যদি এর মান কত?

Explanation:

Another Explanation (5): ```html যদি \( \sin\left(\alpha - \frac{\pi}{6}\right) + \sin\left(\alpha + \frac{5\pi}{6}\right) \) হয়, তবে 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, \( \sin\left(\alpha - \frac{\pi}{6}\right) = \sin \alpha \cos \frac{\pi}{6} - \cos \alpha \sin \frac{\pi}{6} \) এবং \( \sin\left(\alpha + \frac{5\pi}{6}\right) = \sin \alpha \cos \frac{5\pi}{6} + \cos \alpha \sin \frac{5\pi}{6} \) এখন, \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), \( \sin \frac{\pi}{6} = \frac{1}{2} \) এবং \( \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \), \( \sin \frac{5\pi}{6} = \frac{1}{2} \) সুতরাং, \( \sin\left(\alpha - \frac{\pi}{6}\right) = \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2} \) \( \sin\left(\alpha + \frac{5\pi}{6}\right) = \sin \alpha \cdot \left(-\frac{\sqrt{3}}{2}\right) + \cos \alpha \cdot \frac{1}{2} \) এখন যোগ করি: \( \sin\left(\alpha - \frac{\pi}{6}\right) + \sin\left(\alpha + \frac{5\pi}{6}\right) = \left(\sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2}\right) + \left(\sin \alpha \cdot \left(-\frac{\sqrt{3}}{2}\right) + \cos \alpha \cdot \frac{1}{2}\right) \) \( = \sin \alpha \cdot \frac{\sqrt{3}}{2} - \cos \alpha \cdot \frac{1}{2} - \sin \alpha \cdot \frac{\sqrt{3}}{2} + \cos \alpha \cdot \frac{1}{2} \) \( = 0 \) অতএব, \( \sin\left(\alpha - \frac{\pi}{6}\right) + \sin\left(\alpha + \frac{5\pi}{6}\right) = 0 \) 🎉 ```