To solve this problem, we’ll follow these steps:

1. Expand $\cos(x - 60^\circ)$ using the cosine angle subtraction formula.
2. Combine terms so that the expression is in the form $R \cos(x - \alpha)$.
3. Determine the values of $R$ and $\alpha$.

Step-by-Step Solution

Step 1: Expand $\cos(x - 60^\circ)$

The cosine angle subtraction formula is: $\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ$ Since $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get: $\cos(x - 60^\circ) = \cos x \cdot \frac{1}{2} + \sin x \cdot \frac{\sqrt{3}}{2}$ $= \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x$

Step 2: Substitute into the Expression

The given expression is: $2 \cos(x - 60^\circ) + \cos x$ Substitute $\cos(x - 60^\circ)$ from above: $= 2 \left( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \right) + \cos x$ $= \cos x + \sqrt{3} \sin x + \cos x$ $= 2 \cos x + \sqrt{3} \sin x$

Step 3: Express in the Form $R \cos(x - \alpha)$

We want to write $2 \cos x + \sqrt{3} \sin x$ in the form $R \cos(x - \alpha)$, where: $R \cos(x - \alpha) = R (\cos x \cos \alpha + \sin x \sin \alpha)$ Comparing terms, we get: $R \cos \alpha = 2 \quad \text{and} \quad R \sin \alpha = \sqrt{3}$

Step 4: Solve for $R$ and $\alpha$

To find $R$, use the Pythagorean identity: $R = \sqrt{(R \cos \alpha)^2 + (R \sin \alpha)^2}$ $= \sqrt{2^2 + (\sqrt{3})^2}$ $= \sqrt{4 + 3} = \sqrt{7}$

Next, find $\alpha$ by taking the tangent: $\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{\sqrt{3}}{2}$ $\alpha = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$ Using a calculator, we find: $\alpha \approx 30.96^\circ$

Final Answer

The expression $2 \cos(x - 60^\circ) + \cos x$ can be written as: $\sqrt{7} \cos(x - 30.96^\circ)$ where: $R = \sqrt{7} \quad \text{and} \quad \alpha \approx 30.96^\circ$

Would you like further details on any part of the solution?

Related Questions

1. How does the cosine angle subtraction formula work for other angles?
2. Why is the form $R \cos(x - \alpha)$ useful in trigonometric expressions?
3. How do you convert other trigonometric expressions to the form $R \cos(x - \alpha)$?
4. What if we had a different phase shift, like $45^\circ$ or $90^\circ$? How would that affect the calculation?
5. How can we use this technique in solving real-world problems involving oscillations or waveforms?

Tip

To rewrite expressions as $R \cos(x - \alpha)$, always start by expanding and identifying coefficients of $\cos x$ and $\sin x$ for easier comparison.