6. (a) Express $2 \, ext{cos}^2 x - 3 \, ext{sin}^2 x$ in the form $k \text{cos}(x + \alpha)$ where $k > 0$ and $0 \leq \alpha < 360$ - Scottish Highers Maths - Question 6 - 2019

We start by using the identity for the cosine of a sum:

$\text{cos}(x + \alpha) = \text{cos} x \text{cos} \alpha - \text{sin} x \text{sin} \alpha.$
We can express $2 \, \text{cos}^2 x - 3 \, \text{sin}^2 x$ as:

$2 \left(\text{cos}^2 x - \frac{3}{2} \text{sin}^2 x \right) = 2 \left(\text{cos}^2 x - \frac{3}{2}(1 - \text{cos}^2 x) \right)$ $= 2 \left(\text{cos}^2 x - \frac{3}{2} + \frac{3}{2} \text{cos}^2 x \right)$ $= \frac{7}{2} \text{cos}^2 x - 3.$
To compare coefficients, we know: $k \text{cos}(x + \alpha) = k \left(\text{cos} x \cos \alpha - \text{sin} x \sin \alpha \right).$ Thus, we determine:

• The coefficient for $\text{cos} x$ is $k \cos \alpha = 2$.
• The coefficient for $\text{sin} x$ is $k \sin \alpha = -3$.
  We can find $k$:

$k = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}.$
Next, we find $\alpha$:
Using: $\tan \alpha = \frac{-3}{2}, \quad\text{therefore}, \alpha = \text{atan2}(-3, 2) \approx \tan^{-1}(-1.5).$