5 Trigonometry — notes on one page

5.1 · Understand and use the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form ½ab sin C; work with radian measure, including use for arc length and area of sector.

For a unit-circle angle $\theta$ , $\cos\theta$ and $\sin\theta$ are the point's horizontal and vertical coordinates, while $\tan\theta=\sin\theta/\cos\theta$ where $\cos\theta\neq0$ ; this extends the ratios beyond acute angles.
For a triangle, use $a/\sin A=b/\sin B=c/\sin C$ when an opposite side-angle pair is known, $a^2=b^2+c^2-2bc\cos A$ for three sides or an included angle, and $\tfrac12ab\sin C$ for area.
In the ambiguous sine-rule case, test the supplementary angle because $\sin A=\sin(\pi-A)$ , then reject any triangle inconsistent with the given sides and angles.
Radian measure makes arc and sector formulae direct: an angle $\theta$ radians in a circle of radius $r$ gives arc length $s=r\theta$ and sector area $A=\tfrac12r^2\theta$ .
Keep the calculator in the required angle mode and label sides opposite their matching angles; using degrees in $s=r\theta$ or pairing the wrong side and angle is a common error.

A labelled triangle: each lower-case side is opposite its matching upper-case angle.

Worked example

Two sides of a triangular sail are $8\,\text{m}$ and $11\,\text{m}$ , with included angle $0.9$ radians. Calculate the third side and the area of the sail, giving each answer to $3$ significant figures.

1.The cosine rule gives $c^2=8^2+11^2-2(8)(11)\cos(0.9)=75.597\ldots$ , so $c=8.694\ldots\,\text{m}=8.69\,\text{m}$ .
2.The area is $\tfrac12(8)(11)\sin(0.9)=34.466\ldots\,\text{m}^2$ , hence $34.5\,\text{m}^2$ .

Answer: Third side $=8.69\,\text{m}$ Area $=34.5\,\text{m}^2$

Common mistakes

Don't use degrees in $s=r\theta$ or $A=\tfrac12r^2\theta$ , even though these formulae require radians.
Don't pair a side with a non-opposite angle in the sine rule.

Exam tip. Label each side opposite its matching angle and check the calculator angle mode before substituting.

5.2 · Understand and use the standard small angle approximations of sine, cosine and tangent: sin θ ≈ θ, cos θ ≈ 1 − θ²/2, tan θ ≈ θ.

For $|\theta|$ close to zero and measured in radians, $\sin\theta\approx\theta$ , $\tan\theta\approx\theta$ and $\cos\theta\approx1-\theta^2/2$ . Replace each trigonometric function by its stated approximation, simplify algebraically, and retain only a solution whose magnitude is small enough for the approximation to be credible.
For example, $1-\cos\theta\approx\theta^2/2$ , so $(1-\cos\theta)/\theta^2\approx1/2$ for a small non-zero $\theta$ .
The approximations are radian results, not degree results; another common error is to accept a large root created by the approximate polynomial.
The approximation sign matters: these are local approximations near zero, not identities.
Quote the approximation used, substitute the small radian angle, and reject any root whose magnitude is inconsistent with the small-angle assumption.

Worked example

Without using a calculator's trigonometric keys, estimate $\dfrac{1-\cos(0.08)}{(0.08)^2}$ by a small-angle approximation.

1.Use $\cos\theta\approx1-\theta^2/2$ .
2.Then $1-\cos(0.08)\approx(0.08)^2/2$ , so division by $(0.08)^2$ gives $1/2=0.5$ .

Answer: $0.5$

Common mistakes

Don't apply the small-angle approximations to an angle measured in degrees.
Don't accept a large algebraic root even though the approximation is valid only near zero.

Exam tip. State that the angle is in radians and reject any solution that is not small.

5.3 · Understand and use the sine, cosine and tangent functions; their graphs, symmetries and periodicity; know and use exact values of sin, cos and tan for standard angles and their multiples.

