinging side never reaches the base; if a = h it just touches (one right-angled triangle); if h < a < b it crosses twice; if a ≥ b the second crossing is behind A, so one triangle survives.

Worked example (Paper 2). In triangle ABC, angle A = 35°, AC = 12 cm and BC = 8 cm. Find the two possible sizes of angle ABC, and the length AB in each case.

Carry the unrounded B forward in the calculator, and give angles to 3 significant figures (59.4° and 121°).

A check by the cosine rule. With AB = c, a² = b² + c² − 2bc cos A gives 64 = 144 + c² − 24c cos 35°, the quadratic c² − 19.66c + 80 = 0. Its two positive roots, 13.9 and 5.75, are the ambiguous case seen from the other side.

Is it always ambiguous? No. With A = 35°, AC = 8 and BC = 12, the given angle is opposite the longer side: sin B = 8 sin 35° ÷ 12 = 0.382, so B = 22.5° or 157.5°, and 157.5° + 35° > 180°. Only 22.5° survives. The check decides it; always write it.