When do I use the sine rule and when the cosine rule?

Use the sine rule with a side and its opposite angle. Use the cosine rule with two sides and the included angle, or when you know all three sides.

Why must I use radians for arc length and sector area?

The formulas l = r theta and A = half r squared theta only work when theta is measured in radians.

How do I find all the solutions of a trig equation in a range?

Find the principal value, then use the graph or CAST to find the others: 180 - x for sine, 360 - x for cosine, x + 180 for tangent.

How do I prove a trig identity?

Start with one side only, use sin squared plus cos squared equals 1 and tan = sin/cos, and simplify until it equals the other side.

Year 12 trigonometry revision notes to download as PDF

This Year 12 trigonometry revision notes is written for pupils aged 16-17 and follows the National Curriculum. In short, it shows you how to use trig rules, radians, identities and solve trig equations. Before you start, it helps to revise straight lines and circles and binomial expansion and sequences, because this lesson builds on them.

Key points

Sine rule: \(\dfrac a{\sin A}=\dfrac b{\sin B}=\dfrac c{\sin C}\), used with a pair of opposite side and angle; beware the ambiguous case.
Cosine rule: \(a^2=b^2+c^2-2bc\cos A\) and \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\), used with two sides and the included angle or three sides.
Area of a triangle \(=\tfrac12ab\sin C\).
Radians: \(\pi\) radians \(=180^\circ\); arc length \(s=r\theta\); sector area \(=\tfrac12r^2\theta\); segment area \(=\tfrac12r^2(\theta-\sin\theta)\).
Exact values: \(\sin30^\circ=\tfrac12\), \(\cos30^\circ=\tfrac{\sqrt3}2\), \(\tan45^\circ=1\), \(\sin45^\circ=\tfrac{\sqrt2}2\), and \(\tan60^\circ=\sqrt3\); learn the triangles that give them.
Graphs of \(\sin\), \(\cos\) and \(\tan\): period \(360^\circ\) (\(2\pi\)) for sine and cosine, \(180^\circ\) (\(\pi\)) for tangent; \(y=a\sin(bx)+c\) has amplitude \(a\), period \(\tfrac{360^\circ}b\) and midline \(y=c\).

Table of Contents

Year 12 trigonometry revision notes: what you will learn

Year 12 trigonometry has three strands: solving any triangle, working in radians, and using graphs and identities to solve trig equations. Learn the exact values and the CAST idea, and always check whether the question wants degrees or radians.

Starter activity

A sailor leaves a harbour and travels 10 km due east and then 6 km on a bearing of 030\(^\circ\). She wants to sail home in a straight line. Sketch her route as a triangle and mark the angle at the turning point. How far is the journey home, and in what direction must she steer? Think about which rule could help if you know two sides and the angle between them, and which rule you could use afterwards to find the direction.

Sine rule, cosine rule and area

Rule: sine rule, cosine rule and area

For a triangle with sides \(a,b,c\) opposite angles \(A,B,C\):

Sine rule: \(\dfrac a{\sin A}=\dfrac b{\sin B}=\dfrac c{\sin C}\) (a side and its opposite angle are known).
Cosine rule: \(a^2=b^2+c^2-2bc\cos A\), or \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\) (two sides and the angle between them, or three sides).
Area \(=\tfrac12ab\sin C\).

Worked example 1

In triangle \(ABC\), \(AB=9\) cm, \(AC=12\) cm and angle \(A=40^\circ\). Find \(BC\) and the area.

By the cosine rule \(BC^2=81+144-2(9)(12)\cos40^\circ=59.5\), so \(BC=7.72\) cm. The area is \(\tfrac12(9)(12)\sin40^\circ=34.7\) cm\(^2\).

Watch out: the ambiguous case

If you are given two sides and a non-included angle there may be two triangles, because \(\sin\theta=\sin(180^\circ-\theta)\).

Worked example 2

In triangle \(ABC\), \(a=8\), \(b=10\) and \(A=30^\circ\). Find \(B\).

\(\sin B=\dfrac{10\sin30^\circ}{8}=0.625\), so \(B=38.7^\circ\) or \(B=180^\circ-38.7^\circ=141.3^\circ\). Both are valid because \(30^\circ+141.3^\circ<180^\circ\): there are two possible triangles.

Radians

Definition: radian

One radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. \(\pi\) radians \(=180^\circ\). To convert degrees to radians multiply by \(\dfrac\pi{180}\); to convert back multiply by \(\dfrac{180}\pi\). So \(150^\circ=\dfrac{5\pi}6\) and \(\dfrac{3\pi}4=135^\circ\).

Rule: arcs and sectors (\(\theta\) in radians)

Arc length \(l=r\theta\), sector area \(A=\tfrac12r^2\theta\), segment area \(=\tfrac12r^2(\theta-\sin\theta)\).

