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Cambridge IGCSE0606

Trigonometry

Mathematics Additional 0606 Chapter Notes

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Trigonometry
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Trigonometry notes

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In this chapter

  1. 1The six trigonometric functions
  2. 2Amplitude and period
  3. 3Sketching the graphs
  4. 4The identities
  5. 5Solving for a given domain
  6. 6Equations with two functions
  7. 7Equations that need an identity
  8. 8Proving identities

1. The six trigonometric functions

The six functions are sine, cosine, tangent, secant, cosecant and cotangent. Secant goes with cosine, cosecant goes with sine, and cotangent goes with tangent.

They are defined for angles of any magnitude. Find the related acute angle, then give the ratio the sign it has in that quadrant.

To simplify or solve, first write each function in terms of sine and cosine.

secθ = 1/cosθ

cosecθ = 1/sinθ

cotθ = 1/tanθ = cosθ/sinθ

tanθ = sinθ/cosθ

Examiner insight

“It is expected that candidates have a basic knowledge of the graphs of y = sin x and y = cos x, together with the values that these functions have at the quadrant boundaries.” “Candidates who did not write the expression in terms of sin x and cos x initially were less successful with simplification.”

Worked example 13 marks

Do not use a calculator. Find the exact value of each of sec 120^(°), cosec210^(°) and cot 225^(°).

  1. 1

    cos 120^(°) = -cos 60^(°) = -1/2, so sec 120^(°) = -2 [1]

  2. 2

    sin 210^(°) = -sin 30^(°) = -1/2, so cosec210^(°) = -2 [1]

  3. 3

    tan 225^(°) = tan 45^(°) = 1, so cot 225^(°) = 1 [1]

2. Amplitude and period

For y = asin bx + c or y = acos bx + c, the amplitude is a. The c moves the graph up or down and leaves the amplitude alone.

The period is the length of one full cycle. Sine and cosine repeat every 360^(°) divided by b. Tangent repeats every 180^(°) divided by b.

The period may be in degrees or in radians. Answer in the units of the question.

amplitude of asin bx + c = a

period of sin bx, cos bx = (360^(°))/b (or 2π/b)

period of tan bx = (180^(°))/b (or π/b)

Examiner insight

“Sine graphs were occasionally misdrawn, with errors in amplitude, midline shifts, or misunderstanding of the period.” “Some candidates gave their answers in radians.”

Worked example 12 marks

Write down the amplitude of 4cos(x/8) + 2.

Write down the period, in degrees, of 4cos(x/8) + 2.

  1. 1

    amplitude = 4 [1]

  2. 2

    period = 360^(°) ÷ 1/8 = 360 × 8 = 2880^(°) [1]

3. Sketching the graphs

Mark the greatest and least values, c ± a, and the midline y = c. Draw one smooth curve through them. Put in the y-intercept and every x-intercept the domain needs.

For y = atan bx + c, clearly label the x-coordinate of each asymptote.

Examiner insight

“Some candidates did not indicate the curve tending towards a maximum point at the endpoints.” “Asymptotes should have been indicated with either dashed or solid vertical lines and the y-intercept also needed to be clearly marked.”

Worked example 13 marks

Sketch y = tan 2x + 3 for 0^(°) ≤ x ≤ 180^(°). Write down the period and the equation of each asymptote.

  1. 1

    period = 180^(°) ÷ 2 = 90^(°) [1]

  2. 2

    asymptotes where 2x = 90^(°) or 270^(°) [1]

  3. 3

    x = 45^(°) and x = 135^(°) [1]

4. The identities

Three relationships connect the functions. Divide sin² A + cos² A = 1 by cos² A to get the second, and by sin² A to get the third.

They are given in the List of formulas, so the marks are for using them correctly: sec²A = 1 + tan²A, not 1 - tan² A.

