Question

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упростить sin(7п-a)cos(15п/2+b)-sin(19п/2-a)cos(6п-b)

Answer 1

Ответ:

$sin(7\pi -a)\cdot cos\Big(\dfrac{15\pi}{2}+\beta \Big)-sin\Big(\dfrac{19\pi}{2}-a\Big)\cdot cos(6\pi -\beta )=\\\\\\=sin(6\pi +\pi -a)\cdot cos\Big(\dfrac{12\pi +3\pi }{2}+\beta \Big)-sin\Big(\dfrac{16\pi +3\pi }{2}-a\Big)\cdot cos(-\beta )=\\\\\\=sin(\pi -a)\cdot cos\Big(6\pi +\dfrac{3\pi}{2}+\beta \Big)-sin\Big(8\pi +\dfrac{3\pi}{2}-a\Big)\cdot cos(-\beta )=\\\\\\=sina\cdot cos\Big(\dfrac{3\pi}{2}+\beta \Big)-sin\Big(\dfrac{3\pi}{2}-a\Big)\cdot cos\beta =$

$=sina\cdot sin\beta -(-cosa)\cdot cos\beta =sina\cdot sin\beta +cosa\cdot cos\beta =cos(a-\beta )$

Answer 2

Решение:

sin(7π - a) · cos(15π/2 + b) - sin(19π/2 - a) · cos(6π - b) =

= sin a · sin b + cos a · cos b =

= cos (a - b)