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° = \(\sqrt{2}cos18°\) 🥰
```
