costheta=1/2 এবং sindelta=1/sqrt2 হলে, sin(theta+delta).sin(theta-delta)=?  

Explanation:

Another Explanation (5): দেওয়া আছে, \( \cos\theta = \frac{1}{2} \) এবং \( \sin\delta = \frac{1}{\sqrt{2}} \)। আমরা জানি, \( \sin(A+B) \sin(A-B) = \sin^2A - \sin^2B = \cos^2B - \cos^2A \) সুতরাং, \( \sin(\theta+\delta) \sin(\theta-\delta) = \sin^2\theta - \sin^2\delta \) অথবা \( \cos^2\delta - \cos^2\theta \) যেহেতু \( \cos\theta = \frac{1}{2} \), তাই \( \cos^2\theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) 🥳 আবার, \( \sin\delta = \frac{1}{\sqrt{2}} \) হলে, \( \sin^2\delta = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \)। 🤩 সুতরাং, \( \sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{4} = \frac{3}{4} \) তাহলে, \( \sin(\theta+\delta) \sin(\theta-\delta) = \sin^2\theta - \sin^2\delta = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \) 😎 অথবা, যেহেতু \( \sin\delta = \frac{1}{\sqrt{2}} \), তাই \( \cos^2\delta = 1 - \sin^2\delta = 1 - \frac{1}{2} = \frac{1}{2} \)। তাহলে, \( \sin(\theta+\delta) \sin(\theta-\delta) = \cos^2\delta - \cos^2\theta = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4} \) 🥰 অতএব, \( \sin(\theta+\delta) \sin(\theta-\delta) = \frac{1}{4} \)। 🥳✨