Question:easy

The value of \[ \sin22\frac{1}{2}^{\circ} \] is

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For angles like \(22.5^\circ,15^\circ,\) and \(75^\circ\), use half-angle identities: \[ \sin\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{2}} \] and \[ \cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}}. \]
Updated On: Jun 25, 2026
  • \(\sqrt{\dfrac{2+\sqrt2}{4}}\)
  • \(\dfrac{2+\sqrt2}{4}\)
  • \(\sqrt{\dfrac{2-\sqrt2}{4}}\)
  • \(\dfrac{2-\sqrt2}{4}\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Identify the half-angle formula for sine.
We know that $ \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} $ (taking the positive root since the angle $ 22.5^\circ $ is in the first quadrant where sine is positive).
Step 2: Recognise that 22.5 degrees is half of 45 degrees.
$ 22\frac{1}{2}^\circ = \frac{45^\circ}{2} $. So we use the half-angle formula with $ \theta = 45^\circ $.
Step 3: Substitute into the formula.
\[ \sin 22\tfrac{1}{2}^\circ = \sqrt{\frac{1 - \cos 45^\circ}{2}} \]
Step 4: Substitute the value of cos 45 degrees.
$ \cos 45^\circ = \frac{\sqrt{2}}{2} $. So: \[ \sin 22\tfrac{1}{2}^\circ = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} \]
Step 5: Confirm this is a valid simplification.
We combined the fractions step by step: $ 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} $, then dividing by 2 gives $ \frac{2 - \sqrt{2}}{4} $. The square root of that is the answer.
Step 6: State the final answer.
\[ \boxed{\sqrt{\frac{2 - \sqrt{2}}{4}}} \]
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