The value of `cos^(-1)(-1)+sin^(-1)(1)` is

Options

A. `-(3pi)/2`
B. `pi/2`
C. `pi`
D. `(3pi)/2`

Answer

To solve the problem of finding the value of \( \cos^{-1}(-1) + \sin^{-1}(1) \), we can break it down into steps:

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

The cosine inverse function, \( \cos^{-1}(x) \), gives the angle whose cosine is \( x \). The principal value of \( \cos^{-1}(x) \) lies in the interval \( [0, \pi] \).

- We know that \( \cos(\pi) = -1 \).
- Therefore, \( \cos^{-1}(-1) = \pi \).

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

The sine inverse function, \( \sin^{-1}(x) \), gives the angle whose sine is \( x \). The principal value of \( \sin^{-1}(x) \) lies in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

- We know that \( \sin\left(\frac{\pi}{2}\right) = 1 \).
- Therefore, \( \sin^{-1}(1) = \frac{\pi}{2} \).

### Step 3: Combine the results

Now we can add the results from Step 1 and Step 2:

\[ \cos^{-1}(-1) + \sin^{-1}(1) = \pi + \frac{\pi}{2} \]

### Step 4: Simplify the expression

To simplify \( \pi + \frac{\pi}{2} \):

- Convert \( \pi \) to a fraction with the same denominator:
\[ \pi = \frac{2\pi}{2} \]
- Now add:
\[ \frac{2\pi}{2} + \frac{\pi}{2} = \frac{2\pi + \pi}{2} = \frac{3\pi}{2} \]

### Final Answer

Thus, the value of \( \cos^{-1}(-1) + \sin^{-1}(1) \) is:

\[ \frac{3\pi}{2} \]

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