[discia] GCSE Maths › Geometry & Measures › Exact trig values › Practise
  1. Non-calculator 1 mark

    Write down the exact value of \(\sin 45^\circ\).

    You must not use a calculator.

    Show answer
    \(\frac{\sqrt{2}}{2}\)

    These values come from two triangles: half of an equilateral triangle of side 2 gives \(30^\circ\) and \(60^\circ\), and half of a square of side 1 gives \(45^\circ\). The values at \(0^\circ\) and \(90^\circ\) come from the graphs.

    For \(45^\circ\): \(\sin 45^\circ = \frac{\sqrt{2}}{2}\) and \(\cos 45^\circ = \frac{\sqrt{2}}{2}\).

    So \(\sin 45^\circ = \frac{\sqrt{2}}{2}\)

    You may also see this written as \(\frac{1}{\sqrt{2}}\). It is the same number — multiplying top and bottom by \(\sqrt{2}\) turns one into the other — and either form is correct.

  2. Non-calculator 2 marks
    Part of a table of exact trigonometric values is shown below.
    \(x\)\(\sin x\)\(\cos x\)\(\tan x\)
    \(30^\circ\)\(\frac{1}{2}\)
    Complete the row. Give each value exactly. You must not use a calculator.
    Show answer
    \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\tan 30^\circ = \frac{\sqrt{3}}{3}\)

    The whole row for \(30^\circ\) is

    \(\sin 30^\circ = \frac{1}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\tan 30^\circ = \frac{\sqrt{3}}{3}\)

    so the two missing entries are \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\tan 30^\circ = \frac{\sqrt{3}}{3}\).

    The tangent is a useful check on the other two, because \(\tan x = \frac{\sin x}{\cos x}\): dividing the sine by the cosine must give the tangent in the same row.

    The tangent is often written \(\frac{1}{\sqrt{3}}\) instead. That is the same number as \(\frac{\sqrt{3}}{3}\), and either form is correct.

  3. Non-calculator 2 marks

    Work out the exact value of \(\tan 45^\circ + \cos 0^\circ\).

    You must not use a calculator.

    Show answer
    \(2\)

    Write down the two exact values first: \(\tan 45^\circ = 1\) and \(\cos 0^\circ = 1\).

    Now substitute them in.

    \(\tan 45^\circ + \cos 0^\circ = 1 + 1 = 2\)

  4. Non-calculator 2 marks

    \(x\) is a number with \(0 \leq x \leq 90\).

    Solve \(\cos x^\circ = 0\).

    You must not use a calculator.

    Show answer
    \(90\)

    This is one of the exact values that has to be known: \(\cos 90^\circ = 0\).

    In the range \(0 \leq x \leq 90\) no other angle gives that value, so there is only one solution.

    \(x = 90\)

  5. Non-calculator 3 marks

    \(ABC\) is a right-angled triangle.

    Angle \(ABC = 90^\circ\), angle \(BAC = 30^\circ\) and \(AC = 12\) cm.

    (a)

    State the exact value of \(\sin 30^\circ\).

    (b)

    Work out the exact length of \(BC\).

    Show answer
    (a) \(\frac{1}{2}\)
    (b) \(6\) cm

    This is one of the values that has to be known rather than looked up.

    \(\sin 30^\circ = \frac{1}{2}\)

    \(AC\) is the hypotenuse, because it is the side opposite the right angle at \(B\). \(BC\) is opposite the angle at \(A\), so this is the sine ratio.

    \(\sin 30^\circ = \frac{BC}{ 12 }\)

    \(BC = 12 \times \frac{1}{2}\)

    \(BC = 6\) cm

    "Exact" means the surd stays as it is — turning it into a decimal would lose the exactness the question is asking for.

  6. Non-calculator 3 marks

    Show that the value of \(4\tan 60^\circ \times \cos 30^\circ + 2\sin 30^\circ\) is an integer.

    You must show your working. Do not use a calculator.

    Show answer
    Exact values written, the surds cancel: \(4 \times \frac{3}{2} = 6\) and \(1\) is added, so the value is \(7\), an integer

    Write down the exact values: \(\tan 60^\circ = \sqrt{3}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\).

    Combine the two exact values, and the surds cancel:

    \(\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2}\)

    The coefficients give a factor of \(4\), so this part of the expression is \(4 \times \frac{3}{2} = 6\).

    The other part is \(2 \times \frac{1}{2} = 1\).

    Add them: \(6 + 1 = 7\).

    The value is \(7\), which is an integer.

    • M1exact values written: \(\sqrt{3}\), \(\frac{\sqrt{3}}{2}\) and \(\frac{1}{2}\)
    • M1surd pair cancels to \(\frac{3}{2}\) and the coefficients give \(6\)
    • A1adds \(1\) and \(6\) to get \(7\), stated as an integer
  7. Non-calculator 4 marks

    Show that \(\dfrac{ 26 \cos 60^\circ - \tan 45^\circ }{ 8\cos 30^\circ }\) can be written as \(\tan\, (x + 16)^\circ\), where \((x + 16)^\circ\) is an acute angle.

