https://file.aiquickdraw.com/mathbot/file/1727365302454pcl86ryn.jpg

Let's break down the questions and solve them step by step.

### Part (i)
We are asked to express the equation \( \cos x + 3 \sin x \) in the form \( R \cos(x - \alpha) \), where \( R > 0 \) and \( 0^\circ < \alpha < 90^\circ \). To do this, we use the following trigonometric identity:

\[
R \cos(x - \alpha) = R (\cos \alpha \cos x + \sin \alpha \sin x)
\]

Now, comparing this to the original expression \( \cos x + 3 \sin x \), we get:

\[
R \cos \alpha = 1 \quad \text{(coefficient of } \cos x\text{)}
\]
\[
R \sin \alpha = 3 \quad \text{(coefficient of } \sin x\text{)}
\]

To find \( R \), we square both equations and add them:

\[
R^2 = (R \cos \alpha)^2 + (R \sin \alpha)^2
\]
\[
R^2 = 1^2 + 3^2 = 1 + 9 = 10
\]
\[
R = \sqrt{10}
\]

Next, to find \( \alpha \), we use:

\[
\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{3}{1} = 3
\]
\[
\alpha = \tan^{-1}(3) \approx 71.57^\circ
\]

Thus, the exact value of \( R \) is \( \sqrt{10} \), and \( \alpha \approx 71.57^\circ \).

### Part (ii)
We are given the equation:

\[
\cos 2\theta + 3 \sin 2\theta = 2
\]

Using the result from part (i), we can express this in the form \( R \cos(2\theta - \alpha) \), where \( R = \sqrt{10} \) and \( \alpha \approx 71.57^\circ \). The equation becomes:

\[
\sqrt{10} \cos(2\theta - 71.57^\circ) = 2
\]

Now, solve for \( \cos(2\theta - 71.57^\circ) \):

\[
\cos(2\theta - 71.57^\circ) = \frac{2}{\sqrt{10}} \approx 0.6325
\]

Next, we take the inverse cosine:

\[
2\theta - 71.57^\circ = \cos^{-1}(0.6325) \approx 50.51^\circ
\]

Now solve for \( 2\theta \):

\[
2\theta = 50.51^\circ + 71.57^\circ = 122.08^\circ
\]
\[
\theta = \frac{122.08^\circ}{2} \approx 61.04^\circ
\]

Thus, the solution is \( \theta \approx 61.04^\circ \).

Would you like a more detailed explanation or have any further questions?

### Follow-up Questions:
1. How do we derive the expression for \( R \cos(x - \alpha) \)?
2. What is the importance of squaring both equations when solving for \( R \)?
3. Why is \( \tan \alpha = 3 \) used to find \( \alpha \)?
4. How does the solution for part (i) help in solving part (ii)?
5. Can we solve part (ii) without using the result from part (i)?

### Tip:
When solving trigonometric equations, it's useful to transform them into a simpler form using identities like \( R \cos(x - \alpha) \), which helps in finding solutions more easily.

Express cos x + 3 sin x in the form R cos(x − α), where R > 0 and 0° < α < 90°, giving the exact value of R and the value of α correct to 2 decimal places. Hence solve the equation cos 2θ + 3 sin 2θ = 2, for 0° < θ < 90°.

This problem demonstrates how to express cos x + 3 sin x in the form R cos(x − α), solve for R and α, and use the result to solve cos 2θ + 3 sin 2θ = 2 for θ between 0° and 90°. Step-by-step solution included.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation