[tan(pai-a)cos(2pai-a)sin(-a+3pai/2)]/[cos(-a-pai)sin(-pai-a)]=_数学解答

[tan(pai-a)cos(2pai-a)sin(-a+3pai/2)]/[cos(-a-pai)sin(-pai-a)]=

人气：430 ℃　时间：2020-03-28 21:55:04

解答

tan周期是π
cos周期是2π
所以原式=tan(-a)cos(-a)sin(-a+3π/2-2π)/[cos(-a+π)sin(2π-π-a)]
=-tanacosasin(-a-π/2)/[-cosasin(π-a)]
=tanacosasin(a+π/2)/(-cosasina)
=(sina/cosa)cosacosa/(-cosasina)
=cosasina/(-cosasina)
=-1