$\sin x$ and $\cos x$ have range $[-1,1]$ and period $2\pi$ , while $\tan x$ has range $\mathbb{R}$ , period $\pi$ and vertical asymptotes at $x=\pi/2+k\pi$ .
Use $\sin(-x)=-\sin x$ , $\cos(-x)=\cos x$ and $\tan(-x)=-\tan x$ , then reduce an angle by a whole period before using its reference angle and quadrant.
Know exact sine and cosine values at $0$ , $\pi/6$ , $\pi/4$ , $\pi/3$ , $\pi/2$ and $\pi$ , and exact tangent values where defined.
Transformations such as $y=\cos(x+\pi/6)$ shift the graph, while $y=\tan2x$ halves its period.
In $a\cos(bx)+c$ , the amplitude is $|a|$ , period $2\pi/|b|$ , and $c$ moves the midline.

Two cycles of

y=\cos x

, showing range

[-1,1]

and period

2\pi

.

Worked example

For $y=3\cos(2x)-1$ , state the amplitude, period, maximum value and minimum value.

1.The coefficient outside cosine gives amplitude $|3|=3$ .
2.The factor $2$ inside gives period $2\pi/2=\pi$ .
3.Since $-1\leq\cos(2x)\leq1$ , multiplying by $3$ and subtracting $1$ gives $-4\leq y\leq2$ .

Answer: Amplitude $3$ Period $\pi$ Maximum $2$ Minimum $-4$

Common mistakes

Don't use the coefficient outside the trigonometric function to calculate the period instead of the amplitude.
Don't give tangent a period of $2\pi$ or omit its vertical asymptotes.

Exam tip. On a graph question, state amplitude, period, midline and range separately before sketching one complete cycle.

5.4 · Understand and use the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains.

The reciprocal functions are $\sec x=1/\cos x$ , $\cosec x=1/\sin x$ and $\cot x=\cos x/\sin x$ ; their graphs inherit zeros of the denominator as vertical asymptotes. Secant and cosecant have range $(-\infty,-1]\cup[1,\infty)$ and period $2\pi$ , while cotangent has range $\mathbb{R}$ and period $\pi$ .
The inverse graphs are restrictions reflected in $y=x$ .
Their principal ranges are $-\pi/2\leq\arcsin x\leq\pi/2$ , $0\leq\arccos x\leq\pi$ and $-\pi/2<\arctan x<\pi/2$ ; $\arcsin x$ and $\arccos x$ require $-1\leq x\leq1$ .
Angles may be in degrees or radians.
The notation $\sin^{-1}x$ means $\arcsin x$ , not the reciprocal $\cosec x$ .

Worked example

Give the principal values, in radians, of $\arcsin(-1/2)$ and $\arctan(-1)$ .

1.Within the principal sine-inverse range, $\sin(-\pi/6)=-1/2$ , so $\arcsin(-1/2)=-\pi/6$ .
2.Within the principal tangent-inverse range, $\tan(-\pi/4)=-1$ , so $\arctan(-1)=-\pi/4$ .

Answer: $\arcsin(-1/2)=-\pi/6$ $\arctan(-1)=-\pi/4$

Common mistakes

Don't read $\sin^{-1}x$ as the reciprocal $1/\sin x$ instead of the inverse function $\arcsin x$ .
Don't give an inverse-trigonometric answer outside the stated principal range.

Exam tip. Check the input domain and principal output range before giving an inverse-trigonometric value.

5.5 · Understand and use tan θ = sin θ / cos θ; understand and use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ.

The quotient identity is $\tan\theta=\sin\theta/\cos\theta$ , and the three Pythagorean identities are $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta=1+\tan^2\theta$ and $\cosec^2\theta=1+\cot^2\theta$ .
Choose an identity containing the known and required functions, rearrange it, and use the stated quadrant to select the correct sign after taking a square root.
If $\tan\theta=-7/24$ in quadrant II, a reference triangle has side magnitudes $7$ , $24$ and $25$ , so $\sin\theta=7/25$ and $\cos\theta=-24/25$ .
From $\sin^2\theta+\cos^2\theta=1$ one obtains two possible signs; ignoring the quadrant or silently choosing the positive root can change every reciprocal ratio that follows.

Worked example

Given that $\tan\theta=-7/24$ and $\pi/2<\theta<\pi$ , determine $\sin\theta$ , $\cos\theta$ and $\cosec\theta$ .

