高一数学,这两道题怎么做,谢谢了
1个回答
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1)f(x)=sin(π-x)sin(π/2-x)+(cosx)^2
=sinxcosx+(cosx)^2
=1/2sin2x+1/2(1+cos2x)
=√2/2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
T=2π/2=π
2)f(x)=3(sinx)^2+2√3sinxcosx+5(cosx)^2
=3[(sinx)^2+(cosx)^2]+2√3sinxcosx+2(cosx)^2
=3+2√3sinxcosx+2(cosx)^2
=3+√3sin2x+(1+cos2x)
=√3sin2x+cos2x+4
=2(√3/2sin2x+1/2cos2x)+4
=2sin(2x+π/6)+4
T=2π/2=π ,最大值=6
3)a向量=(√3sinwx,coswx) ,b向量=(coswx,coswx) ,
f(x)=a*b=√3sinwx*coswx+(coswx)^2
=√3/2sin2wx+1/2(1+cos2wx)
=√3/2sin2wx+1/2cos2wx+1/2
=sin(2wx+π/6)+1/2
T=2π/2w=π
w=1 ,,最大值=3/2
=sinxcosx+(cosx)^2
=1/2sin2x+1/2(1+cos2x)
=√2/2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
T=2π/2=π
2)f(x)=3(sinx)^2+2√3sinxcosx+5(cosx)^2
=3[(sinx)^2+(cosx)^2]+2√3sinxcosx+2(cosx)^2
=3+2√3sinxcosx+2(cosx)^2
=3+√3sin2x+(1+cos2x)
=√3sin2x+cos2x+4
=2(√3/2sin2x+1/2cos2x)+4
=2sin(2x+π/6)+4
T=2π/2=π ,最大值=6
3)a向量=(√3sinwx,coswx) ,b向量=(coswx,coswx) ,
f(x)=a*b=√3sinwx*coswx+(coswx)^2
=√3/2sin2wx+1/2(1+cos2wx)
=√3/2sin2wx+1/2cos2wx+1/2
=sin(2wx+π/6)+1/2
T=2π/2w=π
w=1 ,,最大值=3/2
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