8. (a) Express $5 \, ext{cos} \, x - 2 \, ext{sin} \, x$ in the form $k \, ext{cos}(x + a)$, where $k > 0$ and $0 < a < 2\pi$ - Scottish Highers Maths - Question 8 - 2016

To express the equation in the form $k \, \text{cos}(x + a)$, we start by using the compound angle formula:

$k \, \text{cos}(x + a) = k \, \text{cos} \, x \cdot \text{cos} \, a - k \, \text{sin} \, x \cdot \text{sin} \, a$

Equating this with $5 \, \text{cos} \, x - 2 \, \text{sin} \, x$, we can compare coefficients:

• For $\text{cos} \, x$: $k \, \text{cos} \, a = 5$
• For $\text{sin} \, x$: $-k \, \text{sin} \, a = -2$

From the first equation, we have: $k = \frac{5}{\text{cos} \, a}$

From the second equation, we find: $k = \frac{2}{\text{sin} \, a}$

Thus, $\frac{5}{\text{cos} \, a} = \frac{2}{\text{sin} \, a} \, \Rightarrow \, 5 \, \text{sin} \, a = 2 \, \text{cos} \, a$\n Dividing both sides by $\text{cos} \, a$, we obtain: $\tan(a) = \frac{2}{5}$

Now that we have found $\tan(a)$, we can find $k$ using: $k = \sqrt{5^2 + 2^2} = \sqrt{29}$

So, the expression can be written as: $\sqrt{29} \, \text{cos}(x + a)$

where $a = \tan^{-1}(\frac{2}{5})$.