Find the value of : `cos{sin("sin"^(-1)(pi)/6)}`

Answer

To find the value of \( \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) \), we will follow these steps:

### Step 1: Understand the Inverse Function
We start with the expression \( \sin^{-1}\left(\frac{\pi}{6}\right) \). The sine inverse function, \( \sin^{-1}(x) \), gives us an angle \( \theta \) such that \( \sin(\theta) = x \).

### Step 2: Check the Validity of \( \frac{\pi}{6} \)
However, we must check if \( \frac{\pi}{6} \) is within the valid range for the sine inverse function. The sine function has a range of \([-1, 1]\). Since \( \frac{\pi}{6} \approx 0.5236 \), which is less than 1, it is valid to use.

### Step 3: Evaluate \( \sin(\sin^{-1}(\frac{\pi}{6})) \)
Using the property of inverse functions:
\[ \sin(\sin^{-1}(x)) = x \quad \text{for } x \in [-1, 1] \]
Thus:
\[ \sin\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} \]

### Step 4: Substitute into the Cosine Function
Now, we need to find:
\[ \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) \]
Using the identity:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \]
we can substitute \( \theta = \sin^{-1}\left(\frac{\pi}{6}\right) \):
\[ \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) = \sqrt{1 - \left(\frac{\pi}{6}\right)^2} \]

### Step 5: Calculate \( \left(\frac{\pi}{6}\right)^2 \)
Calculating \( \left(\frac{\pi}{6}\right)^2 \):
\[ \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{36} \]

### Step 6: Substitute and Simplify
Now substituting back:
\[ \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) = \sqrt{1 - \frac{\pi^2}{36}} \]
This simplifies to:
\[ \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) = \sqrt{\frac{36 - \pi^2}{36}} = \frac{\sqrt{36 - \pi^2}}{6} \]

### Final Answer
Thus, the value of \( \cos\left(\sin^{-1}\left(\frac{\pi}{6}\right)\right) \) is:
\[ \frac{\sqrt{36 - \pi^2}}{6} \]

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