Find the exact value of s in the given interval that has the given circular function value.

[π/2, π] ; sin s = 1/2

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Identify the given interval for the variable \(s\), which is \(\left[\frac{\pi}{2}, \pi\right]\), and the equation \(\sin s = \frac{1}{2}\).

Recall the unit circle values where \(\sin s = \frac{1}{2}\). The sine function equals \(\frac{1}{2}\) at angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) within one full rotation \([0, 2\pi]\).

Determine which of these angles lie within the given interval \(\left[\frac{\pi}{2}, \pi\right]\). Since \(\frac{\pi}{6}\) is less than \(\frac{\pi}{2}\), it is excluded, but \(\frac{5\pi}{6}\) lies within the interval.

Conclude that the exact value of \(s\) in the interval \(\left[\frac{\pi}{2}, \pi\right]\) satisfying \(\sin s = \frac{1}{2}\) is \(s = \frac{5\pi}{6}\).

Verify the solution by substituting \(s = \frac{5\pi}{6}\) back into the sine function to confirm \(\sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}\).

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Unit Circle and Reference Angles

The unit circle is a circle with radius 1 centered at the origin, used to define trigonometric functions. Reference angles help find sine values by relating any angle to an acute angle in the first quadrant, simplifying the determination of exact trigonometric values.

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Sine Function Values and Their Symmetry

The sine function gives the y-coordinate on the unit circle and is positive in the first and second quadrants. Knowing sine’s symmetry helps identify all angles within a given interval that share the same sine value, such as sin s = 1/2.

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Interval Restriction and Solution Selection

When solving trigonometric equations, restricting the solution to a specific interval ensures the answer fits the problem’s domain. For s in [π/2, π], only angles within this range are considered, which helps select the correct exact value of s.

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