যদি sin(alpha-π/6)+Sin (alpha+(5pi)/6)এর মান কত?

Explanation:

Another Explanation (5): দেওয়া আছে, \( \sin(\alpha - \frac{\pi}{6}) + \sin(\alpha + \frac{5\pi}{6}) \) এর মান নির্ণয় করতে হবে। আমরা জানি, \(\sin(A+B) = \sin A \cos B + \cos A \sin B\) এবং \(\sin(A-B) = \sin A \cos B - \cos A \sin B\). তাহলে, \(\sin(\alpha - \frac{\pi}{6}) = \sin \alpha \cos \frac{\pi}{6} - \cos \alpha \sin \frac{\pi}{6}\) \(\sin(\alpha + \frac{5\pi}{6}) = \sin \alpha \cos \frac{5\pi}{6} + \cos \alpha \sin \frac{5\pi}{6}\) আমরা জানি, \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\), \(\sin \frac{\pi}{6} = \frac{1}{2}\), \(\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), \(\sin \frac{5\pi}{6} = \frac{1}{2}\). সুতরাং, \(\sin(\alpha - \frac{\pi}{6}) = \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2}\) \(\sin(\alpha + \frac{5\pi}{6}) = \sin \alpha \cdot (-\frac{\sqrt{3}}{2}) + \cos \alpha \cdot \frac{1}{2}\) এখন, উভয় রাশি যোগ করে পাই: \(\sin(\alpha - \frac{\pi}{6}) + \sin(\alpha + \frac{5\pi}{6}) = (\sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2}) + (\sin \alpha \cdot (-\frac{\sqrt{3}}{2}) + \cos \alpha \cdot \frac{1}{2})\) \( = \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2} - \sin \alpha \cdot \frac{\sqrt{3}}{2} + \cos \alpha \cdot \frac{1}{2}\) \( = (\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}) \sin \alpha + (-\frac{1}{2} + \frac{1}{2}) \cos \alpha\) \( = 0 \cdot \sin \alpha + 0 \cdot \cos \alpha\) \( = 0\) অতএব, \(\sin(\alpha - \frac{\pi}{6}) + \sin(\alpha + \frac{5\pi}{6}) = 0\) সুতরাং, উত্তর 0। ✅🎉