Question

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

• A. \(\sqrt{\dfrac{2+\sqrt2}{4}}\)
• B. \(\dfrac{2+\sqrt2}{4}\)
• C. \(\sqrt{\dfrac{2-\sqrt2}{4}}\)
• D. \(\dfrac{2-\sqrt2}{4}\)

Hint:

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 \[ \cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}}. \]

Answer 1

The Correct Option is C

Step 1: Use the half-angle identity.
We know that \[ \sin\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{2}} \] Here, \[ 22\frac{1}{2}^{\circ}=\frac{45^\circ}{2} \] So, \[ \sin22\frac{1}{2}^{\circ} = \sqrt{\frac{1-\cos45^\circ}{2}} \]

Step 2: Substitute the value of \(\cos45^\circ\).
Since \[ \cos45^\circ=\frac{\sqrt2}{2}, \] we get \[ \sin22\frac{1}{2}^{\circ} = \sqrt{ \frac{ 1-\frac{\sqrt2}{2} }{2} } \] Taking LCM inside the bracket, \[ = \sqrt{ \frac{ \frac{2-\sqrt2}{2} }{2} } \] Therefore, \[ = \sqrt{ \frac{2-\sqrt2}{4} } \]

Step 3: Final conclusion.
Hence, \[ \boxed{ \sqrt{\frac{2-\sqrt2}{4}} } \]