Recall all the exact values that you need to know for the cosine ratio.

$\begin{array}{rcl} \cos (0 °) & = & 1 \\ \cos (30 °) & = & \frac{\sqrt{3}}{2} \\ \cos (45 °) & = & \frac{\sqrt{2}}{2} \\ \cos (60 °) & = & \frac{1}{2} \\ \cos (90 °) & = & 1 \end{array}$

Writing one of these values [1]

As there are two unknowns you need to form two equations. Use the two angles that give the easiest values, 0° and 90°. Substitute these into the given equation.

$\begin{array}{rcl} a \cos (0 °) + b & = & a \times 1 + b = a + b \\ a \cos (90 °) + b = a \times 0 + b = b \end{array}$

Use the table to find the y values when x equals those angles.

$\begin{array}{rcl} x & = & 0, y = 3 \Rightarrow a + b = 3 1 \\ x & = & 90, y = 1 \Rightarrow b = 1 2 \end{array}$

[1]

The second equation gives you the value of b straightaway so substitute that into the first equation to find the value of a.

$\begin{array}{rcl} a + 1 & = & 3 \\ a & = & 2 \end{array}$

[1]

You can now substitute these values into the equation.

$\begin{array}{rcl} y & = & 2 \cos (x °) + 1 \end{array}$

Substitute x = 45 into the equation and use the exact value for the cosine ratio.

$\begin{array}{rcl} y & = & 2 \cos (45 °) + 1 \\ = & 2 \times \frac{\sqrt{2}}{2} + 1 \\ = & \sqrt{2} + 1 \end{array}$

$x = 45 \Rightarrow y = 1 + \sqrt{2}$ [1]
It is acceptable to write: $y = \sqrt{2} + 1$

$x$	$0$	$30$	$60$	$90$	$120$	$150$	$180$
$y$	$3$	$1 + \sqrt{3}$	$2$	$1$	$0$	$1 - \sqrt{3}$	$- 1$

Right-Angled Triangles - Pythagoras & Trigonometry (OCR GCSE Maths)

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