A squared function can be rearranged: sin² A = 1 - cos² A and cot² A = cosec² A - 1.

sin² A + cos² A = 1

sec² A = 1 + tan² A

cosec² A = 1 + cot² A

Examiner insight

“The best responses demonstrated a careful, step-by-step application of trigonometric identities, with clear justification for each transformation.”

Worked example 14 marks

Show that (3 + 5cotθ)² + (3 - 5cotθ)² - 50cosec²θ, where 0 < θ < π/2, can be simplified to a constant.

  1. 1

    (3 + 5cotθ)² = 9 + 30cotθ + 25cot²θ [1]

  2. 2

    (3 - 5cotθ)² = 9 - 30cotθ + 25cot²θ, so the sum is 18 + 50cot²θ - 50cosec²θ [1]

  3. 3

    = 18 - 50(cosec²θ - cot²θ) and cosec²θ - cot²θ = 1 [1]

  4. 4

    = 18 - 50 = -32 [1]

5. Solving for a given domain

Make the function the subject. Use the calculator for the first angle. Then find all the angles in the domain, using the symmetry of the graph: π - α and -π - α for sine, or ±α for cosine.

When the angle is θ - 0.5 or 3x + 1, first change the domain to the domain of that angle. Solve, then undo the shift. Give no other solutions outside the domain.

Give radians to 3 significant figures and degrees to 1 decimal place.

Examiner insight

“Many candidates obtained an answer of – 0.23 radians rather than – 0.231 radians. Angles in radians should be given correct to 3 significant figures.” “Some candidates did not consider all the possible solutions and these usually stated the answer ‘x = 10’ or ‘x = 10, x = 70’ only.”

Worked example 15 marks

Solve the equation 5 - 2cosec(θ - 0.5) = 0 where θ is in radians and -3.5 ≤ θ ≤ 3.5.

  1. 1

    cosec(θ - 0.5) = 5/2, so sin(θ - 0.5) = 0.4 [1]

  2. 2

    θ - 0.5 lies between -4.0 and 3.0. Principal value 0.4115, then π - 0.4115 = 2.7300 and -π - 0.4115 = -3.5531 [1]

  3. 3

    θ = 0.912 [1]

  4. 4

    θ = 3.23 [1]

  5. 5

    θ = -3.05, and no other solutions in the domain [1]

6. Equations with two functions

When two different functions of the same angle appear, write the equation in terms of sine and cosine. Move every term to one side and factorise.

Do not divide by a function that contains the unknown. That throws away every solution where it is zero. Set each factor equal to zero and solve each one.

Examiner insight

“Some candidates understood the need to bring all the terms to one side and factorise so that no solutions were lost.”

Worked example 16 marks

Solve the equation tan(3x + 1) = 5sin(3x + 1) for -π/3 < x < π/3.

  1. 1

    (sin(3x+1))/(cos(3x+1)) = 5sin(3x+1) [1]

  2. 2

    sin(3x+1)[1 - 5cos(3x+1)] = 0, so sin(3x+1) = 0 [1]

  3. 3

    or cos(3x+1) = 1/5 [1]

  4. 4

    3x + 1 lies between -2.14 and 4.14. Find at least one value of 3x+1 [1]

  5. 5

    3x + 1 = 0 or π, so x = -1/3 or 0.714 [1]

  6. 6

    3x + 1 = ± 1.3694, so x = 0.123 or -0.790 [1]

7. Equations that need an identity

An equation with a squared function and another function of the same angle needs an identity. Replace the squared term so that only one function is left.

This gives a quadratic in that function. Solve it as a quadratic, then solve each value. Reject a value the function can never take: secθ is never between -1 and 1, so secθ = 0 gives no solutions.

Examiner insight

“Candidates who used a quadratic equation in terms of cosine obtained incorrect solutions from considering cos (x – 1) = 0. These solutions were not checked in the original equation.”

Worked example 15 marks

Solve the equation 3sec(x - 1) - 1 = tan²(x - 1) where x is in radians and -1 < x < 3.

