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Cambridge IGCSE0607

Trigonometric functions

International Mathematics 0607 Chapter Notes

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Trigonometric functions
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Trigonometric functions notes

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In this chapter

  1. 1The graph of y = sin x
  2. 2The graph of y = cos x
  3. 3The graph of y = tan x
  4. 4Solving sin x = k and cos x = k
  5. 5Solving tan x = k
  6. 6Equations to rearrange first

1. The graph of y = sin x

For 0^(°) leqslant x leqslant 360^(°), the graph of y = sin x is a smooth wave. It starts at 0, rises to a maximum of 1 at 90°, falls through 0 at 180°, reaches a minimum of −1 at 270° and returns to 0 at 360°. The curve never goes above 1 or below −1.

sin 0^(°) = 0, sin 90^(°) = 1, sin 180^(°) = 0, sin 270^(°) = -1, sin 360^(°) = 0

Examiner insight

“Marks were sometimes lost for carelessness when the maxima and minima went too far above or below 1 or −1.”

Worked example 13 marks

Do not use a calculator. For 0^(°) leqslant x leqslant 360^(°), write down the maximum value of sin x and the values of x at which the maximum and the minimum occur.

  1. 1

    maximum value of sin x = 1 [1]

  2. 2

    maximum at x = 90^(°) [1]

  3. 3

    minimum (-1) at x = 270^(°) [1]

2. The graph of y = cos x

The graph of y = cos x is a wave of the same shape as the sine graph, but it starts at 1 when x = 0^(°). It falls through 0 at 90°, reaches a minimum of −1 at 180°, rises through 0 at 270° and is back at 1 at 360°.

The curve crosses the x axis at 90° and 270°. Sketch it over the whole range, from 0° to 360°.

cos 0^(°) = 1, cos 90^(°) = 0, cos 180^(°) = -1, cos 270^(°) = 0, cos 360^(°) = 1

Examiner insight

“Straight lines and parabolas were common wrong answers and not sketching over the whole domain of the graph lost the mark.”

Worked example 13 marks

Do not use a calculator. For 0^(°) leqslant x leqslant 360^(°), write down the value of cos 180^(°) and the two values of x for which cos x = 0.

  1. 1

    cos 180^(°) = -1 [1]

  2. 2

    cos x = 0 where the curve crosses the x axis

  3. 3

    x = 90^(°) [1]

  4. 4

    x = 270^(°) [1]

3. The graph of y = tan x

The graph of y = tan x is not a wave. It has no maximum and no minimum. It rises from 0 at 0° towards infinity as x approaches 90°.

At 90° and at 270° the value of tan x is undefined, and the graph has a break there. Between the breaks the curve rises from far below to far above the axis. It crosses the x axis at 0°, 180° and 360°.

tan 0^(°) = 0, tan 45^(°) = 1, tan 180^(°) = 0, tan 360^(°) = 0

Worked example 13 marks

Do not use a calculator. For 0^(°) leqslant x leqslant 360^(°), write down the two values of x at which tan x is undefined, and the value of tan 45^(°).

  1. 1

    x = 90^(°) [1]

  2. 2

    x = 270^(°) [1]

  3. 3

    tan 45^(°) = 1 [1]

4. Solving sin x = k and cos x = k

Take k between 0 and 1 first. A calculator, or an exact value, gives one angle between 0° and 90°. Call it α. The graph shows two solutions in 0^(°) leqslant x leqslant 360^(°), and symmetry gives the second.

For sin x, the two solutions total 180°. For cos x, they total 360°.

If k is negative, find α from the positive value, then place the angles where the curve is below the axis. For sin x they are 180^(°) + α and 360^(°) - α, which total 540°. For cos x they are 180^(°) - α and 180^(°) + α.

Know the exact values: sin 30^(°) = cos 60^(°) = 0.5, sin 45^(°) = cos 45^(°) = (√2)/2, sin 60^(°) = cos 30^(°) = (√3)/2.

sin x = k > 0 implies x = α and x = 180^(°) - α

cos x = k > 0 implies x = α and x = 360^(°) - α

Examiner insight

“The starting point was being able to rearrange the equation correctly and to know that sin30 = 0.5 or sin (–30) = –0.5.” “A few candidates used a graphical approach, which was successful when a suitable domain was used.”

Worked example 13 marks

Do not use a calculator. Solve sin x = (√3)/2 for 0^(°) leqslant x leqslant 360^(°).

