sin65°+cos65°=?

প্রশ্ন: sin65°+cos65°=?

সমাধান:

আমরা জানি, \(a \cdot sinx + b \cdot cosx = R \cdot cos(x - \alpha)\), যেখানে \(R = \sqrt{a^2 + b^2}\) এবং \(tan\alpha = \frac{a}{b}\)

এখানে, a = 1 এবং b = 1

সুতরাং, \(R = \sqrt{1^2 + 1^2} = \sqrt{2}\)

এবং \(tan\alpha = \frac{1}{1} = 1\), সুতরাং \(\alpha = 45°\)

অতএব, sin65° + cos65° = \(\sqrt{2} \cdot cos(65° - 45°)\)

= \(\sqrt{2} \cdot cos(20°)\)

সুতরাং, sin65°+cos65° = \(\sqrt{2} cos20°\)

✅ উত্তর: \(\sqrt{2} cos20°\) 🎉