1. The problem is to prove or verify the trigonometric identity: $$\sin(\pi - x) = \sin x$$. 2. The formula used here is the sine subtraction identity and the properties of sine function on the unit circle. 3. Recall that sine of an angle in the second quadrant (which is $\pi - x$) is equal to the sine of its reference angle $x$. 4. Using the sine subtraction formula: $$\sin(\pi - x) = \sin \pi \cos x - \cos \pi \sin x$$ 5. Substitute the known values: $$\sin \pi = 0, \quad \cos \pi = -1$$ 6. So, $$\sin(\pi - x) = 0 \cdot \cos x - (-1) \cdot \sin x = \sin x$$ 7. Therefore, the identity $$\sin(\pi - x) = \sin x$$ is true. This shows that sine of $\pi - x$ equals sine of $x$ because they have the same vertical coordinate on the unit circle.