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Edexcel A-level Maths revision notes

Trigonometry

Section 5
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
9 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section 5

Checked against Edexcel 9MA0 section 5. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

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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.

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a unit-circle angle θ\theta, cosθ\cos\theta and sinθ\sin\theta are the point's horizontal and vertical coordinates, while tanθ=sinθ/cosθ\tan\theta=\sin\theta/\cos\theta where cosθ0\cos\theta\neq0; this extends the ratios beyond acute angles.
  • For a triangle, use a/sinA=b/sinB=c/sinCa/\sin A=b/\sin B=c/\sin C when an opposite side-angle pair is known, a2=b2+c22bccosAa^2=b^2+c^2-2bc\cos A for three sides or an included angle, and 12absinC\tfrac12ab\sin C for area.
  • In the ambiguous sine-rule case, test the supplementary angle because sinA=sin(πA)\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 θ\theta radians in a circle of radius rr gives arc length s=rθs=r\theta and sector area A=12r2θA=\tfrac12r^2\theta.
  • Keep the calculator in the required angle mode and label sides opposite their matching angles; using degrees in s=rθs=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 8m8\,\text{m} and 11m11\,\text{m}, with included angle 0.90.9 radians. Calculate the third side and the area of the sail, giving each answer to 33 significant figures.

  1. 1.The cosine rule gives c2=82+1122(8)(11)cos(0.9)=75.597c^2=8^2+11^2-2(8)(11)\cos(0.9)=75.597\ldots, so c=8.694m=8.69mc=8.694\ldots\,\text{m}=8.69\,\text{m}.
  2. 2.The area is 12(8)(11)sin(0.9)=34.466m2\tfrac12(8)(11)\sin(0.9)=34.466\ldots\,\text{m}^2, hence 34.5m234.5\,\text{m}^2.

Answer: Third side =8.69m=8.69\,\text{m} Area =34.5m2=34.5\,\text{m}^2

Common mistakes

  • Don't use degrees in s=rθs=r\theta or A=12r2θA=\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.

Tier 1 · Easy

ORIGINAL

1.

A circular arc has radius 7.5cm7.5\,\text{cm} and subtends 1.21.2 radians at the centre. Find its length.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A sector of a circle has radius 7cm7\,\text{cm} and arc length 11.2cm11.2\,\text{cm}. Find the angle of the sector in radians and its area.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A minor segment is cut from a circle of radius 6cm6\,\text{cm} by a chord whose endpoints subtend 1.41.4 radians at the centre. Determine the perimeter and area of the segment, giving both to 33 significant figures.

(5)

(Total for Question 1 is 5 marks)

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

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.

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For θ|\theta| close to zero and measured in radians, sinθθ\sin\theta\approx\theta, tanθθ\tan\theta\approx\theta and cosθ1θ2/2\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, 1cosθθ2/21-\cos\theta\approx\theta^2/2, so (1cosθ)/θ21/2(1-\cos\theta)/\theta^2\approx1/2 for a small non-zero θ\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 1cos(0.08)(0.08)2\dfrac{1-\cos(0.08)}{(0.08)^2} by a small-angle approximation.

  1. 1.Use cosθ1θ2/2\cos\theta\approx1-\theta^2/2.
  2. 2.Then 1cos(0.08)(0.08)2/21-\cos(0.08)\approx(0.08)^2/2, so division by (0.08)2(0.08)^2 gives 1/2=0.51/2=0.5.

Answer: 0.50.5

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.

Tier 1 · Easy

ORIGINAL

1.

Use a standard small-angle approximation to estimate sin(0.064)\sin(0.064).

(1)

(Total for Question 1 is 1 mark)

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Tier 2 · Standard

ORIGINAL

1.

Use small-angle approximations to estimate the non-zero small positive solution of sinx=5(1cosx)\sin x=5(1-\cos x).

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A small positive angle xx satisfies sinx+cosx=1.08\sin x+\cos x=1.08. Use the standard small-angle approximations to estimate xx, giving 44 decimal places, and explain which algebraic root is admissible.

(4)

(Total for Question 1 is 4 marks)

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.

