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

Substituting the point $\left(\frac{\pi}{3}, 2\sqrt{3}\right)$ into the equation:

$y = a \sin\left(\frac{\pi}{3}\right) + b \cos\left(\frac{\pi}{3}\right)$
This gives: $2\sqrt{3} = a \cdot \frac{\sqrt{3}}{2} + b \cdot \frac{1}{2}$
Simplifying, we get: $2\sqrt{3} = \frac{\sqrt{3}}{2}a + \frac{1}{2}b$
Multiplying by 2: $4\sqrt{3} = \sqrt{3}a + b$

We now have a system of two equations:
1.
$a^2 + b^2 = 16$
2.
$b = 4\sqrt{3} - \sqrt{3}a$
Substituting equation 2 into equation 1 gives:

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

Expanding this, we get:

$a^2 + (48 - 32\sqrt{3}a + 3a^2) = 16$
Combining like terms gives: $4a^2 - 32\sqrt{3}a + 32 = 0$
Dividing through by 4: $a^2 - 8\sqrt{3}a + 8 = 0$
Using the quadratic formula:

$a = \frac{8\sqrt{3} \pm \sqrt{(8\sqrt{3})^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$ This simplifies to: $a = 4\sqrt{3} \pm 4\sqrt{2}$
Calculating the values, we find: $$a = 4$(using $\left(\frac{\pi}{3}, 2\sqrt{3}\right)$). Calculating $b$ using $b = 4\sqrt{3} - \sqrt{3}a$ leads to the values for $b$.