$sin\{\frac{\pi}{3}-sin^{-1}(-\frac{1}{2})\}=$

Q: $sin\{\frac{\pi}{3}-sin^{-1}(-\frac{1}{2})\}=$

To solve the expression $ \sin\left\{\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right\} $, we need to evaluate each component step-by-step. 1. **Evaluate $ \sin^{-1}\left(-\frac{1}{2}\right) $:** The function $ \sin^{-1}(x) $ gives the angle whose sine is $ x $. For $ \sin^{-1}\left(-\frac{1}{2}\right) $, we need to find an angle $ \theta $ such that $ \sin(\theta) = -\frac{1}{2} $. The angle $ \theta $ that satisfies $ \sin(\theta) = -\frac{1}{2} $ in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$ is $ -\frac{\pi}{6} $. 2. **Substitute back into the expression:** Now substitute $ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} $ into the original expression: \[ \sin\left\{\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right\} = \sin\left\{\frac{\pi}{3} + \frac{\pi}{6}\right\} \] 3. **Simplify the angle:** Combine the angles: \[ \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \] 4. **Evaluate the sine of the resulting angle:** $\sin\left(\frac{\pi}{2}\right) = 1$. Thus, the expression evaluates to 1, which corresponds to Option 4. **Explanation for Incorrect Options:** - **Option 1: $\frac{1}{2}$** This would be the sine of $\frac{\pi}{6}$, not $\frac{\pi}{2}$. The calculation shows that the angle simplifies to $\frac{\pi}{2}$, not $\frac{\pi}{6}$. - **Option 2: $\frac{1}{3}$** This value does not correspond to any standard sine value for angles derived from $\frac{\pi}{3}$ or $\frac{\pi}{6}$. - **Option 3: $\frac{1}{4}$** Similarly, this value is not a standard sine value for any angle involved in this calculation. Therefore, the correct answer is indeed Option 4, as the calculation confirms that $\sin\left(\frac{\pi}{2}\right) = 1$.

Question

Accepted Answer

To solve the expression $ \sin\left\{\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right\} $, we need to evaluate each component step-by-step.

1. **Evaluate $ \sin^{-1}\left(-\frac{1}{2}\right) $:**

The function $ \sin^{-1}(x) $ gives the angle whose sine is $ x $. For $ \sin^{-1}\left(-\frac{1}{2}\right) $, we need to find an angle $ \theta $ such that $ \sin(\theta) = -\frac{1}{2} $.

The angle $ \theta $ that satisfies $ \sin(\theta) = -\frac{1}{2} $ in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$ is $ -\frac{\pi}{6} $.

2. **Substitute back into the expression:**

Now substitute $ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} $ into the original expression:

$$
   \sin\left\{\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right\} = \sin\left\{\frac{\pi}{3} + \frac{\pi}{6}\right\}
   $$

3. **Simplify the angle:**

Combine the angles:

$$
   \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}
   $$

4. **Evaluate the sine of the resulting angle:**

$\sin\left(\frac{\pi}{2}\right) = 1$.

Thus, the expression evaluates to 1, which corresponds to Option 4.

**Explanation for Incorrect Options:**

- **Option 1: $\frac{1}{2}$**

This would be the sine of $\frac{\pi}{6}$, not $\frac{\pi}{2}$. The calculation shows that the angle simplifies to $\frac{\pi}{2}$, not $\frac{\pi}{6}$.

- **Option 2: $\frac{1}{3}$**

This value does not correspond to any standard sine value for angles derived from $\frac{\pi}{3}$ or $\frac{\pi}{6}$.

- **Option 3: $\frac{1}{4}$**

Similarly, this value is not a standard sine value for any angle involved in this calculation.

Therefore, the correct answer is indeed Option 4, as the calculation confirms that $\sin\left(\frac{\pi}{2}\right) = 1$.