Worked example 3

A sector has radius 8 cm and area 24 cm\(^2\). Find the angle and the arc length.

\(24=\tfrac12(64)\theta\), so \(\theta=0.75\) rad and the arc length is \(8\times0.75=6\) cm.

Exact values

Key fact: exact values

\(\theta\)	\(0\)	\(30^\circ\ (\tfrac\pi6)\)	\(45^\circ\ (\tfrac\pi4)\)	\(60^\circ\ (\tfrac\pi3)\)	\(90^\circ\ (\tfrac\pi2)\)
\(\sin\theta\)	\(0\)	\(\tfrac12\)	\(\tfrac{\sqrt2}2\)	\(\tfrac{\sqrt3}2\)	\(1\)
\(\cos\theta\)	\(1\)	\(\tfrac{\sqrt3}2\)	\(\tfrac{\sqrt2}2\)	\(\tfrac12\)	\(0\)
\(\tan\theta\)	\(0\)	\(\tfrac1{\sqrt3}\)	\(1\)	\(\sqrt3\)	undefined

Graphs of trig functions

Key fact: sine, cosine and tangent graphs

\(y=\sin x\) and \(y=\cos x\) have period \(360^\circ\) (\(2\pi\)), amplitude 1, and lie between \(-1\) and \(1\). \(y=\tan x\) has period \(180^\circ\) (\(\pi\)) and vertical asymptotes at \(x=90^\circ,270^\circ,\dots\). For \(y=a\sin bx\) the amplitude is \(a\) and the period is \(\dfrac{360^\circ}b\).

Identities

Rule: two identities

\(\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta+\cos^2\theta\equiv1\) (Pythagoras in a right-angled triangle with hypotenuse 1). Rearranged: \(\sin^2\theta=1-\cos^2\theta\) and \(\cos^2\theta=1-\sin^2\theta\).

Worked example 4

Prove that \(\dfrac{\cos\theta}{1-\sin\theta}\equiv\dfrac{1+\sin\theta}{\cos\theta}\).

Multiply the left side by \(\dfrac{1+\sin\theta}{1+\sin\theta}\): \(\dfrac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}=\dfrac{\cos\theta(1+\sin\theta)}{\cos^2\theta}=\dfrac{1+\sin\theta}{\cos\theta}\). \(\blacksquare\)

Watch out

Work on one side only when proving an identity, and state the identity you use. Never work on both sides at once.

Solving trig equations

Method: solving \(\sin\theta=k\), \(\cos\theta=k\) or \(\tan\theta=k\)

Rearrange to \(\sin\theta=k\), \(\cos\theta=k\) or \(\tan\theta=k\) (use identities to get a quadratic in one function first).
Find the principal value \(\alpha\) with your calculator.
Use the graph or CAST to find the other solutions in the range: \(\sin\theta\): \(180^\circ-\alpha\); \(\cos\theta\): \(360^\circ-\alpha\); \(\tan\theta\): \(\alpha+180^\circ\).
For \(\sin2\theta\), solve for \(2\theta\) over the doubled range, then halve.

Worked example 5

Solve \(2\cos^2\theta+\sin\theta-1=0\) for \(0\le\theta\le360^\circ\).

Replace \(\cos^2\theta\) by \(1-\sin^2\theta\): \(2-2\sin^2\theta+\sin\theta-1=0\), so \(2\sin^2\theta-\sin\theta-1=0\) and \((2\sin\theta+1)(\sin\theta-1)=0\). Then \(\sin\theta=1\) gives \(\theta=90^\circ\); \(\sin\theta=-\tfrac12\) gives \(\theta=210^\circ,330^\circ\).

Worked example 6

Solve \(\tan\theta=-1\) for \(0\le\theta\le360^\circ\).

The calculator gives \(-45^\circ\), outside the range. Add \(180^\circ\) to get \(135^\circ\), then another \(180^\circ\) to get \(315^\circ\). So \(\theta=135^\circ\) or \(315^\circ\).

Watch out

Check the calculator mode (degrees or radians) and never use \(l=r\theta\) with \(\theta\) in degrees. Do not cancel \(\sin\theta\) from both sides of an equation: you lose solutions. Do not stop at the calculator answer: find all the solutions in the range.

Worked example 7: The sine rule and two possible angles

In triangle \(PQR\), \(PQ=8\) cm, \(QR=6\) cm and angle \(P=35^\circ\). Find the possible values of angle \(R\).

