设f(sinx+cosx)=sinx×cosx则f(cosπ/6)=?
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设f(sinx+cosx)=sinx×cosx则f(cosπ/6)=?
设f(sinx+cosx)=sinx×cosx则f(cosπ/6)=?
设f(sinx+cosx)=sinx×cosx则f(cosπ/6)=?
2sinx×cosx=(sinx+cosx)*(sinx+cosx)-1,所以
f(sinx+cosx)=(sinx+cosx)*(sinx+cosx)/2-1/2
f(X)=x*x/2-1/2
f(cosπ/6)=-1/8
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=2sin...
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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)
收起
sinxcosx=[(sinx+cosx)^2-1]/2
f(sinx+cosx)=[(sinx+cosx)^2-1]/2
f(x)=(x^2-1)/2
f(cosπ/6)=f(√3/2)=(3/4-1)/2=-1/8
由题可得 sinx+cosx=cosπ/6
∴(sinx+cosx)^2=3/4
∴2sinx*cosx=(sinx+cosx)^2-(sinx^2+cosx^2)=-1/4
f(cosπ/6)=-1/8
设sinx+cosx=cosπ/6=根号3/2
平方得1+2sinx×cosx=3/4
即sinx×cosx=1/8
即f(cosπ/6)=f(sinx+cosx)=sinx×cosx=1/8