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • sinx\sin x and cosx\cos x have range [1,1][-1,1] and period 2π2\pi, while tanx\tan x has range R\mathbb{R}, period π\pi and vertical asymptotes at x=π/2+kπx=\pi/2+k\pi.
  • Use sin(x)=sinx\sin(-x)=-\sin x, cos(x)=cosx\cos(-x)=\cos x and tan(x)=tanx\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 00, π/6\pi/6, π/4\pi/4, π/3\pi/3, π/2\pi/2 and π\pi, and exact tangent values where defined.
  • Transformations such as y=cos(x+π/6)y=\cos(x+\pi/6) shift the graph, while y=tan2xy=\tan2x halves its period.
  • In acos(bx)+ca\cos(bx)+c, the amplitude is a|a|, period 2π/b2\pi/|b|, and cc moves the midline.
Two cycles of y=cosxy=\cos x, showing range [1,1][-1,1] and period 2π2\pi.
Worked example

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

  1. 1.The coefficient outside cosine gives amplitude 3=3|3|=3.
  2. 2.The factor 22 inside gives period 2π/2=π2\pi/2=\pi.
  3. 3.Since 1cos(2x)1-1\leq\cos(2x)\leq1, multiplying by 33 and subtracting 11 gives 4y2-4\leq y\leq2.

Answer: Amplitude 33 Period π\pi Maximum 22 Minimum 4-4

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 2π2\pi or omit its vertical asymptotes.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

Find the exact value of tan(5π/6)\tan(5\pi/6).

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the exact values of sin(7π6)\sin\left(\dfrac{7\pi}{6}\right), cos(5π3)\cos\left(\dfrac{5\pi}{3}\right) and tan(3π4)\tan\left(\dfrac{3\pi}{4}\right).

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Evaluate exactly sin(11π/6)\sin(-11\pi/6), cos(13π/3)\cos(13\pi/3) and tan(7π/4)\tan(7\pi/4).

(4)

(Total for Question 1 is 4 marks)

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.

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The reciprocal functions are secx=1/cosx\sec x=1/\cos x, cosecx=1/sinx\cosec x=1/\sin x and cotx=cosx/sinx\cot x=\cos x/\sin x; their graphs inherit zeros of the denominator as vertical asymptotes. Secant and cosecant have range (,1][1,)(-\infty,-1]\cup[1,\infty) and period 2π2\pi, while cotangent has range R\mathbb{R} and period π\pi.
  • The inverse graphs are restrictions reflected in y=xy=x.
  • Their principal ranges are π/2arcsinxπ/2-\pi/2\leq\arcsin x\leq\pi/2, 0arccosxπ0\leq\arccos x\leq\pi and π/2<arctanx<π/2-\pi/2<\arctan x<\pi/2; arcsinx\arcsin x and arccosx\arccos x require 1x1-1\leq x\leq1.
  • Angles may be in degrees or radians.
  • The notation sin1x\sin^{-1}x means arcsinx\arcsin x, not the reciprocal cosecx\cosec x.
Worked example

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

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

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

Common mistakes

  • Don't read sin1x\sin^{-1}x as the reciprocal 1/sinx1/\sin x instead of the inverse function arcsinx\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.

Tier 1 · Easy

ORIGINAL

1.

Given cosθ=4/5\cos\theta=-4/5, write down secθ\sec\theta.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1.

For 0x2π0\leq x\leq2\pi, state where secx\sec x is undefined and solve secx=2\sec x=-2.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

For y=2secx1y=2\sec x-1, state the period and range, then give all vertical asymptotes in πxπ-\pi\leq x\leq\pi.

(5)

(Total for Question 1 is 5 marks)

5.5

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The quotient identity is tanθ=sinθ/cosθ\tan\theta=\sin\theta/\cos\theta, and the three Pythagorean identities are sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1, sec2θ=1+tan2θ\sec^2\theta=1+\tan^2\theta and cosec2θ=1+cot2θ\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 tanθ=7/24\tan\theta=-7/24 in quadrant II, a reference triangle has side magnitudes 77, 2424 and 2525, so sinθ=7/25\sin\theta=7/25 and cosθ=24/25\cos\theta=-24/25.
  • From sin2θ+cos2θ=1\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 tanθ=7/24\tan\theta=-7/24 and π/2<θ<π\pi/2<\theta<\pi, determine sinθ\sin\theta, cosθ\cos\theta and cosecθ\cosec\theta.

  1. 1.Use sec2θ=1+tan2θ=1+49/576=625/576\sec^2\theta=1+\tan^2\theta=1+49/576=625/576.
  2. 2.In quadrant II cosine and secant are negative, so secθ=25/24\sec\theta=-25/24 and cosθ=24/25\cos\theta=-24/25.
  3. 3.Then sinθ=tanθcosθ=(7/24)(24/25)=7/25\sin\theta=\tan\theta\cos\theta=(-7/24)(-24/25)=7/25, giving cosecθ=25/7\cosec\theta=25/7.

