Scottish Highers Maths Addition Formula: 3. (a) Express 4sin x + 5cos x i

3. (a) Express 4sin x + 5cos x in the form k sin(x + α) where k > 0 and 0 < α < 2π - Scottish Highers Maths - Question 3 - 2022

Question 3

3. (a) Express 4sin x + 5cos x in the form k sin(x + α) where k > 0 and 0 < α < 2π. (b) Hence solve 4sin x + 5cos x = 5.5 for 0 ≤ x < 2π.

Worked Solution & Example Answer:3. (a) Express 4sin x + 5cos x in the form k sin(x + α) where k > 0 and 0 < α < 2π - Scottish Highers Maths - Question 3 - 2022

Step 1

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Answer

To express the function in the required form, we start with the identity:

$k \sin(x + \alpha) = k(\sin x \cos \alpha + \cos x \sin \alpha)$

Matching coefficients with the original expression, we have:

Next, we solve for k using the Pythagorean identity:

$k = \sqrt{(4^2) + (5^2)} = \sqrt{16 + 25} = \sqrt{41}$

Now, using the equations for $\cos \alpha$ and $\sin \alpha$ we get:

$\cos \alpha = \frac{4}{\sqrt{41}}, \quad \sin \alpha = \frac{5}{\sqrt{41}}$

Thus, we can rewrite the original expression as:

$4 \sin x + 5 \cos x = \sqrt{41} \sin(x + \alpha),$

where, $\alpha = \arctan\left(\frac{5}{4}\right)$ .

Step 2

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Answer

Using the established expression, we rewrite the equation:

$\sqrt{41} \sin(x + \alpha) = 5.5$

Dividing by $\sqrt{41}$ gives:

$\sin(x + \alpha) = \frac{5.5}{\sqrt{41}}$

Next, we compute:

$\frac{5.5}{\sqrt{41}} \approx 0.8596$

For the equation $\sin \theta = 0.8596$ , we find:

Calculating these provides:

This yields two solutions which we need to evaluate within the range $[0, 2\pi]$ .