设f(sinx+cosx)=sinxcosx,则f(cosπ/6)=

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设f(sinx+cosx)=sinxcosx,则f(cosπ/6)=
设f(sinx+cosx)=sinxcosx,则f(cosπ/6)=

设f(sinx+cosx)=sinxcosx,则f(cosπ/6)=
那好,
因为
sinx+cosx=√2*[(√2)/2*sinx+(√2)/2*cosx] (注:(√2)/2=1/(√2))
=√2*(cosπ/4*sinx+sinπ/4*cosx) (注:sinπ/4=cosπ/4=(√2)/2)
=√2*(x+π/4) (注:三角函数两角和公式:sina*cosb+sinb*cosa=sin(a+b))
而sinxcosx=2sin2x(二倍角公式:sin2x=2sinxcosx)
且[√2*sin(x+π/4)]^2=2sin(x+π/4)^2
=-cos(2x+π/2)+1
(注:把(x+π/4)看成一个角,则由(cosx)^2+(sinx)^2=1和(cosx)^2-(sinx)^2=cos2x——2倍角公式得2sin(x+π/4)^2=1-{[cos(x+π/4)]^2-[sin(x+π/4)]^2}=...)
=sin2x+1(注:有诱导公式cos(x+π/2)=-sinx)
所以f(x)=2x^2-2,(注:对比发现:2*[√2*sin(x+π/4)]^2-2=2(sin2x+1)-2=sin2x),
于是:
f(cosπ/6)=2*(√3/2)^2-2=-1/2