Question

use the half - angle formulas to find the exact value of the trigonometric function \\( \sin 112.5 ^ { \circ } \\).

\\( \sin 112.5 ^ { \circ } = \square \\)
(type an exact answer, using radicals as needed.)

Explanation:

Step1: Identify the half - angle formula

The half - angle formula for sine is \(\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}\). Since \(112.5^{\circ}=\frac{225^{\circ}}{2}\) and \(112.5^{\circ}\) is in the second quadrant (\(\sin\theta>0\) in the second quadrant), we use \(\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}\) with \(\alpha = 225^{\circ}\).

Step2: Find the value of \(\cos225^{\circ}\)

We know that \(\cos225^{\circ}=\cos(180^{\circ}+45^{\circ})=-\cos45^{\circ}=-\frac{\sqrt{2}}{2}\)

Step3: Substitute \(\cos\alpha\) into the half - angle formula

Substitute \(\alpha = 225^{\circ}\) (so \(\cos\alpha=-\frac{\sqrt{2}}{2}\)) into \(\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}\)

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Answer:

\(\frac{\sqrt{2+\sqrt{2}}}{2}\)