By the sine rule \(\dfrac{\sin R}{8}=\dfrac{\sin35^\circ}{6}\), so \(\sin R=\dfrac{8\sin35^\circ}{6}=0.7648\). Then \(R=49.9^\circ\) or \(R=180^\circ-49.9^\circ=130.1^\circ\). Check whether each fits in the triangle: with \(R=130.1^\circ\) the angle sum is \(35^\circ+130.1^\circ=165.1^\circ<180^\circ\), so it is valid; and \(35^\circ+49.9^\circ=84.9^\circ<180^\circ\) is also valid. Both triangles exist because the side opposite the given angle (6 cm) is shorter than the other given side (8 cm). This is the ambiguous case of the sine rule.

Worked example 8: Using the cosine rule to find an angle

A triangle has sides 7 cm, 9 cm and 12 cm. Find all three angles.

The largest angle is opposite 12 cm: \(\cos C=\dfrac{7^2+9^2-12^2}{2\times7\times9}=\dfrac{-14}{126}=-0.1111\), so \(C=96.4^\circ\). The angle opposite 9 cm: \(\cos B=\dfrac{7^2+12^2-9^2}{2\times7\times12}=\dfrac{112}{168}=0.6667\), so \(B=48.2^\circ\). Then \(A=180^\circ-96.4^\circ-48.2^\circ=35.4^\circ\). Check with the cosine rule for \(A\): \(\cos A=\dfrac{9^2+12^2-7^2}{2\times9\times12}=\dfrac{176}{216}=0.8148\), giving \(35.4^\circ\) ✓.

Worked example 9: Radians and arc length in a context

A pendulum of length 80 cm swings through an angle of 0.35 radians. Find the length of the arc traced by the bob and the area swept out by the string.

Arc length \(s=r\theta=80\times0.35=28\) cm. Area swept \(=\tfrac12r^2\theta=\tfrac12\times6400\times0.35=1\,120\) cm\(^2\). In degrees the swing is \(0.35\times\dfrac{180}{\pi}=20.05^\circ\), so about \(20^\circ\). Using radians makes the formulae \(s=r\theta\) and \(A=\tfrac12r^2\theta\) much simpler than the degree versions \(\dfrac{\theta}{360}\times2\pi r\) and \(\dfrac\theta{360}\times\pi r^2\).

Worked example 10: Exact values from special triangles

Use an equilateral triangle of side 2 to show that \(\sin30^\circ=\tfrac12\) and \(\cos30^\circ=\tfrac{\sqrt3}2\), and use a right-angled isosceles triangle to find \(\tan45^\circ\).

Draw an equilateral triangle with side 2 and split it down the middle. This gives a right-angled triangle with hypotenuse 2, shortest side 1 and height \(\sqrt{4-1}=\sqrt3\). The angle at the apex is \(30^\circ\), so \(\sin30^\circ=\tfrac12\), \(\cos30^\circ=\tfrac{\sqrt3}2\), \(\tan30^\circ=\tfrac1{\sqrt3}\), and for the angle \(60^\circ\), \(\sin60^\circ=\tfrac{\sqrt3}2\), \(\cos60^\circ=\tfrac12\) and \(\tan60^\circ=\sqrt3\). A right-angled isosceles triangle with equal sides 1 has hypotenuse \(\sqrt2\), so \(\sin45^\circ=\cos45^\circ=\tfrac1{\sqrt2}\) and \(\tan45^\circ=1\).

Worked example 11: Solving with a multiple angle in radians

Solve \(2\cos\left(3x-\dfrac\pi4\right)=\sqrt2\) for \(0\le x\le\pi\).

\(\cos\left(3x-\dfrac\pi4\right)=\dfrac{\sqrt2}2\). Let \(u=3x-\dfrac\pi4\); as \(x\) runs from 0 to \(\pi\), \(u\) runs from \(-\dfrac\pi4\) to \(\dfrac{11\pi}4\). \(\cos u=\dfrac{\sqrt2}2\) when \(u=\pm\dfrac\pi4+2n\pi\). The values in range are \(u=-\dfrac\pi4,\ \dfrac\pi4,\ \dfrac{7\pi}4,\ \dfrac{9\pi}4\). Then \(3x=u+\dfrac\pi4\) gives \(3x=0,\ \dfrac\pi2,\ 2\pi,\ \dfrac{5\pi}2\), so \(x=0,\ \dfrac\pi6,\ \dfrac{2\pi}3,\ \dfrac{5\pi}6\).

Worked example 12: Proving an identity by working on one side

Prove that \((\sin\theta+\cos\theta)^2\equiv1+2\sin\theta\cos\theta\) and use it to find \(\sin\theta\cos\theta\) when \(\sin\theta+\cos\theta=\tfrac75\).

Expand the bracket: \((\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=1+2\sin\theta\cos\theta\) using \(\sin^2\theta+\cos^2\theta=1\). If \(\sin\theta+\cos\theta=\tfrac75\) then \(\tfrac{49}{25}=1+2\sin\theta\cos\theta\), so \(2\sin\theta\cos\theta=\tfrac{24}{25}\) and \(\sin\theta\cos\theta=\tfrac{12}{25}\). These values agree with \(\sin\theta=\tfrac35\) and \(\cos\theta=\tfrac45\) since \(\tfrac35+\tfrac45=\tfrac75\) and \(\tfrac35\times\tfrac45=\tfrac{12}{25}\) ✓.

