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

Now substituting the point into the equation: $y = a \sin\left(\frac{\pi}{3}\right) + b \cos\left(\frac{\pi}{3}\right)$ We know: $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ Substituting these values in: $2\sqrt{3} = a \cdot \frac{\sqrt{3}}{2} + b \cdot \frac{1}{2}$ Multiplying through by 2 gives: $4\sqrt{3} = a\sqrt{3} + b$

Therefore, we can write: 2. $a\sqrt{3} + b = 4\sqrt{3}$

Now we have the two equations:

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

From equation 2, we can express $b$ in terms of $a$: $b = 4\sqrt{3} - a\sqrt{3}$

Substituting this into equation 1: $a^2 + (4\sqrt{3} - a\sqrt{3})^2 = 16$ Expanding: $a^2 + (48 - 8a + a^2) = 16$ $2a^2 - 8a + 32 = 0$ Dividing through by 2: $a^2 - 4a + 16 = 0$ Using the quadratic formula: $a = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$ This gives: $a = 4 ext{ or } a = 0$ However, substitute back to get the values for $b$.

We get:

• If $a = 4$, then $b = 0$ (checking in the earlier equations).
• If $a = 0$, then we solve: $0 + b = 4\sqrt{3} \Rightarrow b = 4\sqrt{3}$

Thus the exact values of $a$ and $b$ are:

1. $a = 4$, $b = 0$ or
2. $a = 0$, $b = 4\sqrt{3}$