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Cambridge IGCSE0580

Exact trigonometric values

Mathematics 0580 Chapter Notes

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Exact trigonometric values
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In this chapter

  1. 1Exact values of sine and cosine
  2. 2Exact values of tangent
  3. 3Using exact values in a right-angled triangle
  4. 4Giving the answer in exact form

1. Exact values of sine and cosine

An exact value is written as a fraction or a surd, never as a rounded decimal. Learn the values of sin x and cos x for x = 0^(°), 30^(°), 45^(°), 60^(°) and 90^(°).

The sine values rise: 0, 1/2, (√2)/2, (√3)/2, 1. The cosine values are the same list in reverse order.

The value (√2)/2 is also written 1/(√2). If you forget a value, rebuild it from the 90^(°), 60^(°), 30^(°) triangle.

sin 0^(°) = 0, cos 0^(°) = 1

sin 30^(°) = 1/2, cos 30^(°) = (√3)/2

sin 45^(°) = (√2)/2, cos 45^(°) = (√2)/2

sin 60^(°) = (√3)/2, cos 60^(°) = 1/2

sin 90^(°) = 1, cos 90^(°) = 0

Examiner insight

“Some candidates listed a table of trigonometric ratios for 30°, 45° and 60° angles, but often did not identify the one that they intended to use.” “In other instances, incorrect values were quoted for the relevant trigonometric ratios.”

Worked example 11 mark

Complete this statement using a value from the list: 1/(√3), 1/2, 1/(√2), √3, (√3)/2.

cos 45^(°) =

  1. 1

    cos 45^(°) = (√2)/2, and (√2)/2 = 1/(√2)

  2. 2

    cos 45^(°) = 1/(√2) [1]

2. Exact values of tangent

Learn tan x for x = 0^(°), 30^(°), 45^(°) and 60^(°). The list stops at 60^(°).

You can build any of them from the sine and cosine of the same angle. Divide the sine by the cosine.

The values rise: 0, 1/(√3), 1, √3. So √3 belongs to 60^(°) and 1/(√3) belongs to 30^(°).

tan x = (sin x)/(cos x)

tan 0^(°) = 0, tan 30^(°) = 1/(√3)

tan 45^(°) = 1, tan 60^(°) = √3

Examiner insight

“However, it was evident that many candidates did not have a method of deriving tan60 without a calculator.”

Worked example 12 marks

y = sin 60^(°) × tan 30^(°).

Find the value of y.

  1. 1

    sin 60^(°) = (√3)/2 and tan 30^(°) = 1/(√3) [1]

  2. 2

    y = (√3)/2 × 1/(√3) = 1/2 [1]

3. Using exact values in a right-angled triangle

Choose the ratio from the sides you have and the side you want. Write the equation first. Then replace the ratio by its exact value.

Rearrange with care. To find a hypotenuse from the adjacent side, divide by the cosine. Do not multiply.

The shortest distance from a point to a line is the perpendicular. It makes a new right-angled triangle.

sin = opposite/hypotenuse

cos = adjacent/hypotenuse

tan = opposite/adjacent

Examiner insight

“Some used an incorrect value for cos 30 in the correct equation.” “Quite a few found the length of AB and gave this as the answer for AC.”

Worked example 17 marks

In triangle ABC, angle B = 90^(°), BC = 12 cm and angle C = 30^(°).

(a) Find the length of AC. Give your answer in the form a√b.

(b) Find the shortest distance from B to AC.

  1. 1

    (a) AC is the hypotenuse and BC is adjacent to 30^(°): cos 30^(°) = 12/AC [1]

  2. 2

    (√3)/2 = 12/AC [1]

  3. 3

    AC = 12 ÷ (√3)/2 = 24/(√3) [1]

  4. 4

    AC = (24√3)/3 = 8√3 cm [1]

  5. 5

    (b) The shortest distance is the perpendicular BD. In triangle BDC, BC is the hypotenuse [1]

  6. 6

    sin 30^(°) = BD/12 [1]

  7. 7

    BD = 12 × 1/2 = 6 cm [1]

4. Giving the answer in exact form

When a question says "exact value" or "in the form a√b", keep the surd. Never use a rounded decimal in the working.

A surd left in a denominator is not in the form asked. Multiply top and bottom by that surd to rationalise.

Use the exact value of each ratio. Then simplify until no trigonometric ratio is left.

9/(√3) = (9√3)/3 = 3√3

Examiner insight

“Others reached a correct expression but didn’t simplify it by eliminating the trigonometrical ratios, such as 2x sin60.” “The most common error was an inexact answer that arose from premature approximation of the evaluation of each of the surds separately prior to multiplying.”

Worked example 14 marks

A right-angled triangle has an angle of 60^(°). The side opposite this angle is 9 cm. The side adjacent to this angle is x cm.

Find the exact value of x. Give your answer in the form a√b.

  1. 1

    tan 60^(°) = 9/x [1]

  2. 2

    √3 = 9/x, so x = 9/(√3) [1]

  3. 3

    x = (9√3)/3 [1]

  4. 4

    x = 3√3 [1]

Recap

  • Extended: sin x and cos x at 0^(°), 30^(°), 45^(°), 60^(°), 90^(°) are exact: 0, 1/2, (√2)/2, (√3)/2, 1 for sine, in reverse for cosine.
  • Extended: tan 0^(°) = 0, tan 30^(°) = 1/(√3), tan 45^(°) = 1, tan 60^(°) = √3.
  • tan x = (sin x)/(cos x) rebuilds a tangent from a sine and a cosine.
  • Write the ratio equation first, then put in the exact value.
  • To find a hypotenuse from the adjacent side, divide by the cosine.
  • Keep the surd. Rationalise any surd left in a denominator. Never round part-way.

End-of-chapter exercise

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

  1. Write down the exact value of cos 60^(°) and of cos 90^(°).2 marks
  2. Write down the exact value of tan 45^(°) and of tan 60^(°).2 marks
  3. Do not use a calculator. Work out sin 60^(°) × cos 30^(°).2 marks
  4. Do not use a calculator. A right-angled triangle has hypotenuse 10 cm and an angle of 30^(°). Find the exact length of the side opposite the 30^(°) angle.2 marks
  5. Do not use a calculator. Work out (sin 45^(°))² + (cos 45^(°))².3 marks
  6. A right-angled triangle has an angle of 60^(°). The side adjacent to this angle is 5 cm. Find the exact length of the side opposite the 60^(°) angle. Then give this length correct to 3 significant figures.3 marks
  7. Do not use a calculator. A right-angled triangle has an angle of 45^(°). The side adjacent to this angle is 6 cm. Find the exact length of the hypotenuse. Give your answer in the form a√b.3 marks
  8. A vertical mast stands on level ground. A cable runs from the top of the mast to the ground. The cable makes an angle of 60^(°) with the ground. The foot of the cable is 10 m from the foot of the mast. Find the exact height of the mast and the exact length of the cable.4 marks
  9. Do not use a calculator. A triangular field PQR has a right angle at Q. Angle QPR = 60^(°) and PQ = 8 m. A water pipe runs from Q to the fence PR by the shortest route and meets the fence at D. Find the exact length of the pipe QD and the exact length of DR.6 marks
  10. Do not use a calculator. A sail ABCD is made from two right-angled triangles joined along AC. In triangle ABC, angle ABC = 90^(°), angle BAC = 30^(°) and BC = 6 m. In triangle ACD, angle ACD = 90^(°) and angle CAD = 30^(°). B and D are on opposite sides of AC. Find the exact perimeter of the sail. Give your answer in the form p + q√3.7 marks

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