\(\sin^2 10^\circ + \sin^2 20^\circ + ... + \sin^2 80^\circ + \sin^2 90^\circ\) এর মান-

Another Explanation (5):

সমাধান:

প্রশ্নটি হলো: \[ \sin^2 10^\circ + \sin^2 20^\circ + \sin^2 30^\circ + \sin^2 40^\circ + \sin^2 50^\circ + \sin^2 60^\circ + \sin^2 70^\circ + \sin^2 80^\circ + \sin^2 90^\circ \] আমরা জানি, \[ \sin^2 \theta + \cos^2 \theta = 1 \] এবং \(\sin(90^\circ - \theta) = \cos \theta\), ফলে \[ \sin^2 (90^\circ - \theta) = \cos^2 \theta \] এখন, দেখা যাক: \[ \sin^2 10^\circ + \sin^2 80^\circ \] এখানে, \(\sin 80^\circ = \cos 10^\circ\), তাই \[ \sin^2 10^\circ + \cos^2 10^\circ = 1 \] অর্থাৎ, \[ \sin^2 10^\circ + \sin^2 80^\circ = 1 \] একইভাবে, \[ \sin^2 20^\circ + \sin^2 70^\circ = 1 \] কারণ, \(\sin 70^\circ = \cos 20^\circ\), \[ \sin^2 30^\circ + \sin^2 60^\circ \] এগুলো নির্ণয় করি: \[ \sin 30^\circ = \frac{1}{2}, \quad \sin^2 30^\circ = \frac{1}{4} \] \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin^2 60^\circ = \frac{3}{4} \] অতএব, \[ \sin^2 30^\circ + \sin^2 60^\circ = \frac{1}{4} + \frac{3}{4} = 1 \] পরবর্তী, \[ \sin^2 40^\circ + \sin^2 50^\circ \] এগুলোও একই পদ্ধতিতে: \[ \sin 50^\circ = \cos 40^\circ \] অতএব, \[ \sin^2 40^\circ + \cos^2 40^\circ = 1 \] অবশেষে, \(\sin^2 90^\circ\): \[ \sin 90^\circ = 1 \Rightarrow \sin^2 90^\circ = 1 \] সবগুলো যোগ করলে: \[ (\sin^2 10^\circ + \sin^2 80^\circ) + (\sin^2 20^\circ + \sin^2 70^\circ) + (\sin^2 30^\circ + \sin^2 60^\circ) + (\sin^2 40^\circ + \sin^2 50^\circ) + \sin^2 90^\circ \] = \(1 + 1 + 1 + 1 + 1\) = **5**

অতএব, উত্তরের মান হলো:

"5"