The table shows some values of x and y that satisfy the equation y = a cos(x) + b | x | 0 | 30 | 60 | 90 | 120 | 150 | 180 | |-----|---|-----------|----|----|-----|-----|-----| | y | 3 | 1 + \sqrt{3} | 2 | 1 | 0 | 1 - \sqrt{3} | -1 | Find the value of y when x = 45. - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

From the table, we can substitute x = 0, 30, and 60 to formulate equations in terms of a and b. For example, we can take y values corresponding to known x values to create equations:

1. When x = 0: [ y = a \cdot \cos(0) + b = 3 \Rightarrow a + b = 3 ]
2. When x = 60: [ y = a \cdot \cos(60) + b = 2 \Rightarrow \frac{a}{2} + b = 2 ]

We now have two equations:

1. ( a + b = 3 )
2. ( \frac{a}{2} + b = 2 )

From the first equation, we can express b in terms of a: ( b = 3 - a ). Substituting into the second equation, we get: [ \frac{a}{2} + (3 - a) = 2 ]

Solving this gives: [ \frac{a}{2} - a + 3 = 2 \Rightarrow -\frac{a}{2} + 3 = 2 \Rightarrow -\frac{a}{2} = -1 \Rightarrow a = 2 ]

Substituting the value of a back into the equation for b: [ b = 3 - a = 3 - 2 = 1 ]

Now that we have the values of a and b, we can substitute them into the original equation for x = 45: [ y = 2 \cdot \cos(45) + 1 = 2 \cdot \frac{1}{\sqrt{2}} + 1 = \sqrt{2} + 1 ]

Hence, the value of y when x = 45 is ( y = \sqrt{2} + 1 ).