Answer: sinθ=7/25\sin\theta=7/25 cosθ=24/25\cos\theta=-24/25 cosecθ=25/7\cosec\theta=25/7

Common mistakes

  • Don't treat cosecθ\cosec\theta as sinθ\sin\theta rather than its reciprocal.
  • Don't use tanθ=cosθ/sinθ\tan\theta=\cos\theta/\sin\theta instead of sinθ/cosθ\sin\theta/\cos\theta.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

An acute angle θ\theta satisfies sinθ=5/13\sin\theta=5/13. Find tanθ\tan\theta and secθ\sec\theta exactly.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Prove that 11sinx+11+sinx=2sec2x\dfrac{1}{1-\sin x}+\dfrac{1}{1+\sin x}=2\sec^2x for values of xx for which both sides are defined.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

An angle θ\theta lies in the first quadrant and satisfies secθ+tanθ=5\sec\theta+\tan\theta=5. Find sinθ\sin\theta exactly.

(5)

(Total for Question 1 is 5 marks)

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).

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The compound-angle formulae are sin(A±B)=sinAcosB±cosAsinB\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B, cos(A±B)=cosAcosBsinAsinB\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B and tan(A±B)=(tanA±tanB)/(1tanAtanB)\tan(A\pm B)=(\tan A\pm\tan B)/(1\mp\tan A\tan B); a geometrical proof can calculate the same unit-circle chord by coordinates and by the cosine rule.
  • Setting A=B=θA=B=\theta gives sin2θ=2sinθcosθ\sin2\theta=2\sin\theta\cos\theta, cos2θ=cos2θsin2θ=12sin2θ=2cos2θ1\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1 and tan2θ=2tanθ/(1tan2θ)\tan2\theta=2\tan\theta/(1-\tan^2\theta).
  • Rearranging a double-angle identity also gives half-angle results, for example sin2θ=(1cos2θ)/2\sin^2\theta=(1-\cos2\theta)/2.
  • To write acosθ+bsinθ=Rcos(θα)a\cos\theta+b\sin\theta=R\cos(\theta-\alpha), compare coefficients to obtain Rcosα=aR\cos\alpha=a, Rsinα=bR\sin\alpha=b, hence R=a2+b2R=\sqrt{a^2+b^2} with the quadrant of α\alpha set by the signs.
  • For a compound-angle expression, the sign in the cosine formula reverses; for an RR-form, expanding the proposed form before choosing α\alpha prevents a wrong sign.
Worked example

Express 5cosθ12sinθ5\cos\theta-12\sin\theta as Rcos(θ+α)R\cos(\theta+\alpha), where R>0R>0 and 0<α<π/20<\alpha<\pi/2. Hence state its maximum and minimum values.

  1. 1.Expand Rcos(θ+α)=RcosθcosαRsinθsinαR\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha.
  2. 2.Comparing coefficients gives Rcosα=5R\cos\alpha=5 and Rsinα=12R\sin\alpha=12, so R=25+144=13R=\sqrt{25+144}=13 and tanα=12/5\tan\alpha=12/5.
  3. 3.Since cosine ranges from 1-1 to 11, the expression ranges from 13-13 to 1313.

Answer: 13cos(θ+α)13\cos(\theta+\alpha), where α=arctan(12/5)\alpha=\arctan(12/5) Maximum 1313 Minimum 13-13

Common mistakes

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

Exam tip

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

Tier 1 · Easy

ORIGINAL

1.

Use a compound-angle formula to find the exact value of sin(75)\sin(75^\circ).

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Given that sinθ=35\sin\theta=\dfrac35 and 0<θ<π20<\theta<\dfrac{\pi}{2}, find the exact values of sin2θ\sin2\theta, cos2θ\cos2\theta and tan2θ\tan2\theta.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Use two points on the unit circle and the cosine rule to prove geometrically that cos(AB)=cosAcosB+sinAsinB\cos(A-B)=\cos A\cos B+\sin A\sin B.

