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

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

Question 3

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

Step 1

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Answer

To express the given equation in the form r sin(x + α), we start by using the compound angle formula:

$r \sin(x + \alpha) = r \sin x \cos \alpha + r \cos x \sin \alpha.$
By comparing coefficients with the original expression, we can write:

Next, we find r:

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

Now, we can find ( \alpha ) using the following:

$\tan \alpha = \frac{5}{4}$
From this, we have:

$\alpha = \tan^{-1}(\frac{5}{4}) \approx 0.896.$

Thus, we can express the equation as:

$4 \sin x + 5 \cos x = \sqrt{41} \sin(x + 0.896).$

Step 2

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Answer

Using the result from part (a), we substitute into the equation:

$\sqrt{41} \sin(x + 0.896) = 5.5.$

Now dividing both sides by ( \sqrt{41} ):

$\sin(x + 0.896) = \frac{5.5}{\sqrt{41}} \approx 0.860.$

Next, we find the general solutions for ( x + 0.896 ):

( x + 0.896 = \sin^{-1}(0.860) ) which gives the primary angle, let's denote this as ( \beta ) and calculate it:
( \beta \approx 1.063. )
The second solution in the interval is given by:
( x + 0.896 = \pi - \beta \approx \pi - 1.063. )

Now solve for ( x ):

Thus, the solutions for ( 0 \leq x < 2\pi ) are approximately:
( x \approx 0.167 ) and ( x \approx 3.278. )