A curve has equation y = a sin x + b cos x where a and b are constants - AQA - A-Level Maths Mechanics - Question 6 - 2019 - Paper 2

The curve passes through the point ( \left( \frac{\pi}{3}, 2\sqrt{3} \right) ). Substitute ( x = \frac{\pi}{3} ) into the equation:

$$y = a \sin \left(\frac{\pi}{3}\right) + b \cos \left(\frac{\pi}{3}\right)$$

Using ( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} ) and ( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} ):

$$2\sqrt{3} = a \left(\frac{\sqrt{3}}{2}\right) + b \left(\frac{1}{2}\right).$$

Multiplying through by 2 to eliminate the fractions:

$$4\sqrt{3} = a\sqrt{3} + b.$$

Now we have the two equations:

1. ( a^2 + b^2 = 16 )
2. ( 4\sqrt{3} = a\sqrt{3} + b )

Using the first equation, we can express one variable in terms of the other. Let's solve for b:

$$b = \sqrt{16 - a^2}.$$

Substituting this into the second equation:

$$4\sqrt{3} = a\sqrt{3} + \sqrt{16 - a^2}.$$

Squaring both sides leads to:

$$(4\sqrt{3} - a\sqrt{3})^2 = 16 - a^2.$$

This results in a quadratic equation in a. Solve this to find the exact values of a and subsequently b.