(5)

(Total for Question 1 is 5 marks)

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.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Solve first for the trigonometric ratio, use a reference angle and the signs in each quadrant, then list only solutions in the stated interval.
  • For a quadratic in one trigonometric function, substitute a temporary variable, factorise or use the quadratic formula, and reject any ratio outside its possible range before solving each remaining branch.
  • When the equation involves kxkx, transform the given interval for xx into the corresponding interval for kxkx, find every solution there, and divide only at the end.
  • Inverse-trigonometric buttons return a principal value rather than the full solution set; endpoints, excluded endpoints and degree-versus-radian mode must all be checked explicitly.
Worked example

Determine all xx satisfying 2cos2x3cosx+1=02\cos^2x-3\cos x+1=0 for 0x<2π0\leq x<2\pi.

  1. 1.Factorise to (2cosx1)(cosx1)=0(2\cos x-1)(\cos x-1)=0.
  2. 2.Hence cosx=1/2\cos x=1/2 or cosx=1\cos x=1.
  3. 3.In the interval, cosx=1/2\cos x=1/2 at x=π/3x=\pi/3 and 5π/35\pi/3, while cosx=1\cos x=1 at x=0x=0; 2π2\pi is excluded.

Answer: x=0, π/3, 5π/3x=0,\ \pi/3,\ 5\pi/3

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.

Tier 1 · Easy

ORIGINAL

1.

Solve sinθ=0.4\sin\theta=0.4 for 0θ2π0\leq\theta\leq2\pi, giving solutions to 33 decimal places.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Solve sin(2x)=32\sin(2x)=\dfrac{\sqrt3}{2} for 0x2π0\leq x\leq2\pi.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Find every solution of tan(2x)=3\tan(2x)=-\sqrt3 in the interval π/2xπ-\pi/2\leq x\leq\pi.

(5)

(Total for Question 1 is 5 marks)

5.8

Construct proofs involving trigonometric functions and identities.

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 1sin2x=cos2x1-\sin^2x=\cos^2x or 1cos2x=sin2x1-\cos^2x=\sin^2x and complete the proof cleanly.
  • Never cancel terms across addition, and record domain restrictions: algebra such as division by sinx\sin x is valid only where that denominator is non-zero.
Worked example

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

  1. 1.Apply the double-angle forms to the left-hand side: (1cos2x)/sin2x=(2sin2x)/(2sinxcosx)=sinx/cosx=tanx(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 1cos(2x)=2sin2x1-\cos(2x)=2\sin^2x and sin(2x)=2sinxcosx\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 sinx\sin x or cosx\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.

Tier 1 · Easy

ORIGINAL

1.

Prove that sinxcotx=cosx\sin x\cot x=\cos x wherever the left-hand side is defined.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Prove that cos(A+B)cos(AB)=cos2Asin2B\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that 11sinx11+sinx=2tanxsecx\dfrac{1}{1-\sin x}-\dfrac{1}{1+\sin x}=2\tan x\sec x wherever the expressions exist.

(5)

(Total for Question 1 is 5 marks)

5.9

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Resolve a vector of magnitude VV at angle θ\theta to the positive horizontal into components VcosθV\cos\theta and VsinθV\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: 8N8\,\text{N} due east and 11N11\,\text{N} at 6060^\circ north of east. Find the magnitude and direction of their resultant, to 33 significant figures and the nearest degree respectively.

  1. 1.The east component is 8+11cos60=13.5N8+11\cos60^\circ=13.5\,\text{N} and the north component is 11sin60=9.526N11\sin60^\circ=9.526\ldots\,\text{N}.
  2. 2.Hence R=13.52+9.5262=16.523NR=\sqrt{13.5^2+9.526^2}=16.523\ldots\,\text{N}.
  3. 3.Its direction is arctan(9.526/13.5)=35.21\arctan(9.526/13.5)=35.21\ldots^\circ north of east.

Answer: Magnitude =16.5N=16.5\,\text{N} Direction 3535^\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.

Tier 1 · Easy

ORIGINAL

1.

A drone travels at 12m s112\,\text{m s}^{-1} on a path 3535^\circ above the horizontal. Calculate its horizontal and vertical velocity components to 33 significant figures.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

A hiker walks 24km24\,\text{km} on a bearing of 040040^\circ and then 18km18\,\text{km} on a bearing of 130130^\circ. Find the hiker's distance and bearing from the starting point, giving the bearing to the nearest degree.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A ring is in equilibrium under its weight of 18N18\,\text{N}, a tension PP directed 2525^\circ above the horizontal, and a horizontal tension QQ acting oppositely. Calculate PP and QQ to 33 significant figures.

(5)

(Total for Question 1 is 5 marks)

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