Question

三角化简cos(b-a)sinb-sin(a-b)cosb

Answer 1

cos(b-a)sinb-sin(a-b)cosb
=sinbcos(b-a)-cosbsin(a-b)
=sinbcos(a-b)-cosbsin(a-b)(cosx是偶函数)
=sin(b-(b-a))(两角和与差的正弦公式)
=sina

Answer 2

这样

追答

Answer 3

解：e^ia=cosa+isina，e^ia*e^ib=e^i(a+b)，(cosa+isina)(cosb+isinb)=cos(a+b)+isin(a+b)，(cosacosb-sinasinb)+i(cosasinb+sinacosb)=cos(a+b)+isin(a+b)，cos(a+b)=cosacosb-sinasinb，sin(a+b)=(cosasinb+sinacosb)。

Answer 4

=sⅰnbcos(b-a)+cosbsⅰn(b-a)
=sⅰn(2b-a)

Answer 5

cos(b-a)sinb-sin(a-b)cosb
=cos(b-a)sinb+sin(b-a)cosb
=sin(b-a+b)
=sin(2b-a)

追答

或
cos(b-a)sinb-sin(a-b)cosb
=cos(a-b)sinb-sin(a-b)cosb
=sin[b-(a-b)]
=sin(2b-a)

Answer 6

原式
=sinbCos(a-b)-cosbsin(a-b)
=sin[b-(a-b)]
=sin(2b-a)