1.Use $\sec^2\theta=1+\tan^2\theta=1+49/576=625/576$ .
2.In quadrant II cosine and secant are negative, so $\sec\theta=-25/24$ and $\cos\theta=-24/25$ .
3.Then $\sin\theta=\tan\theta\cos\theta=(-7/24)(-24/25)=7/25$ , giving $\cosec\theta=25/7$ .

Answer: $\sin\theta=7/25$ $\cos\theta=-24/25$ $\cosec\theta=25/7$

Common mistakes

Don't treat $\cosec\theta$ as $\sin\theta$ rather than its reciprocal.
Don't use $\tan\theta=\cos\theta/\sin\theta$ instead of $\sin\theta/\cos\theta$ .

Exam tip. Use the interval to fix the signs of sine and cosine before taking reciprocals.

5.6 · Understand and use double angle formulae; formulae for sin(A ± B), cos(A ± B), tan(A ± B) with geometrical proofs; express a cos θ + b sin θ in the form r cos(θ ± α) or r sin(θ ± α).

Definition. Specification 5.6: understand and use double angle formulae; use of formulae for $\sin(A\pm B)$ , $\cos(A\pm B)$ and $\tan(A\pm B)$ ; understand geometrical proofs of these formulae; understand and use expressions for $a\cos\theta+b\sin\theta$ in the equivalent forms of $r\cos(\theta\pm\alpha)$ or $r\sin(\theta\pm\alpha)$ .

The compound-angle formulae are $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ , $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ and $\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ ; note the sign reversal in cosine.
Putting $A=B$ gives $\sin2\theta=2\sin\theta\cos\theta$ , $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$ and $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$ , and rearranging the cosine forms gives $\cos^2\theta=\frac12(1+\cos2\theta)$ for integration.
To write $a\cos\theta+b\sin\theta$ as $R\cos(\theta\pm\alpha)$ , expand the target, match coefficients to get $R\cos\alpha=a$ and $R\sin\alpha=b$ , so $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\dfrac ba$ with $\alpha$ in the stated range and in the stated units.
The $R$ form then gives the maximum $R$ , the minimum $-R$ , the values of $\theta$ where they occur, and solves $a\cos\theta+b\sin\theta=c$ in one step.

Worked example

1.

Express

5\cos\theta-12\sin\theta

in the form

R\cos(\theta+\alpha)

, where

R>0

and

0<\alpha<\dfrac{\pi}{2}

. Give the value of

\alpha

to 3 decimal places.

(3)

(Total for Question 1 is 3 marks)

Working

1.Expand: $R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$ .
2.Match coefficients: $R\cos\alpha=5$ and $R\sin\alpha=12$ , so $R=\sqrt{5^2+12^2}=13$ .
3.Divide: $\tan\alpha=\dfrac{12}{5}$ , so $\alpha=\arctan\dfrac{12}{5}=1.176$ (radians, in the required range).

Answer: $13\cos(\theta+1.176)$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	$R=13$	B1
	$\tan\alpha=\pm\dfrac{12}{5}$ or $\tan\alpha=\pm\dfrac{5}{12}$	M1
	$\alpha=$ awrt $1.176$	A1
Notes
M1: Condone $\cos\alpha=\frac{5}{13}$ or $\sin\alpha=\frac{12}{13}$
A1: Degrees ( $67.4^\circ$ ) do not earn this mark

(3 marks)

Mark-scheme wording to use

“K = 500 B1 tan α = 480/140 ⇒ α = ... M1 α = awrt 73.74° or 500cos(θ + 73.74)° A1 … B1: Correct value for K. Condone R = 500 M1: Award for tan α = ±480/140 ⇒ α = ..., tan α = ±140/480 ⇒ α = ..., sin α = ±480/"500" ⇒ α = ... or cos α = ±140/"500" ⇒ α = ... … A1: α = awrt 73.74{°} or correct expression 500cos(θ + 73.74){°}” — Pearson 9MA0, June 2024 Paper 1, Q12(a)
“R = 1000 + 500cos(30t + 73.74)° or R = 1000 + 140cos(30t)° − 480sin(30t)° B1ft (b)(ii) {R_min =}500 B1ft … B1ft: Correct equation of the model in either form including the R = following through on their numerical K (0 < K ≤ 750) and their numerical α. … B1ft: 500 or follow through on (their A – their K) or (1500 – 2 × their K) provided it is non-negative and less than 1500. It must be clear this is their answer to (b)(ii) so expect to see e.g. (b) or R_min = or an indication it is the minimum.” — Pearson 9MA0, June 2024 Paper 1, Q12(b)
“B1 for the exact value R = sqrt(109) … M1 for tan(alpha) = 3/10 or an equivalent comparison. A1 for alpha = 16.70 degrees and the correct completed form” — Pearson 9MA0, Sample Paper 2, Q13(a)