Worked example 13: A graph transformation question in trigonometry

Sketch \(y=2\cos(x-60^\circ)\) for \(0^\circ\le x\le360^\circ\), stating the coordinates of the maximum, the minimum and the \(y\)-intercept.

Start with \(y=\cos x\), which has a maximum at \((0^\circ,1)\). The factor 2 stretches the graph vertically so the amplitude is 2. Replacing \(x\) by \(x-60^\circ\) translates it \(60^\circ\) to the right. So the maximum is at \((60^\circ,2)\) and a second maximum would be at \(420^\circ\), outside the range. The minimum is at \((240^\circ,-2)\). The \(y\)-intercept is \(2\cos(-60^\circ)=2\times\tfrac12=1\), so \((0,1)\). The curve crosses the \(x\)-axis where \(x-60^\circ=90^\circ\) and \(270^\circ\), that is at \(x=150^\circ\) and \(x=330^\circ\).

Worked example 14: Using the three area formulae together

In triangle \(ABC\) the area is \(30\) cm\(^2\), \(AB=10\) cm and \(AC=8\) cm. Angle \(A\) is acute. Find angle \(A\) and the length of \(BC\).

Area \(=\tfrac12\times10\times8\sin A=40\sin A=30\), so \(\sin A=0.75\) and \(A=48.6^\circ\) (acute). Now the cosine rule: \(BC^2=10^2+8^2-2\times10\times8\cos48.6^\circ=164-160\times0.6614=58.2\), so \(BC=7.63\) cm. A useful check is the converse: the area with sides 10, 8 and included angle \(48.6^\circ\) is \(40\sin48.6^\circ=30.0\) cm\(^2\) ✓. If the question had allowed an obtuse angle, the second solution would have been \(A=131.4^\circ\) with \(BC^2=164+160\times0.6614=269.8\) and \(BC=16.4\) cm.

Note: choosing between degrees and radians

Degrees are convenient for bearings and for angles in diagrams of triangles. Radians are used whenever arc lengths, sector areas, small-angle approximations or calculus appear: the formulae \(s=r\theta\) and \(A=\tfrac12r^2\theta\) only hold when \(\theta\) is in radians. Read the question carefully. If an interval is given as \(0\le x\le2\pi\) your answers must be in radians, preferably exact multiples of \(\pi\) such as \(\tfrac\pi6\) or \(\tfrac{5\pi}3\). Before every trigonometry calculation check the mode of the calculator: a wrong mode is the most common source of lost marks.

Method: solving a trigonometric equation

Rewrite the equation so that a single trigonometric function appears (use \(\sin^2\theta+\cos^2\theta=1\) or \(\tan\theta=\sin\theta/\cos\theta\) if necessary).
If it is a quadratic in \(\sin\theta\), \(\cos\theta\) or \(\tan\theta\), factorise and solve each factor; reject values outside \(-1\) to \(1\) for sine and cosine.
Find the principal value on a calculator and use the symmetry of the graph (or a CAST diagram) to find all the solutions in the range.
For \(\sin(n\theta)\) and similar, widen the range first, solve, then divide by \(n\).
Never divide through by \(\sin\theta\) or \(\cos\theta\) unless you check separately the case where it is zero.

Watch out: common trigonometry mistakes

Having the calculator in degrees when the question uses radians, or vice versa.
Using the sine rule in the ambiguous case without checking whether \(180^\circ-\theta\) also works.
Writing \(\sin^2\theta\) as \(\sin\theta^2\) or \(2\sin\theta\) for \(\sin2\theta\). They are different.
Dividing by \(\sin\theta\) or \(\cos\theta\) when solving, which loses solutions.
Rounding too early; keep full calculator values until the final answer.

Did you know?

The word sine comes from a mistranslation. Indian mathematicians called the half-chord jya; Arab scholars wrote it as jiba, and when this was translated into Latin in the 12th century, it was confused with jaib (a fold or bay) and became sinus. Radians were first named by James Thomson in 1873, but the idea of measuring angles by arc length goes back to Roger Cotes in 1714. GPS and satellite navigation use the sine and cosine rules constantly.

Otto's tip

Sketch the graph first. The picture tells you how many solutions to expect in the range. Keep full calculator values until the last line and give 3 significant figures.

How to use this Year 12 trigonometry revision notes

First, read the key points so that you know where the Year 12 trigonometry revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the trigonometry worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the trigonometry test and mark it with the test answers.

Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 12 trigonometry revision notes gives you everything you need for the chapter.

To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.