sin27°+cos27° এর মান কত?

Explanation:

Another Explanation (5): ```html উদ্দিষ্ট মান: sin27°+cos27° 🧐 আমরা জানি, \(a\sin x + b\cos x = \sqrt{a^2+b^2} \cos(x-\alpha)\), যেখানে \(\cos \alpha = \frac{b}{\sqrt{a^2+b^2}}\) এবং \(\sin \alpha = \frac{a}{\sqrt{a^2+b^2}}\). এখানে, a = 1 এবং b = 1. সুতরাং, \(\sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2}\). এখন, \(\cos \alpha = \frac{1}{\sqrt{2}}\) এবং \(\sin \alpha = \frac{1}{\sqrt{2}}\). সুতরাং, \(\alpha = 45°\). অতএব, sin27°+cos27° = \(\sqrt{2} \cos(27°-45°) = \sqrt{2} \cos(-18°) = \sqrt{2} \cos(18°)\) 🥳 যেহেতু \(\cos(-x) = \cos(x)\). সুতরাং, sin27°+cos27° = \(\sqrt{2}cos18°\) 🥰 ```