What the examiner wants. The right ratio in the right units. The June 2024 Paper 1 report says the $R$ -form part was mostly done correctly, and that the common mistakes among those who dropped marks were using radians instead of degrees and calculating $\arctan\dfrac{140}{480}$ instead of $\arctan\dfrac{480}{140}$ .

Common mistakes

Don't use the same sign in $\cos(A\pm B)$ instead of reversing it in the expansion.
Don't choose the sign of $\alpha$ in an $R$ -form without expanding and comparing both coefficients.

Exam tip. Expand the proposed $R$ -form, compare coefficients, then state $R>0$ and the required range for $\alpha$ .

5.7 · Solve simple trigonometric equations in a given interval, including quadratic equations in sin, cos and tan and equations involving multiples of the unknown angle.

Definition. Specification 5.7: solve simple trigonometric equations in a given interval, including quadratic equations in $\sin$ , $\cos$ and $\tan$ , and equations involving multiples of the unknown angle.

Rearrange to a single ratio, find the principal value from the calculator, then use the CAST diagram or the graph to generate every solution in the interval: for $\cos$ the second solution is $360^\circ-\theta$ or $2\pi-\theta$ , for $\sin$ it is $180^\circ-\theta$ , for $\tan$ add $180^\circ$ .
A quadratic in one ratio is factorised or solved by the formula, and any root outside $[-1,1]$ for $\sin$ or $\cos$ is rejected before solving each branch; a Pythagorean identity converts a mixed equation into a quadratic in one ratio first.
For a multiple angle, transform the interval, so $0<x<90^\circ$ becomes $0<2x<180^\circ$ , find every value of $2x$ in that range, and divide at the end.
Never divide by $\sin\theta$ or $\cos\theta$ , which loses solutions; factorise instead.
Check the mode, degrees or radians, against the interval, and check endpoints against strict or non-strict inequalities.

Worked example

1.

Solve, for

0\leq x<2\pi

, the equation

2\cos^2x-3\cos x+1=0

.

(4)

(Total for Question 1 is 4 marks)

Working

1.Factorise as a quadratic in $\cos x$ : $(2\cos x-1)(\cos x-1)=0$ .
2.So $\cos x=\dfrac12$ or $\cos x=1$ .
3. $\cos x=\dfrac12$ gives $x=\dfrac{\pi}{3}$ and $x=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}$ ; $\cos x=1$ gives $x=0$ only, since $2\pi$ is excluded.

Answer: $x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Attempts to solve the three-term quadratic in $\cos x$ , eg $(2\cos x-1)(\cos x-1)$	M1
	$\cos x=\dfrac12$ and $\cos x=1$	A1
	Obtains $x=\dfrac{\pi}{3}$ and a second solution from $2\pi-$ their principal value	M1
	$x=0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}$ and no extra solutions in the interval	A1
Notes
A1: Degrees score A0

(4 marks)