  1. 1

    sin 60^(°) = (√3)/2, so x = 60^(°) [1]

  2. 2

    the sine graph is symmetrical about x = 90^(°), so the second solution is 180^(°) - 60^(°) [1]

  3. 3

    x = 60^(°) and x = 120^(°) [1]

5. Solving tan x = k

The tangent graph repeats every 180°, so there are two solutions in 0^(°) leqslant x leqslant 360^(°), 180° apart.

Find the angle α, between 0° and 90°, from the positive value of k. If k is positive, the solutions are α and α + 180^(°). If k is negative, they are 180^(°) - α and 360^(°) - α.

Know the exact values: tan 45^(°) = 1 and tan 60^(°) = √3.

tan x = k > 0 implies x = α and x = α + 180^(°)

tan x = k < 0 implies x = 180^(°) - α and x = 360^(°) - α

Examiner insight

“Using knowledge of the tan graph, candidates should be expecting two solutions to the given equation.” “Candidates then used the CAST diagram or sketches of the tan graph to identify the other angle of 240 in the required range.”

Worked example 13 marks

Do not use a calculator. Solve 2tan x - 2√3 = 0 for 0^(°) leqslant x leqslant 360^(°).

  1. 1

    rearrange: tan x = √3 [1]

  2. 2

    tan 60^(°) = √3, so x = 60^(°) [1]

  3. 3

    the second solution is 180° later: x = 240^(°) [1]

6. Equations to rearrange first

Rearrange until a single sin x, cos x or tan x stands alone. Then solve as before and give both solutions in the range.

When cos w is in a denominator, it cannot be zero. That is why the question excludes 90° and 270°. Take the reciprocal of both sides only when the whole side is one fraction.

1/(cos w) = k implies cos w = 1/k

Worked example 13 marks

f(w) = 1/(cosw) + 2 for values of w between 0° and 360°, w ≠ 90^(°) and w ≠ 270^(°). Solve f(w) = 0.

  1. 1

    1/(cosw) = -2, so cos w = -1/2 [1]

  2. 2

    cos 60^(°) = 1/2, and cosine is negative, so w = 180^(°) - 60^(°) = 120^(°) [1]

  3. 3

    the second solution is 180^(°) + 60^(°) = 240^(°) [1]

Recap

  • sin x and cos x are waves between −1 and 1. tan x has no maximum or minimum.
  • sin x: 0 at 0°, 180°, 360°. Maximum 1 at 90°. Minimum −1 at 270°.
  • cos x: 1 at 0° and 360°. 0 at 90° and 270°. Minimum −1 at 180°.
  • tan x: 0 at 0°, 180°, 360°. Undefined at 90° and 270°.
  • For positive k: sin x = k gives α and 180^(°) - α. cos x = k gives α and 360^(°) - α.
  • tan x = k: two solutions 180° apart.
  • Rearrange first. Exact values: sin 30^(°) = 0.5, cos 60^(°) = 0.5, sin 45^(°) = cos 45^(°) = (√2)/2, tan 45^(°) = 1, tan 60^(°) = √3.

End-of-chapter exercise

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

  1. State the maximum value of cos x.1 mark
  2. Do not use a calculator. For 0^(°) leqslant x leqslant 360^(°), write down the values of x for which sin x = 0.2 marks
  3. State the two values of x, between 0° and 360°, at which tan x is undefined.2 marks
  4. Do not use a calculator. Solve cos x = 1/2 for 0^(°) leqslant x leqslant 360^(°).2 marks
  5. Solve sin x = -1/2 for 0^(°) leqslant x leqslant 360^(°).3 marks
  6. Do not use a calculator. Solve tan x = -1 for 0^(°) leqslant x leqslant 360^(°).3 marks
  7. Use a graphic display calculator. Solve 5cos x + 2 = 0 for 0^(°) leqslant x leqslant 360^(°). Give each answer correct to 1 decimal place.3 marks
  8. Do not use a calculator. Solve 4/(sin x) + 4√2 = 0 for 0^(°) < x < 360^(°), x ≠ 180^(°).4 marks
  9. Use a graphic display calculator. Solve (3sin x + 2)(sin x - 4) = 0 for 0^(°) leqslant x leqslant 360^(°). Give each answer correct to 1 decimal place.4 marks
  10. Do not use a calculator. A wheel makes one full turn. When it has turned through x^(°), the height, h metres, of a seat above the ground is h = 12 - 10cos x, for 0^(°) leqslant x leqslant 360^(°). (a) Find the greatest height of the seat and the value of x at which it occurs. (b) Find the two values of x for which the seat is 17 m above the ground. (c) Find the number of degrees the wheel turns through while the seat is more than 17 m above the ground.6 marks

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