যদি x=rsin(theta+45°)ও y=rsin(theta-45°) হয় তবে x²+y²=?

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।

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

সুতরাং,

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

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

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

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

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

অতএব,

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

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

সুতরাং,

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

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

অতএব,

\(x^2 + y^2 = r^2 [\sin^2(\theta + 45^\circ) + \sin^2(\theta - 45^\circ)] = r^2 \cdot 1 = r^2\). 💖

সুতরাং, \(x^2 + y^2 = r^2\). ✅