Mark-scheme wording to use

“M1: Correct order of operations from tan² 2x = α or sin² 2x = β or cos² 2x = γ or cos 4x = k or equivalents e.g. cosec² 2x = 1/β where α > 0 or 0 < β < 1 or 0 < γ < 1 or 0 < k < 1 leading to at least one value for x e.g. square roots, finds inverse tan/sin/cos/cosec and divides by 2 or inverse cos and divides by 4 … A1: Correct values. Allow awrt 17.6° and awrt 72.4°. The degrees symbol is not required. … if there are extra angles in range score A0. Answers in radians score A0.” — Pearson 9MA0, June 2024 Paper 2, Q8(b)
“M1 for using cos^2(x) = 1 - sin^2(x) to form a quadratic in sin(x). A1 for 12sin^2(x) + sin(x) - 1 = 0. M1 for attempting to solve that quadratic. A1 for sin(x) = 1/4 and -1/3. M1 for inverse-trigonometric work producing two correct values. A1 for all four values 14.48, 165.52, -19.47 and -160.53 degrees” — Pearson 9MA0, Sample Paper 2, Q12(a)
“B1 for explaining that squaring introduced an extraneous solution or lost the sign information” — Pearson 9MA0, Sample Paper 2, Q2(b)(ii)

What the examiner wants. Every in-range solution and nothing else. The June 2024 Paper 2 scheme withholds the final mark for radians in a degrees question or for extra in-range angles, and the Paper 1 report on the trigonometric model praises candidates who identified that degrees, not radians, were required before solving.

Common mistakes

Don't give only the principal value returned by the inverse-trigonometric button.
Don't include an excluded endpoint or miss solutions created by the multiple angle.

Exam tip. Solve for the trig value first, then use symmetry and periodicity to list every solution in the stated interval.

5.8 · Construct proofs involving trigonometric functions and identities.

A trigonometric identity is true for every value in its domain, so a proof transforms one side into the other using exact algebra and established identities rather than testing selected angles.
Usually begin with the more complicated side, replace secant, cosecant, cotangent or tangent by sine and cosine when helpful, and factor or take a common denominator before cancelling.
Multiplying numerator and denominator by a conjugate can expose $1-\sin^2x=\cos^2x$ or $1-\cos^2x=\sin^2x$ and complete the proof cleanly.
Never cancel terms across addition, and record domain restrictions: algebra such as division by $\sin x$ is valid only where that denominator is non-zero.

Worked example

Prove that $\dfrac{1-\cos(2x)}{\sin(2x)}=\tan x$ for values at which both sides are defined.

1.Apply the double-angle forms to the left-hand side: $(1-\cos2x)/\sin2x=(2\sin^2x)/(2\sin x\cos x)=\sin x/\cos x=\tan x$ , with cancellation only where the original expressions are defined.

Answer: Use $1-\cos(2x)=2\sin^2x$ and $\sin(2x)=2\sin x\cos x$ .

Common mistakes

Don't cancel a term across addition while manipulating a trigonometric fraction.
Don't divide by $\sin x$ or $\cos x$ without recording where that factor is zero.

Exam tip. For “prove”, transform one side only with named identities until it exactly matches the other side.

5.9 · Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces.

Resolve a vector of magnitude $V$ at angle $\theta$ to the positive horizontal into components $V\cos\theta$ and $V\sin\theta$ , changing signs to match its actual direction.
For resultant or equilibrium problems, form separate equations in two perpendicular directions; a zero resultant requires both component sums to equal zero.
Trigonometric models can describe wave motion, a point on a vertical circular wheel or changing hours of sunlight.
Interpret every solution using the stated time interval, units and physical constraints.
A calculator angle without a quadrant check can point in the opposite direction; draw and label a diagram, then state bearings or directions in the form the context requests.

Perpendicular components combine to give a resultant whose direction is measured from east.

Worked example

Two horizontal forces act on a crate: $8\,\text{N}$ due east and $11\,\text{N}$ at $60^\circ$ north of east. Find the magnitude and direction of their resultant, to $3$ significant figures and the nearest degree respectively.

1.The east component is $8+11\cos60^\circ=13.5\,\text{N}$ and the north component is $11\sin60^\circ=9.526\ldots\,\text{N}$ .
2.Hence $R=\sqrt{13.5^2+9.526^2}=16.523\ldots\,\text{N}$ .
3.Its direction is $\arctan(9.526/13.5)=35.21\ldots^\circ$ north of east.

Answer: Magnitude $=16.5\,\text{N}$ Direction $35^\circ$ north of east

Common mistakes

Don't use an inverse-tangent calculator value without checking the resultant's quadrant.
Don't state a bare angle without the bearing or directional wording required by the context.

Exam tip. Draw and label component directions, then give the final magnitude with units and the direction in contextual form.