Question

已知函数f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)

f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3sin(x+π/2)+sinx
这一步是怎么来的

Answer 1

f(x)=2√3(sinx/2cosπ/4+cosx/2sinπ/4)(cosx/2cosπ/4-sinx/2sinπ/4) - (sinxcosπ-cosxsinπ)
<sinπ=0,cosπ=1>
=2√3(√2/2sinx/2+√2/2cosx/2)(√2/2cosx/2-√2/2sinx/2)+sinπ
=2√3[1/2(cosx/2的平方－sinx/2的平方)+sinx
=√3[ (1+cosx)/2 - (1-cocx)/2 ]+sinx
=√3cosx+sinx
你可以写下来看

Answer 2

x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
f(x)=根号3sin[2(x/2+π/4)]+sinx
f(x)=根号3sin(x+π/2)+sinx

追问

不明白

追答

sin(2x)=2sinxcosx 这是公式
所以：2√3sin(x/2+π/4)cos(x/2+π/4)
=根号3sin[2(x/2+π/4)]

sin(x+π)=-sinx 也是公式
-sin(x+π)=-（-sinx)=sinx

=√3sin(x+π/2)+sinx
=-根号3cosx+sinx=2(-根号3/2cosx+1/2sinx)
=2(cos5π/6cosx+sin5π/6sinx)
=2cos(x-5π/6)

Answer 3

2sin(x/2+π/4)cos(x/2+π/4)=sin2(x/2+π/4)=sin(x+π/2)，用的是公式sin2x=2sinxcosx

sin(x+π/2)=cos[π/2-(x+π/2)]=cos(-x)=cosx，用的是公式sinx=cos(π/2-x)和cosx=cos(-x)

Answer 4

sin2x=2sinxcosx

sin(x+π)=-sinx

追问

那√3sin(x+π/2)+sinx=√3cosx+sinx是怎么来的

追答

sin(x+π/2)=cosx
即sin(x+π/2)=sinxcos(π/2)+cosxsinπ/2)sinx*0+cosx*1=cosx

Answer 5

sin2x=2*Sinx*Cosx

追问

=√3sin(x+π/2)+sinx
后面的+sinx又怎么来的

追答

-sin(x+π)=sin(-π-x)=sin(π-x)=sinx