  1. 1

    use tan²(x-1) = sec²(x-1) - 1 [1]

  2. 2

    sec²(x-1) - 3sec(x-1) = 0 [1]

  3. 3

    sec(x-1)[sec(x-1) - 3] = 0 and sec(x-1) ≠ 0, so sec(x-1) = 3 [1]

  4. 4

    x - 1 = 1.23 or -1.23 [1]

  5. 5

    x = -0.231 or 2.23 [1]

8. Proving identities

Start with one side, usually the left-hand side. Change it step by step until it matches the right-hand side. Do not work on both sides at once.

Write each function in terms of sine and cosine, join fractions over a common denominator, then use an identity from 10.4. Keep the angle symbol on every line.

Examiner insight

“It was essential that necessary detail be shown.” “The omission of the angle symbol θ throughout the proof was common, and reduced clarity.”

Worked example 13 marks

Show that cosecx - sin x can be written as (cos x)/(tan x).

  1. 1

    cosecx - sin x = 1/(sinx) - sin x [1]

  2. 2

    = (1 - sin² x)/(sinx) [1]

  3. 3

    = (cos² x)/(sinx) = cos x × (cos x)/(sinx) = (cos x)/(tanx) [1]

Recap

  • Secant, cosecant and cotangent are the reciprocals of cosine, sine and tangent. tan θ = sin θ ÷ cos θ.
  • For a sin bx + c, the amplitude is a. The period is 360° ÷ b (or 2π ÷ b). For a tan bx + c the period is 180° ÷ b (or π ÷ b).
  • Sketch with the greatest and least values, the midline, the intercepts and a smooth curve. Label every asymptote.
  • sin²A + cos²A = 1, sec²A = 1 + tan²A, cosec²A = 1 + cot²A.
  • Solving: change the domain to that of the whole angle, find every solution, give no others. Radians to 3 significant figures, degrees to 1 decimal place.
  • Factorise, never divide by a function of the unknown. Use an identity to leave one function, then solve the quadratic.
  • Proofs: start with one side, show every step, keep the angle symbol.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Write secθ, cosecθ and cotθ in terms of sinθ and cosθ.3 marks
  2. Do not use a calculator. Find the exact value of cosec300^(°).2 marks
  3. State the amplitude and the period, in degrees, of y = 5sin 3x - 2.2 marks
  4. State the period, in radians, of y = 2tan(2x/3) + 1.2 marks
  5. Do not use a calculator. For y = 3cos 2x + 1 with 0^(°) ≤ x ≤ 360^(°), write down the greatest value of y, the least value of y, and the values of x at which the greatest value occurs.4 marks
  6. Do not use a calculator. Write down the equations of the asymptotes of y = tan 3x + 1 for 0^(°) < x < 120^(°).2 marks
  7. Do not use a calculator. Given that tanθ = 2 and that θ is acute, find the exact value of secθ and of cosθ.3 marks
  8. Show that sinθ/(1 + cosθ) + (1 + cosθ)/sinθ = 2cosecθ.4 marks
  9. Do not use a calculator. (a) Show that sec x - cos x = sin xtan x. (b) Hence solve the equation sec 2θ - cos 2θ = sin 2θ for 0 ≤ θ ≤ π, where θ is in radians.8 marks
  10. Two markers on a machine move up and down. After t seconds their heights above a fixed line are p cm and q cm, where p = 6sin²(0.4t + 0.2) and q = 5cos(0.4t + 0.2). The angle is in radians and 0 ≤ t ≤ 15. (a) Show that the markers are at the same height when 6cos² u + 5cos u - 6 = 0, where u = 0.4t + 0.2. (b) Find the two values of t at which the markers are at the same height. (c) When t = 5 the first marker is the higher one. Find the length of time, within the 15 seconds, for which the first marker is higher than the second.9 marks

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