যদি  x=p sin(theta+45^@) এবং  y=p sin(theta-45^@) হয়  তবে  x^2+y^2=? 

Explanation:

Another Explanation (5): দেওয়া আছে, \(x = p \sin(\theta + 45^\circ)\) এবং \(y = p \sin(\theta - 45^\circ)\)। আমাদের \(x^2 + y^2\) এর মান নির্ণয় করতে হবে। \(x^2 = p^2 \sin^2(\theta + 45^\circ)\) \(y^2 = p^2 \sin^2(\theta - 45^\circ)\) সুতরাং, \(x^2 + y^2 = p^2 \sin^2(\theta + 45^\circ) + p^2 \sin^2(\theta - 45^\circ)\) \(= p^2 [\sin^2(\theta + 45^\circ) + \sin^2(\theta - 45^\circ)]\) \(= p^2 [\frac{1 - \cos(2(\theta + 45^\circ))}{2} + \frac{1 - \cos(2(\theta - 45^\circ))}{2}]\) \(= p^2 [\frac{1 - \cos(2\theta + 90^\circ)}{2} + \frac{1 - \cos(2\theta - 90^\circ)}{2}]\) \(= p^2 [\frac{2 - (\cos(2\theta + 90^\circ) + \cos(2\theta - 90^\circ))}{2}]\) আমরা জানি, \(\cos(A+B) + \cos(A-B) = 2\cos A \cos B\) অতএব, \(\cos(2\theta + 90^\circ) + \cos(2\theta - 90^\circ) = 2\cos(2\theta) \cos(90^\circ) = 2\cos(2\theta) \cdot 0 = 0\) তাহলে, \(x^2 + y^2 = p^2 [\frac{2 - 0}{2}]\) \(= p^2 [\frac{2}{2}]\) \(= p^2 \cdot 1\) \(= p^2\) 🎉 সুতরাং, \(x^2 + y^2 = p^2\) 🥳