Question

高一数学急！！！！

Answer 1

f(x)=2cos(x/2)(√3cos(x/2)-sin(x/2))
=2√3cos²(x/2)-2cos(x/2)sin(x/2)
=√3(cosx+1)-sinx
=√3cosx-sinx+√3=2cos(x+π/6)+√3
(1)f(θ)=2cos(θ+π/6)+√3=√3+1, 2cos(θ+π/6)=1
cos(θ+π/6)=1/2,∴θ+π/6=π/3 θ+π/6=-π/3
θ=π/6 或 θ=-π/2
(2) f(C)=2cos(C+π/6)+√3=√3+1, 2cos(C+π/6)=1
∴C+π/6=π/3, C=π/6
S=1/2*absinC=1/4*ab=√3/2, ab=2√3
1=a²+b²-2abcos30=a²+b²-√3ab=a²+b²-6
a²+b² =1+6=7
(a+b)²=a²+b²+2ab=7+4√3=(2+√3)²
a+b=2+√3 而 c/sin30=2R, ∴2R=2
2RsinA+2RsinB=2R(sinA+sinB)=2+√3
sinA+sinB=（2+√3）/2R=（2+√3）/2

Answer 2

f(x)=2cosx/2(√3cosx/2-sinx/2)=4cosx/2(√3/2cosx/2-1/2sinx/2)
=4cosx/2(cosx/2sinπ/3-sinx/2cosπ/3)=4cosx/2sin(π/3-x/2)
=2[sinπ/3-sin(x-π/3)]=√3-2sin(x-π/3)
(1). 为方便用 x 表示, x∈[-π/2,π/2], x-π/3∈[-5π/6,π/6]
f(x)=√3-2sin(x-π/3)=√3+1
sin(x-π/3)=-1/2=-sinπ/6=sin(-π/6)
x-π/3=-π/6, x=π/3-π/6=π/6
(2). 同上, C∈(0,π), C-π/3∈(-π/3,2π/3)
sin(C-π/3)=-sinπ/6=sin(-π/6), C-π/3=-π/6, C=π/6
AB=c=1
S=1/2absinC=1/2absinπ/6=ab=√3/2
cosC=(a²+b²-c²)/(2ab), a²+b²-1=2*√3/2*√3/2=3/2
a²+b²=5/2, a²+2ab+b²-2ab=5/2, (a+b)²=5/2+√3, a+b=√(10+4√3)/2
a/sinA=b/sinB=c/sinC=1/sinπ/6=2
sinA=a/2, sinB=b/2
sinA+sinB=a/2+b/2=(a+b)/2=√(10+√3)/4