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"
