f(x)=sinx^2+根号3*sinx*cosx+2cos^2 x属于R 该怎样化简?

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f(x)=sinx^2+根号3*sinx*cosx+2cos^2 x属于R 该怎样化简?
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f(x)=sinx^2+根号3*sinx*cosx+2cos^2 x属于R 该怎样化简?
f(x)=sinx^2+根号3*sinx*cosx+2cos^2 x属于R 该怎样化简?

f(x)=sinx^2+根号3*sinx*cosx+2cos^2 x属于R 该怎样化简?
答:
利用公式:
(sinx)^2+(cosx)^2=1
cos2x=2(cosx)^2-1=1-2(sinx)^2
sin2x=2sinxcosx
sin(a+b)=sinacosb+cosasinb
f(x)=(sinx)^2+√3sinxcosx+2(cosx)^2
=1+(cosx)^2+(√3/2)*2sinxcosx
=(1/2)*(cos2x+1)+(√3/2)sin2x+1
=(√3/2)sin2x+(1/2)cos2x+3/2
=sin(2x+π/6)+3/2

前面应该错了吧,sin^2x
f(x)=(sin²x+cos²x)+√3sinxcosx+cos²x
=1+√3/2*sin2x+(1+cos2x)/2
=√3/2sin2x+1/2*cos2x+3/2
=sin2xcosπ/6+cos2xsinπ/6+3/2
=sin(2x+π/6)+3/2

f(x)=sin²x+√3sinxcosx+2cos²x=0.5√3sin2x+0.5(1+cos2x)+1=0.5[√3sin2x+cos2x]+1.5=0.5sin(2x+π/6)+1.5