We are asked to find the value of $\sin \left( 2 \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) \right)$.

Step-by-step solution:

1. Let $\theta = \cos^{-1} \left( \frac{\sqrt{2}}{2} \right)$.
   This implies that $\cos(\theta) = \frac{\sqrt{2}}{2}$, and we are asked to find $\sin(2\theta)$.

2. Use the double-angle identity for sine: $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ We already know that $\cos(\theta) = \frac{\sqrt{2}}{2}$, so now we need to find $\sin(\theta)$.

3. Find $\sin(\theta)$ using the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ Substituting $\cos(\theta) = \frac{\sqrt{2}}{2}$: $\sin^2(\theta) + \left( \frac{\sqrt{2}}{2} \right)^2 = 1$ $\sin^2(\theta) + \frac{2}{4} = 1$ $\sin^2(\theta) + \frac{1}{2} = 1$ $\sin^2(\theta) = 1 - \frac{1}{2} = \frac{1}{2}$ $\sin(\theta) = \frac{\sqrt{2}}{2}$ (Note that $\sin(\theta)$ is positive because $\theta$ is in the range $[0, \pi]$, given that $\cos^{-1}$ produces values within this range.)

4. Now calculate $\sin(2\theta)$: $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ Substituting the values of $\sin(\theta)$ and $\cos(\theta)$: $\sin(2\theta) = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}$ $\sin(2\theta) = 2 \times \frac{2}{4} = 1$

Final Answer:

The value of $\sin \left( 2 \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) \right)$ is 1.

Would you like any additional details or clarifications?

Related Questions:

1. How do we derive the double angle identities for trigonometric functions?
2. What is the relationship between inverse trigonometric functions and the unit circle?
3. How does the range of inverse cosine (cos⁻¹) affect the value of $\sin(\theta)$?
4. What is the significance of the Pythagorean identity in trigonometry?
5. How do you find $\cos^{-1}$ of values greater than 1 or less than -1?

Tip: Always ensure to check the domain and range of inverse functions to avoid errors when calculating trigonometric values.