    You must not use a calculator. State the value of \(x\).

    Show answer
    \(44\)

    Use the exact values \(\cos 60^\circ = \frac{1}{2}, \tan 45^\circ = 1\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\).

    Numerator: \(26 \times \frac{1}{2} - 1 = \frac{ 26 }{2} - \frac{ 2 }{2} = \frac{ 24 }{2}\)

    Denominator: \(8 \times \frac{\sqrt{3}}{2} = \frac{ 8\sqrt{3} }{2}\)

    Divide, the halves cancel: \(\dfrac{ 24 }{ 8\sqrt{3} } = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}\)

    This is a value from the table of exact values: \(\tan 60^\circ = \sqrt{3}\).

    So the expression can be written as \(\tan\, 60^\circ\). So \(x + 16 = 60\), giving \(x = 60 - 16 = 44\).

    • M1numerator \(26 \times \frac{1}{2} - 1 = \frac{ 24 }{2}\)
    • M1denominator \(8 \times \frac{\sqrt{3}}{2} = \frac{ 8\sqrt{3} }{2}\)
    • A1quotient simplified to \(\sqrt{3}\)
    • A1identified \(\tan 60^\circ = \sqrt{3}\), so \(x + 16 = 60\) and \(x = 44\)
  8. Non-calculator 3 marks

    \(x\) and \(y\) are acute angles.

    \(\cos x = \frac{\sqrt{3}}{2}\)

    \(\tan y = \frac{\sqrt{3}}{3}\)

    \(w = 6x - 4y\)

    Work out the exact value of \(4 \tan w\).

    You must show your working.

    Show answer
    \(4\sqrt{3}\)

    Recognise each angle from its exact value.

    \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), so \(x = 30^\circ\).

    \(\tan 30^\circ = \frac{\sqrt{3}}{3}\), so \(y = 30^\circ\).

    Substitute into \(w\).

    \(w = 6 \times 30^\circ - 4 \times 30^\circ = 180^\circ - 120^\circ = 60^\circ\)

    Now use the exact value at \(60^\circ\).

    \(4 \tan w = 4 \tan 60^\circ = 4 \times \sqrt{3} = 4\sqrt{3}\)

    • B1\(x = 30^\circ\) from \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
    • B1\(y = 30^\circ\) from \(\tan 30^\circ = \frac{\sqrt{3}}{3}\)
    • A1\(w = 60^\circ\), then \(4 \tan w = 4\sqrt{3}\)
  9. Non-calculator 5 marks
    8√3 cmDCBA60°Diagram NOT accurately drawn

    \(ABCD\) is a trapezium. \(AB\) is parallel to \(DC\).

    Angle \(BCD = 60^\circ\) and \(BC = 8\sqrt{3}\) cm.

    The area of the trapezium is \(180\) cm².

    \(AB : DC = 2 : 3\)

    Find the length of \(DC\).

    You must not use a calculator.

    Show answer
    \(18\ \text{cm}\)

    The height is not on the diagram. Drop a perpendicular from \(B\) to \(DC\): it is the opposite side to the angle of \(60^\circ\), and \(BC\) is the hypotenuse.

    \(h = BC \times \sin 60^\circ = 8\sqrt{3} \times \frac{\sqrt{3}}{2}\)

    \(h = \frac{ 24 }{2} = 12\) cm

    Let one part of the ratio be \(x\) cm, so \(AB = 2x\) and \(DC = 3x\).

    Area of a trapezium \(= \frac{1}{2}(AB + DC) \times h\):

    \(\frac{1}{2}(2x + 3x) \times 12 = 180\)

    \(\frac{ 60 }{2}x = 180\)

    \(x = \frac{ 360 }{ 60 } = 6\)

    \(DC = 3 \times 6 = 18\) cm

    • B1height \(BC \sin 60^\circ = 12\) cm
    • M1parallel sides written as \(2x\) and \(3x\) for one part \(x\)
    • M1area equation \(\frac{1}{2}(2x + 3x) \times 12 = 180\)
    • A1one part \(x = 6\)
    • A1length of \(DC\) is \(18\) cm
  10. Non-calculator 1 mark

    Write down the exact value of \(\cos 60^\circ\).

    You must not use a calculator.

    Show answer
    \(\frac{1}{2}\)

    These values come from two triangles: half of an equilateral triangle of side 2 gives \(30^\circ\) and \(60^\circ\), and half of a square of side 1 gives \(45^\circ\). The values at \(0^\circ\) and \(90^\circ\) come from the graphs.

    For \(60^\circ\): \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) and \(\cos 60^\circ = \frac{1}{2}\).

    So \(\cos 60^\circ = \frac{1}{2}\)