To evaluate the expression \(\sin^{-1}\left(\frac{5}{6}\right) + \sin^{-1}\left(\frac{\sqrt{11}}{6}\right)\), we can use the following identity for the sum of two inverse sine functions:

\[
\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)
\]

where \( -1 \leq x, y \leq 1 \).

### Step 1: Check if the values fall within the allowed range
For both \(\sin^{-1}(5/6)\) and \(\sin^{-1}(\sqrt{11}/6)\), the arguments must be between \(-1\) and \(1\). However, \(\frac{\sqrt{11}}{6} \approx 0.55\), so it is within the correct range.

### Step 2: Apply the identity
Let \(x = \frac{5}{6}\) and \(y = \frac{\sqrt{11}}{6}\).

Now compute the values inside the inverse sine:

\[
\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(\frac{5}{6} \cdot \sqrt{1 - \left(\frac{\sqrt{11}}{6}\right)^2} + \frac{\sqrt{11}}{6} \cdot \sqrt{1 - \left(\frac{5}{6}\right)^2}\right)
\]

### Step 3: Simplify the square roots
1. For \(1 - \left(\frac{\sqrt{11}}{6}\right)^2\):
\[
1 - \left(\frac{\sqrt{11}}{6}\right)^2 = 1 - \frac{11}{36} = \frac{36 - 11}{36} = \frac{25}{36}
\]
So \(\sqrt{1 - \left(\frac{\sqrt{11}}{6}\right)^2} = \frac{5}{6}\).

2. For \(1 - \left(\frac{5}{6}\right)^2\):
\[
1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36}
\]
So \(\sqrt{1 - \left(\frac{5}{6}\right)^2} = \frac{\sqrt{11}}{6}\).

### Step 4: Substitute the values
Now substitute back into the formula:

\[
\sin^{-1}\left(\frac{5}{6} \cdot \frac{5}{6} + \frac{\sqrt{11}}{6} \cdot \frac{\sqrt{11}}{6}\right) = \sin^{-1}\left(\frac{25}{36} + \frac{11}{36}\right)
\]

\[
\sin^{-1}\left(\frac{25 + 11}{36}\right) = \sin^{-1}\left(\frac{36}{36}\right) = \sin^{-1}(1)
\]

### Step 5: Evaluate \(\sin^{-1}(1)\)
\(\sin^{-1}(1)\) is the angle whose sine is 1. This angle is \(\frac{\pi}{2}\).

Thus, the final answer is:

\[
\sin^{-1}\left(\frac{5}{6}\right) + \sin^{-1}\left(\frac{\sqrt{11}}{6}\right) = \frac{\pi}{2}
\]

Would you like further details or explanations?

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Here are five related questions to expand on this concept:

1. What is the domain of the inverse sine function \(\sin^{-1}(x)\)?
2. How can the sum of two inverse trigonometric functions be simplified?
3. What is the significance of the identity \(\sin^{-1}(x) + \sin^{-1}(y)\)?
4. Can you use a similar approach to simplify \(\cos^{-1}(x) + \cos^{-1}(y)\)?
5. How do we interpret \(\sin^{-1}(x)\) geometrically?

**Tip:** When dealing with inverse trigonometric functions, always check if the arguments fall within the allowed domain!

evaluate sin^-1(5/6) + sin^-1(sqrt(11)/6)

Learn how to evaluate sin^-1(5/6) + sin^-1(sqrt(11)/6) using the inverse sine addition formula. This detailed solution walks through each step, simplifying the expression and arriving at the final answer of π/2. Suitable for students in Grades 10-12 studying trigonometry.

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