Question

Помогите пожалуйста
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Answer 1

Ответ:

$1)\ \ sin0,8\pi =sin\dfrac{4}{5}\pi =sin\Big(\pi -\dfrac{\pi}{5}\Big)=sin\dfrac{\pi}{5}\\\\cos2,4\pi =cos\dfrac{12\pi}{5}=cos\Big(2\pi +\dfrac{2\pi}{5}\Big)=cos\dfrac{2\pi}{5}=cos\Big(\dfrac{\pi}{2}-\dfrac{\pi}{10}\Big)=sin\dfrac{\pi}{10} \\\\tg3,7\pi =tg\Big(3\pi +\dfrac{7\pi}{10}\Big)=tg\dfrac{7\pi}{10}=tg\Big(\dfrac{\pi}{2}+\dfrac{\pi}{5}\Big)=-ctg\dfrac{\pi}{5}\\\\ctg5,4\pi =ctg\Big(5\pi +\dfrac{2\pi}{5}\Big)=ctg\dfrac{2\pi}{5}=ctg\Big(\dfrac{\pi}{2}-\dfrac{\pi}{10}\Big)=tg\dfrac{\pi}{10}$

$2)\ \ sin(-1,2\pi )=sin(-\pi-\dfrac{\pi}{5}\Big)=sin\dfrac{\pi}{5}\\\\cos3,6\pi =cos\Big(3\pi +\dfrac{3\pi}{5}\Big)=-sin\dfrac{3\pi}{5}=-sin\Big(\dfrac{\pi}{2}+\dfrac{\pi}{10}\Big)=-cos\dfrac{\pi}{10}\\\\tg1,9\pi =tg\Big(2\pi -\dfrac{\pi}{10}\Big)=-tg\dfrac{\pi}{10}\\\\ctg(-1,7\pi )=ctg\Big(-2\pi +\dfrac{3\pi}{10}\Big)=ctg\dfrac{3\pi}{10}=ctg\Big(\dfrac{\pi}{2}-\dfrac{\pi}{5}\Big)=tg\dfrac{\pi}{5}$