Question

ПОМОГИТЕ СРОЧНО С АЛГЕБРОЙ ДАЮ 20 БАЛЛОВ!!!

Напишите в виде произведения. Ещё один пример
$cos \frac{11\pi}{12} + cos \frac{3\pi}{4}$
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Answer 1

Ответ:

$5)sin\frac{\pi }{6} -sin\frac{\pi }{9} =2*sin(\frac{\pi }{6} }-\frac{\pi }{9} )*cos(\frac{\pi }{6} +\frac{\pi }{9})=2*sin\frac{\pi }{36} *cos\frac{5\pi }{36} .\\6)sin\alpha -sin(\alpha +\frac{\pi }{3} )=2*sin( -\frac{\pi }{6} )*cos(\alpha +\frac{\pi }{6} )=\\=2*(-\frac{1}{2} )*cos(\alpha +\frac{\pi }{6} )=-cos(\alpha +\frac{\pi }{6}). \\7)cos\frac{11\pi }{12} +cos\frac{3\pi }{4} =2*cos(\frac{5\pi }{6} )*cos(\frac{\pi }{12} )=-2*\frac{\sqrt{3} }{2} *cos\frac{\pi }{12} =-\sqrt{3}*cos\frac{\pi }{12} .$

$3)cos(\frac{\pi }{3 } -\alpha )+cos\alpha =2*cos(\frac{\frac{\pi }{3} -\alpha+\alpha }{2})* cos(\frac{\frac{\pi }{3} -\alpha-\alpha }{2})=2*cos\frac{\pi }{6}*cos(\frac{\pi }{6} -\alpha )=\\=2*\frac{\sqrt{3} }{2} *cos(\frac{\pi }{6}-\alpha )=\sqrt{3} *cos(\frac{\pi }{6}-\alpha ).\\4)sin(\frac{\pi }{6 }+\alpha )-sin(\frac{\pi }{6} -\alpha )=2*sin\alpha *cos\frac{\pi }{3}=2*sin\alpha *\frac{1}{2}=sin\alpha .$