Find the value of ` sin^(-1) ( cos. ( 33pi)/5)` .

Options

A. `-pi/5`
B. `-pi/10`
C. `pi/5`
D. `pi/10`

Answer

To find the value of \( \sin^{-1} \left( \cos \left( \frac{33\pi}{5} \right) \right) \), we can follow these steps:

### Step 1: Simplify the angle
First, we need to simplify the angle \( \frac{33\pi}{5} \). We can express it in terms of a standard angle by subtracting multiples of \( 2\pi \).

\[ \frac{33\pi}{5} - 6\pi = \frac{33\pi}{5} - \frac{30\pi}{5} = \frac{3\pi}{5} \]

Thus, we have:

\[ \cos \left( \frac{33\pi}{5} \right) = \cos \left( \frac{3\pi}{5} \right) \]

**Hint for Step 1:** Use the periodicity of cosine, which is \( 2\pi \), to simplify the angle.

### Step 2: Use the identity for cosine
Next, we can use the identity for cosine, which states that:

\[ \cos(2\pi + \theta) = \cos(\theta) \]

Since \( \frac{3\pi}{5} \) is already in the range of \( [0, 2\pi] \), we can directly use it.

### Step 3: Find \( \sin^{-1} \)
Now, we need to find \( \sin^{-1} \left( \cos \left( \frac{3\pi}{5} \right) \right) \).

Using the co-function identity, we know that:

\[ \cos \left( \frac{3\pi}{5} \right) = \sin \left( \frac{\pi}{2} - \frac{3\pi}{5} \right) = \sin \left( \frac{\pi}{2} - \frac{3\pi}{5} \right) = \sin \left( \frac{\pi}{10} \right) \]

### Step 4: Apply the inverse sine function
Now we can apply the inverse sine function:

\[ \sin^{-1} \left( \cos \left( \frac{3\pi}{5} \right) \right) = \sin^{-1} \left( \sin \left( \frac{\pi}{10} \right) \right) \]

### Step 5: Determine the final value
Since \( \frac{\pi}{10} \) is within the range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we have:

\[ \sin^{-1} \left( \sin \left( \frac{\pi}{10} \right) \right) = \frac{\pi}{10} \]

Thus, the final answer is:

\[ \sin^{-1} \left( \cos \left( \frac{33\pi}{5} \right) \right) = \frac{\pi}{10} \]

**Final Answer:** \( \frac{\pi}{10} \)