Trigonometry: A-level Maths OCR revision notes

1.05

Radians, arcs and sectors

Radians and conversion 1.05

Definitions

Radian: the angle subtended at the centre of a circle by an arc equal in length to the radius.

Key results

A full turn is $2 π$ radians, so $180° = π$ radians.
Degrees to radians: $θ_{rad} = θ_{deg} \times \frac{π}{180}$ ; radians to degrees: $θ_{deg} = θ_{rad} \times \frac{180}{π}$ .
Common conversions: $30° = \frac{π}{6}$ , $45° = \frac{π}{4}$ , $60° = \frac{π}{3}$ , $90° = \frac{π}{2}$ .

Notes

Radians are the natural unit for calculus involving trigonometric functions — the derivative of $sin θ$ is only exactly $cos θ$ when $θ$ is measured in radians, which is one reason A-level mathematics moves to radians for anything beyond elementary triangle geometry.
Set the calculator to the mode the question uses; an answer of the right shape in the wrong mode is the most common avoidable error in this topic.
An answer 'in terms of $π$ ' (e.g. $\frac{π}{3}$ ) is expected to stay exact — converting it to a decimal first and then back loses the exact form a question may specifically ask for.

Arc length, sector area and segment area 1.05

Key results

Arc length: $s = r θ$ .
Sector area: $A = \frac{1}{2} r^{2} θ$ .
Area of the triangle formed by the two radii and the chord: $\frac{1}{2} r^{2} sin θ$ .
Segment area, being the sector minus that triangle: $\frac{1}{2} r^{2} (θ - sin θ)$ .

Method

Convert the angle to radians if it is given in degrees.
Apply $s = r θ$ or $A = \frac{1}{2} r^{2} θ$ as required.
For a perimeter, remember to add the two radii to the arc; for a segment, subtract the triangle from the sector.

Notes

Both formulas require $θ$ in radians — using degrees inflates the answer by a factor of $\frac{180}{π}$ .
In $\frac{1}{2} r^{2} (θ - sin θ)$ the $θ$ is a pure number in radians while $sin θ$ must be evaluated in radian mode; mixing the two is the usual cause of a wrong segment area.
A sector's perimeter includes the two straight radii as well as the arc — a common slip is quoting only the arc length $r θ$ as if it were the whole perimeter.

Worked example

A sector of a circle of radius $8$ cm subtends an angle of $0.6$ radians at the centre.

Find

(a) the arc length,

(b) the perimeter of the sector,

(d) the area of the corresponding minor segment, to 3 significant figures where rounding is needed.

Show worked solution

(a):

s = r θ = 8 \times 0.6 = 4.8

(b) The perimeter is the arc plus the two radii:

4.8 + 8 + 8 = 20.8

(c):

A = \frac{1}{2} r^{2} θ = \frac{1}{2} \times 64 \times 0.6 = 19.2

$^{2}$ .

(d) The triangle formed by the two radii has area:

\frac{1}{2} r^{2} sin θ = \frac{1}{2} \times 64 \times sin (0.6) = 32 \times 0.564642 = 18.0686

$^{2}$ , so the segment area is:

19.2 - 18.0686 = 1.13

$^{2}$ (3 s.f.).

Note $sin (0.6)$ is evaluated in radian mode; as a sense check, $0.6$ radians is:

0.6 \times \frac{180}{π} = 34.4°

a narrow sector, so a segment much smaller than the sector is expected.

1.05

Trigonometric graphs and identities

Graphs, periods and exact values 1.05

Key results

$y = sin θ$ and $y = cos θ$ both have period $360°$ ( $2 π$ ) and range $- 1 \leq y \leq 1$ ; the cosine graph is the sine graph translated left by $90°$ .
$y = tan θ$ has period $180°$ ( $π$ ), takes all real values, and has vertical asymptotes where $cos θ = 0$ , that is at odd multiples of $\frac{π}{2}$ .
Symmetries: $sin (- θ) = - sin θ$ and $cos (- θ) = cos θ$ ; $sin (180° - θ) = sin θ$ ; $cos (360° - θ) = cos θ$ .
Exact values: $sin 30° = \frac{1}{2}$ , $cos 30° = \frac{3}{2}$ , $tan 30° = \frac{1}{3}$ .
Exact values: $sin 45° = cos 45° = \frac{2}{2}$ , $tan 45° = 1$ .
Exact values: $sin 60° = \frac{3}{2}$ , $cos 60° = \frac{1}{2}$ , $tan 60° = 3$ .

