Question

Известно ,что sin(п/6+t)+sin)п/6-t=p.Найдите sin(п/6+t)sin(п/6-t)СРОЧНО!

Answer 1

$\sin\left(\dfrac{\pi}{6}+t\right)+\sin\left(\dfrac{\pi}{6}-t\right)=2\sin\dfrac{\dfrac{\pi}{6}+t+\dfrac{\pi}{6}-t}{2}\cos\dfrac{\dfrac{\pi}{6}+t-\dfrac{\pi}{6}+t}{2}=\\ \\ =2\sin\dfrac{\pi}{6}\cos t=2\cdot\dfrac{1}{2}\cos t=\cos t;~~~~~\boxed{\cos t=p}$

$\sin\left(\dfrac{\pi}{6}+t\right)\sin\left(\dfrac{\pi}{6}-t\right)=\dfrac{\cos\left(\dfrac{\pi}{6}+t-\dfrac{\pi}{6}+t\right)-\cos\left(\dfrac{\pi}{6}+t+\dfrac{\pi}{6}-t\right)}{2}=\\ \\ \\ =\dfrac{\cos2t-\cos\dfrac{\pi}{3}}{2}=\dfrac{\cos2t-\dfrac{1}{2}}{2}=\dfrac{2\cos2t-1}{4}=\dfrac{2(2\cos^2t-1)-1}{4}=\\ \\ \\ =\dfrac{4\cos^2t-2-1}{4}=\dfrac{4\cos^2t-3}{4}=\boxed{\dfrac{4p^2-3}{4}}$