Question

Докажите тождество:
sin^2(45°+a)-sin^2(30°-a) -sin15°*cos(15°+2a) =sin2a

Answer 1

$displaystyle sin^2(45а+ alpha )-sin^2(30а- alpha )-sin15аcos(15а+2 alpha )=\ \ = frac{1-cos(90а+2 alpha )}{2} - frac{1-cos(60а-2 alpha )}{2} - frac{sin(30а+2 alpha )-sin2 alpha }{2} =\ \ = frac{1+sin2 alpha -1+cos(90а-(30а+2 alpha ))-sin(30а+2 alpha )+sin2 alpha }{2} =\ \ = frac{2sin2 alpha +sin(30а+2 alpha )-sin(30а+2 alpha )}{2}= frac{2sin2 alpha }{2} =sin2 alpha$