Question

Дано cosα=0,6; 3π/2<α<2π. Обчисліть cos(π/3-α)

Answer 1

$sin \alpha =- \sqrt{1-cos^2 \alpha }=- \sqrt{1- \frac{9}{25} } =- \sqrt{ \frac{16}{25} } = -\frac{4}{5} \\ cos( \frac{\pi}{3}- \alpha )=cos\frac{\pi}{3}cos \alpha +sin\frac{\pi}{3}sin \alpha \\ = \frac{1}{2} * \frac{3}{5} - \frac{ \sqrt{3} }{2} * \frac{4}{5} = \frac{3}{10} - \frac{4 \sqrt{3} }{10} = \frac{3-4 \sqrt{3} }{10}$