To evaluate $\sin(\arccos(-1/3))$, we can follow these steps:

Step 1: Understanding the Problem

We are asked to find the sine of the angle whose cosine is $-\frac{1}{3}$.

Let: $\theta = \arccos\left(-\frac{1}{3}\right)$ This means: $\cos(\theta) = -\frac{1}{3}$ Now, we need to find $\sin(\theta)$.

Step 2: Use the Pythagorean Identity

The Pythagorean identity states that: $\sin^2(\theta) + \cos^2(\theta) = 1$ We already know $\cos(\theta) = -\frac{1}{3}$, so we can plug this value into the identity: $\sin^2(\theta) + \left(-\frac{1}{3}\right)^2 = 1$ $\sin^2(\theta) + \frac{1}{9} = 1$ $\sin^2(\theta) = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$

Step 3: Solve for $\sin(\theta)$

Now, take the square root of both sides: $\sin(\theta) = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}$

Step 4: Determine the Sign of $\sin(\theta)$

Since $\arccos(x)$ returns values in the range $[0, \pi]$, where sine is always non-negative, we take the positive value: $\sin(\theta) = \frac{2\sqrt{2}}{3}$

Final Answer:

$\sin(\arccos(-1/3)) = \frac{2\sqrt{2}}{3}$

Would you like further clarification on any of the steps, or do you have another question?

Additional Questions:

1. What is the value of $\cos(\arcsin(x))$?
2. How can we derive the Pythagorean identity?
3. What is the general form of $\sin(\arccos(x))$ for any $x$?
4. Can you explain the relationship between the range of the arccosine function and the sign of sine?
5. How can you calculate $\tan(\arcsin(x))$?

Tip: To remember the relationship between the inverse trigonometric functions and the Pythagorean identity, think of the right triangle formed by the sine and cosine values and use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to solve for the unknown.