sin65° + cos65° =?

দেওয়া আছে: sin65° + cos65°

আমরা জানি, \(a\sin\theta + b\cos\theta = r\cos(\theta - \alpha)\), 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° = \(\sqrt{2}\)cos(65° - 45°)

= \(\sqrt{2}\)cos(20°)

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

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