The value of `sin^(-1)sin17+cos^(-1)cos10` is equal to

Options

A. 27
B. `-27`
C. `17-5pi`
D. `9pi-27`

Answer

To solve the expression \( \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) \), we will follow these steps:

### Step 1: Evaluate \( \sin^{-1}(\sin 17) \)

The function \( \sin^{-1}(x) \) (the inverse sine function) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( 17 \) is outside this range, we need to find an equivalent angle within the range.

To do this, we can express \( 17 \) in terms of \( \pi \):
\[ 17 \text{ radians} \approx 5.4 \pi \text{ (since } \pi \approx 3.14\text{)} \]
We can write:
\[ 17 = 5\pi + (17 - 5\pi) \]
Calculating \( 17 - 5\pi \):
\[ 5\pi \approx 15.7 \quad \Rightarrow \quad 17 - 15.7 = 1.3 \]
Thus, we have:
\[ \sin^{-1}(\sin 17) = \sin^{-1}(\sin(5\pi + 1.3)) \]
Using the sine function's periodicity and properties:
\[ \sin(5\pi + x) = -\sin(x) \quad \Rightarrow \quad \sin(5\pi + 1.3) = -\sin(1.3) \]
Therefore,
\[ \sin^{-1}(\sin 17) = -1.3 \]

### Step 2: Evaluate \( \cos^{-1}(\cos 10) \)

The function \( \cos^{-1}(x) \) has a range of \( [0, \pi] \). Since \( 10 \) is outside this range, we will also express \( 10 \) in terms of \( \pi \):
\[ 10 \text{ radians} \approx 3.18 \pi \]
We can write:
\[ 10 = 3\pi + (10 - 3\pi) \]
Calculating \( 10 - 3\pi \):
\[ 3\pi \approx 9.42 \quad \Rightarrow \quad 10 - 9.42 = 0.58 \]
Thus, we have:
\[ \cos^{-1}(\cos 10) = \cos^{-1}(\cos(3\pi + 0.58)) \]
Using the cosine function's periodicity and properties:
\[ \cos(3\pi + x) = -\cos(x) \quad \Rightarrow \quad \cos(3\pi + 0.58) = -\cos(0.58) \]
Therefore,
\[ \cos^{-1}(\cos 10) = \pi - 0.58 \]

### Step 3: Combine the Results

Now we can combine the results from Step 1 and Step 2:
\[ \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) = -1.3 + (\pi - 0.58) \]
Calculating this gives:
\[ = -1.3 + \pi - 0.58 \]
\[ = \pi - 1.3 - 0.58 = \pi - 1.88 \]

### Final Result

Thus, the value of \( \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) \) is:
\[ \pi - 1.88 \]