যদি x=rsin(θ+45°) এবং y=rsin(θ–45°) হয়, তবে x²+y²= কত?

দেওয়া আছে, \( x = r \sin(\theta + 45^\circ) \) এবং \( y = r \sin(\theta - 45^\circ) \)।

আমাদের \( x^2 + y^2 \) এর মান নির্ণয় করতে হবে।

তাহলে, \( x^2 = r^2 \sin^2(\theta + 45^\circ) \) এবং \( y^2 = r^2 \sin^2(\theta - 45^\circ) \)।

সুতরাং, \( x^2 + y^2 = r^2 \sin^2(\theta + 45^\circ) + r^2 \sin^2(\theta - 45^\circ) \)

\(= r^2 [\sin^2(\theta + 45^\circ) + \sin^2(\theta - 45^\circ)] \)

আমরা জানি, \( \sin(A+B) = \sin A \cos B + \cos A \sin B \) এবং \( \sin(A-B) = \sin A \cos B - \cos A \sin B \)।

সুতরাং,

\( \sin(\theta + 45^\circ) = \sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ = \frac{1}{\sqrt{2}} (\sin \theta + \cos \theta) \)

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

তাহলে,

\( \sin^2(\theta + 45^\circ) = \frac{1}{2} (\sin \theta + \cos \theta)^2 = \frac{1}{2} (\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta) \)

\( \sin^2(\theta - 45^\circ) = \frac{1}{2} (\sin \theta - \cos \theta)^2 = \frac{1}{2} (\sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta) \)

যোগ করে পাই,

\( \sin^2(\theta + 45^\circ) + \sin^2(\theta - 45^\circ) = \frac{1}{2} (\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta) + \frac{1}{2} (\sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta) \)

\(= \frac{1}{2} (2 \sin^2 \theta + 2 \cos^2 \theta) = \sin^2 \theta + \cos^2 \theta = 1 \)

অতএব, \( x^2 + y^2 = r^2 \cdot 1 = r^2 \)। 🎉