https://file.aiquickdraw.com/mathbot/file/1727112985200k8q7pr0i.jpg

Let's break down the problem from the image:

### Part (a):
We are asked to show that the expression \( 2 \cos(x - 60^\circ) + \cos(x) \) can be written in the form \( R \cos(x - \alpha) \), where \( R > 0 \) and \( 0^\circ < \alpha < 90^\circ \), and to determine the values of \( R \) and \( \alpha \).

#### Step 1: Expand \( \cos(x - 60^\circ) \)
We use the angle subtraction identity for cosine:
\[
\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ
\]
Substituting the values \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we get:
\[
\cos(x - 60^\circ) = \cos x \cdot \frac{1}{2} + \sin x \cdot \frac{\sqrt{3}}{2}
\]

#### Step 2: Substitute into the expression
Now substitute this into the given expression \( 2 \cos(x - 60^\circ) + \cos x \):
\[
2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right) + \cos x
\]
Simplifying:
\[
= \cos x + \sqrt{3} \sin x + \cos x
\]
\[
= 2 \cos x + \sqrt{3} \sin x
\]

#### Step 3: Express in the form \( R \cos(x - \alpha) \)
We now have the expression \( 2 \cos x + \sqrt{3} \sin x \), which is in the form \( R \cos(x - \alpha) \). To rewrite this expression in the desired form, we need to find \( R \) and \( \alpha \).

We use the following identities:
\[
R \cos(x - \alpha) = R \cos \alpha \cos x + R \sin \alpha \sin x
\]
By comparing coefficients with \( 2 \cos x + \sqrt{3} \sin x \), we get:
\[
R \cos \alpha = 2 \quad \text{and} \quad R \sin \alpha = \sqrt{3}
\]

#### Step 4: Solve for \( R \)
We can find \( R \) using the Pythagorean identity:
\[
R = \sqrt{(R \cos \alpha)^2 + (R \sin \alpha)^2}
\]
Substituting the values:
\[
R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}
\]

#### Step 5: Solve for \( \alpha \)
Now, to find \( \alpha \), we take the ratio:
\[
\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{\sqrt{3}}{2}
\]
Thus:
\[
\alpha = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) \approx 40.89^\circ
\]

So, the values are:
- \( R = \sqrt{7} \approx 2.65 \)
- \( \alpha \approx 40.89^\circ \)

### Part (b):
We are asked to find the value of \( x \) in the interval \( 0^\circ < x < 360^\circ \) for which \( 2 \cos(x - 60^\circ) + \cos x \) takes its least possible value.

Since the expression has now been rewritten as \( R \cos(x - \alpha) \), the minimum value of a cosine function occurs when \( \cos(x - \alpha) = -1 \).

So, we solve:
\[
x - \alpha = 180^\circ \quad \Rightarrow \quad x = 180^\circ + \alpha
\]
Substitute \( \alpha \approx 40.89^\circ \):
\[
x \approx 180^\circ + 40.89^\circ = 220.89^\circ
\]

Thus, the value of \( x \) for which the expression takes its least possible value is approximately \( 220.89^\circ \).

---

Would you like more details or further clarification on any part? 

Here are 5 related questions to explore:
1. How can the sum of cosine and sine terms be represented as a single cosine function?
2. Why do we use the Pythagorean identity to find \( R \) in such problems?
3. What is the geometric interpretation of the minimum value of a cosine function?
4. How does the phase shift \( \alpha \) affect the graph of \( \cos(x - \alpha) \)?
5. How would the solution change if the coefficient of \( \cos(x - 60^\circ) \) was different?

**Tip**: When solving trigonometric equations, always remember that cosine and sine terms can often be combined using a single-phase shifted cosine function, simplifying the expression.

By first expanding cos(x - 60°), show that the expression 2 cos(x - 60°) + cos(x) can be written in the form R cos(x - α), where R > 0 and 0° < α < 90°. Give the exact value of R and the value of α correct to 2 decimal places.

Explore how to expand and simplify trigonometric expressions, including the cosine subtraction formula. Learn how to rewrite expressions in the form R cos(x - α), and solve for minimum values using trigonometric identities. Suitable for students in Grades 10-12.

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