Question

设f(a)=2sin(-a)cos(π+a)-cso(π-a)/1+sin²(π+a)+cos(3π/2+a)-sin²(π/2+a)(1+2sina≠0)

求f(π/6)的值。我的算的是0，不知道对不对

Answer 1

f(a)=2sin(-a)cos(π+a)-cso(π-a)/1+sin²(π+a)+cos(3π/2+a)-sin²(π/2+a)

=(-2sina)(-cosa)-(-cosa)/(1+sin²a)+sina-cos²a
=2sinacosa+cosa/(1+sin²a)+sina-cos²a
=sin2a+cosa/(1+sin²a)+sina-cos²a
f(π/6)=√3/2+(√3/2)/[1+(1/2)²]+1/2-(√3/2)²
=√3/2+(√3/2)/(5/4)+1/2-3/4
=√3/2+4/5×√3/2-1/4
=9/5×√3/2-1/4
=9√3/10-1/4