Question

ТРИГОНОМЕТРИЯ!!! ПОМОГИТЕ!!! ДАЮ 73 БАЛЛА!!!

Answer 1

405
$\frac{sin( \alpha + \beta )+sin( \alpha - \beta )}{sin( \alpha + \beta )-sin( \alpha - \beta )} = \frac{sin \alpha cos \beta +cos \alpha sin \beta +sin \alpha cos \beta -cos \alpha sin \beta }{sin \alpha cos \beta +cos \alpha sin \beta -sin \alpha cos \beta +cos \alpha sin \beta } = \\ \\ =\frac{2sin \alpha cos \beta }{2cos \alpha sin \beta } = tg \alpha ctg \beta = \frac{tg \alpha }{tg \beta }$
406
$\frac{sin( 45^o+\alpha)-cos( 45^o+\alpha)}{sin( 45^o+\alpha)+cos( 45^o+\alpha)} = \frac{sin45^ocos\alpha+cos45^osin \alpha -cos 45^ocos\alpha+sin45^osin \alpha }{sin45^ocos\alpha+cos45^osin \alpha +cos 45^ocos\alpha-sin45^osin \alpha} = \\ \\ = \frac{ \frac{ \sqrt{2} }{2}(cos \alpha +sin \alpha -cos \alpha +sin \alpha)}{\frac{ \sqrt{2} }{2}(cos \alpha +sin \alpha +cos \alpha -sin \alpha )}= \frac{2sin \alpha }{2cos \alpha }=tg \alpha$
407
$\frac{cos65^ocos40^o+sin65^osin40^o}{sin37^ocos12^o-cos37^osin12^o}= \frac{cos(65^o-40^o)}{sin(37^o-12^o)}= \frac{cos25^o}{sin25^o}=ctg25^o$
408
cos3αcosα+sin3αsinα=cos(3α-α)=cos2α
409
sin5αcos2α-sin2αcos5α=sin(5α-2α)=sin3α