Question

Выражение cos(a+b)+cos(a-b)/sin(a+b)+sin(a-b)=?

Answer 1

Объяснение:

Воспользуемся формулами :

$cos (\alpha +\beta ) = cos\alpha cos\beta -sin\alpha sin\beta ;\\cos (\alpha -\beta ) = cos\alpha cos\beta +sin\alpha sin\beta ;\\sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta ;\\sin(\alpha -\beta )=sin\alpha cos\beta -cos\alpha sin\beta ;$

$\frac{cos(\alpha+\beta)+cos(\alpha-\beta ) }{sin(\alpha +\beta ) +sin(\alpha -\beta) } =\frac{cos\alpha cos\beta -sin\alpha sin\beta + cos\alpha cos\beta +sin\alpha sin\beta }{sin\alpha cos\beta +cos\alpha sin\beta +sin\alpha cos\beta -cos\alpha sin\beta } =\\\\\frac{2cos\alpha cos\beta }{2sin\alpha cos\beta } =\frac{cos\alpha }{sin\alpha } =ctg\alpha .$

Воспользуемся формулами :

$cos\alpha +cos\beta = 2cos\frac{\alpha +\beta }{2} *cos\frac{\alpha-\beta }{2} ;\\\\sin\alpha +sin\beta =2 sin\frac{\alpha+\beta }{2} * cos\frac{\alpha-\beta }{2} .$

$\frac{cos(\alpha+\beta)+cos(\alpha-\beta ) }{sin(\alpha +\beta ) +sin(\alpha -\beta) } =\frac{2 cos\frac{\alpha+\beta +\alpha -\beta }{2}*cos\frac{\alpha+\beta -\alpha +\beta }{2} }{2sin \frac{\alpha+\beta+\alpha-\beta }{2}*cos \frac{\alpha +\beta -\alpha +\beta }{2} } }{} =\frac{cos\alpha *cos\beta }{sin\alpha* cos\beta } = \frac{cos\alpha }{sin\alpha } =ctg\alpha .$