Another Explanation (5): প্রশ্ন: \( 2\cos^2 15^\circ \) এর মান কত?
উত্তর: \(\frac{2 + \sqrt{3}}{2}\)
সমাধান:
প্রথমে, \(\cos 15^\circ\) এর মান নির্ণয় করি।
আমরা জানি,
\[
\cos 15^\circ = \cos (45^\circ - 30^\circ)
\]
ব্যবহার করি,
\[
\cos (A - B) = \cos A \cos B + \sin A \sin B
\]
অর্থাৎ,
\[
\cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ
\]
প্রতিটি মান জানি:
\[
\cos 45^\circ = \frac{\sqrt{2}}{2}
\]
\[
\cos 30^\circ = \frac{\sqrt{3}}{2}
\]
\[
\sin 45^\circ = \frac{\sqrt{2}}{2}
\]
\[
\sin 30^\circ = \frac{1}{2}
\]
সুতরাং,
\[
\cos 15^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)
\]
\[
= \frac{\sqrt{2} \sqrt{3}}{4} + \frac{\sqrt{2}}{4}
\]
\[
= \frac{\sqrt{6} + \sqrt{2}}{4}
\]
এখন,
\[
\cos^2 15^\circ = \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right)^2
\]
\[
= \frac{(\sqrt{6} + \sqrt{2})^2}{16}
\]
বর্গের নিয়ম অনুসারে,
\[
(\sqrt{6} + \sqrt{2})^2 = (\sqrt{6})^2 + 2 \cdots \sqrt{6} \cdot \sqrt{2} + (\sqrt{2})^2 = 6 + 2 \sqrt{12} + 2
\]
\[
= 6 + 2 \times 2 \sqrt{3} + 2 = 6 + 4 \sqrt{3} + 2 = 8 + 4 \sqrt{3}
\]
অতএব,
\[
\cos^2 15^\circ = \frac{8 + 4 \sqrt{3}}{16} = \frac{8}{16} + \frac{4 \sqrt{3}}{16} = \frac{1}{2} + \frac{\sqrt{3}}{4}
\]
এখন,
\[
2 \cos^2 15^\circ = 2 \times \left( \frac{1}{2} + \frac{\sqrt{3}}{4} \right) = 2 \times \frac{1}{2} + 2 \times \frac{\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2}
\]
উপসংহার:
\[
2 \cos^2 15^\circ = 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}
\]
অতএব, উত্তর হলো:
\[
\boxed{\frac{2 + \sqrt{3}}{2}}
\]
