The expression in your image is $2 \cos x - 5 \sin x$ and needs to be rewritten in the form $k \cos (x + a)$, where $k$ is a constant and $a$ is the phase shift in degrees.

Solution

We can use the following identity to express a linear combination of sine and cosine as a single cosine function:

$R \cos (x + \alpha) = R \cos \alpha \cos x - R \sin \alpha \sin x$

Comparing this with $2 \cos x - 5 \sin x$, we have:

$2 \cos x - 5 \sin x = R \cos \alpha \cos x - R \sin \alpha \sin x$

This gives us the system of equations:

$R \cos \alpha = 2 \quad \text{(1)}$ $R \sin \alpha = 5 \quad \text{(2)}$

Step 1: Find $R$

To find $R$, we use the Pythagorean identity:

$R = \sqrt{(R \cos \alpha)^2 + (R \sin \alpha)^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$

So, $R = \sqrt{29}$.

Step 2: Find $\alpha$

Next, we find $\alpha$ by taking the ratio of equation (2) to equation (1):

$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{5}{2}$

$\alpha = \tan^{-1} \left( \frac{5}{2} \right)$

Using a calculator, we find:

$\alpha \approx 68.2^\circ$

Thus, the expression $2 \cos x - 5 \sin x$ can be rewritten as:

$\sqrt{29} \cos (x + 68.2^\circ)$

Final Answer

The expression in the form $k \cos(x + a)$ is:

$\boxed{\sqrt{29} \cos(x + 68.2^\circ)}$

Let me know if you want more details or have any questions!

Related questions:

1. Can you explain why we use the identity $R \cos(x + \alpha)$ to solve this?
2. How would the solution change if the coefficients of $\cos x$ and $\sin x$ were different?
3. What if the expression was in terms of sine instead of cosine?
4. How do we convert radians to degrees when working with angles in trigonometry?
5. Can we express this in terms of sine instead of cosine, and how?

Tip:

Always check your angle mode (degrees or radians) when using trigonometric functions on a calculator!