高一 数学 求f(a)的值 请详细解答,谢谢! (18 22:20:33)
4个回答

1、a是第三象限角,f(a)=sin(a-派/2)cos(3/2派+a)tan(派-a)

tan(-a-派)sin(-派-a)

(1)化简f(a)(2)若cos(a-3/2派)=1/5,求f(a)的值

f(a)=sin(a-π/2)cos(3π/2+a)tan(π-a)

tan(-a-π)sin(-π-a)

=[sin(π/2-a)][sin a][-tan a]

[-tan a][-sin(π+a)]

=[cos a][sin a][tan a][tan a]

[sin a]

=[tan a][sin a]^3

cos(a-3π/2)=1/5

cos(3π/2-a)=1/5

-sin a=1/5

f(a)=[tan a][sin a]^3

=[sin a]^4/cosa

=-[1/625]/[(2根6)/5

=-(根6)/1500