To find the minimum and maximum values of the expression $f(x) = 5 \sin x - 2 \cos x$, we can rewrite it in a form that makes it easier to identify the extreme values.

Step 1: Express in the form $R \sin(x + \alpha)$

We can rewrite $5 \sin x - 2 \cos x$ as a single trigonometric function of the form $R \sin(x + \alpha)$, where $R$ is a constant amplitude and $\alpha$ is a phase shift.

To do this, we use the following trigonometric identity: $R \sin(x + \alpha) = R \left( \sin x \cos \alpha + \cos x \sin \alpha \right)$ By comparing this with $5 \sin x - 2 \cos x$, we identify: $R \cos \alpha = 5 \quad \text{and} \quad R \sin \alpha = -2$

Step 2: Solve for $R$ and $\alpha$

To find $R$, use the Pythagorean identity: $R^2 = (R \cos \alpha)^2 + (R \sin \alpha)^2$ $R^2 = 5^2 + (-2)^2 = 25 + 4 = 29$ $R = \sqrt{29}$

Now, to find $\alpha$, we use the relation: $\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{-2}{5}$ $\alpha = \tan^{-1} \left( \frac{-2}{5} \right)$ Thus, $\alpha = \tan^{-1} \left( -\frac{2}{5} \right)$.

Step 3: Find the maximum and minimum values

The maximum and minimum values of $R \sin(x + \alpha)$ occur when $\sin(x + \alpha) = 1$ and $\sin(x + \alpha) = -1$, respectively. Therefore:

• Maximum value: $R \times 1 = \sqrt{29}$
• Minimum value: $R \times (-1) = -\sqrt{29}$

Thus, the maximum value of $5 \sin x - 2 \cos x$ is $\sqrt{29}$ and the minimum value is $-\sqrt{29}$.

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Related Questions:

1. How do we convert a trigonometric expression into the form $R \sin(x + \alpha)$?
2. What are other methods to find the maximum and minimum values of a trigonometric expression?
3. How do we graph the function $f(x) = 5 \sin x - 2 \cos x$?
4. What is the general method to find the amplitude and phase shift of a trigonometric function?
5. Can we find the maximum and minimum values of this function without using the phase shift method?

Tip: When combining sine and cosine terms, rewriting them in the form $R \sin(x + \alpha)$ simplifies finding the amplitude and extreme values.