求解这道高中数学题
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a^2=(sinx)^2+(cosx)^2=1
ab=sinxcosx+√ 3(cosx)^2
f(x)=a*(a+b)=a^2+ab
=1+sinxcosx+√ 3(cosx)^2
=1+1/2*sin2x+√ 3/2(1+cos2x)
=(1/2*sin2x+√ 3/2cos2x)+(2+√ 3)/2
=sin(2x+π/3)+(2+√ 3)/2
T=2π/2=π
2)acosB+bcosA=2ccosC
sinAcosB+sinBcosA=2sinCcosC
sin(A+B)=2sinCcosC
sinC=2sinCcosC
cosC=1/2
C=π/3 ,B=2π/3-A
f(B/2)=sin(B+π/3)+(2+√ 3)/2, 0<B<2π/3
B+π/3=π/2
B=π/6
A=π/2
b=√ 3*tanB=1
三角形ABC是直角三角形 ,c=√ 3 ,b=1
S=1/2*bc=1/2*√ 3*1=√ 3/2
ab=sinxcosx+√ 3(cosx)^2
f(x)=a*(a+b)=a^2+ab
=1+sinxcosx+√ 3(cosx)^2
=1+1/2*sin2x+√ 3/2(1+cos2x)
=(1/2*sin2x+√ 3/2cos2x)+(2+√ 3)/2
=sin(2x+π/3)+(2+√ 3)/2
T=2π/2=π
2)acosB+bcosA=2ccosC
sinAcosB+sinBcosA=2sinCcosC
sin(A+B)=2sinCcosC
sinC=2sinCcosC
cosC=1/2
C=π/3 ,B=2π/3-A
f(B/2)=sin(B+π/3)+(2+√ 3)/2, 0<B<2π/3
B+π/3=π/2
B=π/6
A=π/2
b=√ 3*tanB=1
三角形ABC是直角三角形 ,c=√ 3 ,b=1
S=1/2*bc=1/2*√ 3*1=√ 3/2
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