Notes

Recognising these shapes is essential for sketching transformations and for judging how many solutions a trigonometric equation has on a given interval.
$y = a sin (b θ) + c$ has amplitude $∣ a ∣$ , period $\frac{360°}{b}$ and is centred on the line $y = c$ — read those three numbers straight off the equation rather than plotting points.
$tan θ$ 's asymptotes are easy to forget when sketching — mark them first (at $90°$ , $270°$ , etc.) before drawing the curve, so the branches are never drawn crossing them.

Reciprocal functions and the Pythagorean identities 1.05

Definitions

$sec θ = \frac{1}{cos θ}$ , $csc θ = \frac{1}{sin θ}$ , $cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}$ , each undefined wherever its denominator is zero.

Key results

$sin^{2} θ + cos^{2} θ \equiv 1$ , which is Pythagoras' theorem applied to the unit circle.
Dividing that identity by $cos^{2} θ$ : $1 + tan^{2} θ \equiv sec^{2} θ$ .
Dividing it instead by $sin^{2} θ$ : $1 + cot^{2} θ \equiv csc^{2} θ$ .
$tan θ \equiv \frac{sin θ}{cos θ}$ , the identity that converts between the three basic functions.

Notes

There is no need to memorise the two derived identities separately: divide $sin^{2} θ + cos^{2} θ \equiv 1$ by $cos^{2} θ$ or $sin^{2} θ$ and read off the result in a few seconds.
An identity ( $\equiv$ ) holds for every value of $θ$ for which both sides are defined, whereas an equation ( $=$ ) holds only for particular values — the distinction matters when a question says 'prove' rather than 'solve'.
$sec θ$ , $csc θ$ and $cot θ$ are undefined wherever their defining fraction's denominator is zero — state this restriction alongside any identity involving them, exactly as with any other fraction.

1.05

Trigonometric equations

Solving trigonometric equations over a given range 1.05

Key results

If $sin θ = k$ has principal value $θ_{0}$ , the other solution in $0° \leq θ \leq 360°$ is $180° - θ_{0}$ .
If $cos θ = k$ has principal value $θ_{0}$ , the other solution in that range is $360° - θ_{0}$ .
If $tan θ = k$ has principal value $θ_{0}$ , further solutions are $θ_{0} + 180° n$ for integer $n$ .

Method

If the equation mixes different trigonometric functions, use an identity to reduce it to a single function first, then factorise it as a quadratic in that function if necessary.
Isolate $sin θ$ , $cos θ$ or $tan θ$ and take the inverse function to get the principal value.
Use the symmetry of the relevant graph, or the CAST diagram, to generate the other solution in the first full turn.
Add or subtract whole periods ( $360°$ for sine and cosine, $180°$ for tangent) to collect every solution lying in the stated range.
Discard any solutions outside the range and state the full list.

Notes

Solving over a specified range means finding every solution in that range, not just the principal value a calculator returns.
A quadratic in $sin θ$ may factorise to give two values of $sin θ$ ; reject any with $∣ sin θ ∣ > 1$ before searching for angles, since no real angle produces them.
Never divide an equation through by $sin θ$ or $cos θ$ : that discards the solutions where the divisor is zero. Factorise instead.
Sketching the relevant graph across the given range before solving algebraically gives a quick check on how many solutions to expect, catching a missed or duplicated answer immediately.

Equations in $k θ$ or $θ + α$ : rescaling the range 1.05

Method

Substitute $u = k θ$ (or $u = θ + α$ ).
Apply the same operation to both ends of the given range for $θ$ to get the corresponding range for $u$ .
Solve the equation for $u$ over that adjusted range, collecting every solution.
Convert each solution back to $θ$ by dividing by $k$ (or subtracting $α$ ).

Notes

Solving for $u$ over the original, unadjusted range is a very common source of missing or extra solutions — an equation in $2 θ$ over $0° \leq θ \leq 360°$ needs solutions for $u$ across $0° \leq u \leq 720°$ , and so typically has twice as many.
As a quick check, an equation in $k θ$ over a full turn usually has $k$ times as many solutions as the same equation in $θ$ .
Always convert the final answers for $u$ back to $θ$ before finishing — leaving the answer in terms of $u$ is a common way to lose marks despite otherwise correct working.

