sin 22 1^@/2 =?

প্রশ্ন: \( \sin 22\frac{1}{2}^\circ = ? \)

উত্তর: \( \frac{\sqrt{2-\sqrt{2}}}{2} \)

ব্যাখ্যা:

আমরা জানি, \( \cos 2\theta = 1 - 2\sin^2 \theta \)

সুতরাং, \( \sin \theta = \sqrt{\frac{1-\cos 2\theta}{2}} \)

এখানে, \( \theta = 22\frac{1}{2}^\circ = 22.5^\circ \)

তাহলে, \( 2\theta = 45^\circ \)

অতএব, \( \sin 22.5^\circ = \sqrt{\frac{1-\cos 45^\circ}{2}} \)

আমরা জানি, \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

সুতরাং, \( \sin 22.5^\circ = \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}} \)

\( = \sqrt{\frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2}} \)

\( = \sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}} \)

\( = \sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}} \)

\( = \sqrt{\frac{2-\sqrt{2}}{4}} \)

\( = \frac{\sqrt{2-\sqrt{2}}}{2} \)

সুতরাং, \( \sin 22\frac{1}{2}^\circ = \frac{\sqrt{2-\sqrt{2}}}{2} \) 🥳