Question

5 sin 11п/12 cos 11п/12

Answer 1

$5\cdot sin\dfrac{11\pi}{12} \cdot cos\dfrac{11\pi}{12} =2,5\cdot \Big(2\cdot sin\dfrac{11\pi}{12} \cdot cos\dfrac{11\pi}{12}\Big) =\\\\=2,5\cdot sin\Big(2\cdot \dfrac{11\pi}{12}\Big)=2,5\cdot sin\dfrac{11\pi}{6}=\\\\=2,5\cdot sin\Big(\dfrac{12\pi-\pi}{6}\Big)=2,5\cdot sin\Big(2\pi-\dfrac{\pi}{6}\Big)=\\\\=2,5\cdot sin \Big(-\dfrac{\pi}{6}\Big)=-2,5\cdot sin\dfrac{\pi}{6}=-2,5\cdot \dfrac{1}{2}=\boxed{\boldsymbol{-1,25}}$

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Использованы формулы

$2\cdot sin~\alpha \cdot cos~\alpha = sin (2\alpha)\\\\sin(2\pi-\alpha)=sin(-\alpha)=-sin~\alpha$