Worked examples

Worked example

Solve $2 sin θ - 1 = 0$ for $0° \leq θ \leq 360°$ .

Show worked solution

Rearranging: $sin θ = 0.5$ .

The principal value is $θ = 30°$ .

Since sine is also positive in the second quadrant, the second solution is:

θ = 180° - 30° = 150°

So $θ = 30°$ or $θ = 150°$ .

Worked example

Solve:

sin 2 θ = \frac{3}{2}

for $0° \leq θ \leq 360°$ .

Show worked solution

Substitute $u = 2 θ$ ; the range for $θ$ rescales to $0° \leq u \leq 720°$ .

The principal value of:

sin u = \frac{3}{2}

is $u = 60°$ , and the second solution in the first turn is:

u = 180° - 60° = 120°

Adding a full period of $360°$ to each gives $u = 420°$ and $u = 480°$ , both still within $720°$ .

Dividing each by $2$ :

θ = 30°, 60°, 210°, 240°

Check one:

sin (2 \times 210°) = sin 420° = sin 60° = \frac{3}{2}

Worked example

Solve $2 cos^{2} θ + sin θ = 1$ for $0° \leq θ \leq 360°$ .

Show worked solution

Use $cos^{2} θ = 1 - sin^{2} θ$ to write everything in terms of $sin θ$ :

2 (1 - sin^{2} θ) + sin θ = 1

so:

2 - 2 sin^{2} θ + sin θ - 1 = 0

giving:

2 sin^{2} θ - sin θ - 1 = 0

Factorising:

(2 sin θ + 1) (sin θ - 1) = 0

so:

sin θ = - \frac{1}{2}

or $sin θ = 1$ .

From:

sin θ = - \frac{1}{2}

the related acute angle is $30°$ and sine is negative in the third and fourth quadrants, giving $θ = 210°$ and $θ = 330°$ .

From $sin θ = 1$ , $θ = 90°$ .

So:

θ = 90°, 210°, 330°

Check $θ = 210°$ :

cos 210° = - \frac{3}{2}

so:

2 (\frac{3}{4}) + (- \frac{1}{2}) = 1.5 - 0.5 = 1

1.05

Addition and double-angle formulae

The addition formulae 1.05

Key results

$sin (A \pm B) = sin A cos B \pm cos A sin B$ .
$cos (A \pm B) = cos A cos B \mp sin A sin B$ — note the reversed sign on the right.
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$ .
These give exact values for non-standard angles, e.g. $cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30° = \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} = \frac{6 + 2}{4}$ .

Notes

$sin (A + B)$ is not $sin A + sin B$ ; the formulae exist precisely because the naive version is false.
The cosine formula takes the opposite sign to the one in the bracket, which is worth checking against a known case such as $cos (90° + 90°) = - 1$ .
These formulae also let you find exact values of $sin$ , $cos$ or $tan$ for any angle built from $30°$ , $45°$ and $60°$ by addition or subtraction — a faster route than a calculator when an exact surd answer is required.

The double-angle formulae 1.05

Key results

Setting $A = B = θ$ in the addition formulae gives $sin 2 θ = 2 sin θ cos θ$ .
$cos 2 θ = cos^{2} θ - sin^{2} θ$ , and applying $sin^{2} θ + cos^{2} θ \equiv 1$ turns this into $cos 2 θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ .
$tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ .
Rearranging the cosine versions gives the forms used for integration: $cos^{2} θ = \frac{1 + cos 2 θ}{2}$ and $sin^{2} θ = \frac{1 - cos 2 θ}{2}$ .

Notes

Choose the version of $cos 2 θ$ that matches the rest of the equation: use $1 - 2 sin^{2} θ$ when the other terms involve $sin θ$ , and $2 cos^{2} θ - 1$ when they involve $cos θ$ , so that everything reduces to a single function.
There is no need to memorise the double-angle formulae separately — each is one substitution away from the addition formulae.
An equation mixing $sin θ$ and $cos 2 θ$ (or $sin 2 θ$ ) almost always needs a double-angle substitution first to reduce everything to a single angle and a single function before it can be solved.

Worked example

Prove that:

\frac{sin 2 θ}{1 + cos 2 θ} \equiv tan θ

stating where the identity is valid.

Show worked solution

Take the left-hand side and apply the double-angle formulae, choosing the version of $cos 2 θ$ that simplifies the denominator: the numerator is $2 sin θ cos θ$ , and the denominator is:

1 + (2 cos^{2} θ - 1) = 2 cos^{2} θ

So the left-hand side is:

\frac{2 sin θ cos θ}{2 cos ^{2} θ} = \frac{sin θ}{cos θ} = tan θ

as required.

The identity holds for all $θ$ with $cos θ \neq = 0$ , that is $θ \neq = 90° + 180° n$ .

Check at $θ = 30°$ : the left-hand side is:

\frac{sin 60°}{1 + cos 60°} = \frac{3 /2}{1.5} = \frac{1}{3} = tan 30°

1.05

The harmonic form R sin(θ + α)

The harmonic form $R sin (θ + α)$ 1.05

Key results

Any expression $a sin θ + b cos θ$ can be written as a single wave $R sin (θ + α)$ with $R = a^{2} + b^{2}$ , $R > 0$ .
The angle satisfies $cos α = \frac{a}{R}$ and $sin α = \frac{b}{R}$ , hence $tan α = \frac{b}{a}$ .
The maximum value of the expression is $R$ , occurring when $sin (θ + α) = 1$ , and the minimum is $- R$ , occurring when $sin (θ + α) = - 1$ .
The same expression can equally be written as $R cos (θ - β)$ with the same $R$ ; only the phase angle changes.

Method

Compute $R = a^{2} + b^{2}$ .
Find $α$ from $tan α = \frac{b}{a}$ , checking the signs of $cos α$ and $sin α$ to place $α$ in the correct quadrant.
Rewrite the equation in terms of the single function, then solve it using the range-rescaling method for $θ + α$ .
Expand $R sin (θ + α)$ back out as a check that the original coefficients are recovered.

Notes

This form turns an otherwise awkward equation into one with a single trigonometric function to solve, and makes the maximum and minimum immediately visible without any calculus.
Keep $α$ to more decimal places than the final answer needs while working, since rounding it early can shift a final angle by a tenth of a degree.
$R$ is always taken as the positive square root — a negative $R$ would just be absorbed into a shifted $α$ , so by convention $R > 0$ always.

Worked example

(a) Express $3 sin θ + 4 cos θ$ in the form $R sin (θ + α)$ , where $R > 0$ and:

0° < α < 90°

giving $α$ to one decimal place.

(b) Hence solve $3 sin θ + 4 cos θ = 2$ for $0° \leq θ \leq 360°$ , giving answers to one decimal place.

Show worked solution

(a):

R = 3^{2} + 4^{2} = 9 + 16 = 25 = 5

Then:

cos α = \frac{3}{5}

and:

sin α = \frac{4}{5}

so:

tan α = \frac{4}{3}

giving $α = 53.1°$ .

Therefore:

3 sin θ + 4 cos θ = 5 sin (θ + 53.1°)

As a check, expanding the right-hand side gives:

5 (sin θ cos 53.1° + cos θ sin 53.1°) = 5 (0.6 sin θ + 0.8 cos θ) = 3 sin θ + 4 cos θ

confirming the result.

(b) The equation becomes $5 sin (θ + α) = 2$ , so:

sin (θ + α) = 0.4

Put $u = θ + α$ with $α = 53.130°$ kept to three decimals while working; the range rescales to:

53.130° \leq u \leq 413.130°

The principal value $u = 23.578°$ lies below the range, so the solutions in range are:

u = 180° - 23.578° = 156.422°

and:

u = 360° + 23.578° = 383.578°

Subtracting $α$ : $θ = 103.3°$ or $θ = 330.4°$ .

Check $θ = 103.3°$ :

3 sin 103.3° + 4 cos 103.3° = 2.919 - 0.922 = 2.00

θ + 53.130° = 90°

i.e.

$θ = 36.9°$ (and the minimum is $- 5$ at $θ = 216.9°$ ).

Radians, arcs and sectors

Radians and conversion 1.05

Arc length, sector area and segment area 1.05

Trigonometric graphs and identities

Graphs, periods and exact values 1.05

Reciprocal functions and the Pythagorean identities 1.05

Trigonometric equations

Solving trigonometric equations over a given range 1.05

Equations in kθ or θ+α: rescaling the range 1.05

Addition and double-angle formulae

The addition formulae 1.05

The double-angle formulae 1.05

The harmonic form R sin(θ + α)

The harmonic form Rsin(θ+α) 1.05

Equations in $k θ$ or $θ + α$ : rescaling the range 1.05

The harmonic form $R sin (θ + α)$ 1.05