Question

已知函数f(x)=2cosx/2(根号3cosx/2-sinx/2)在三角形ABC中c=1f(c)=根3+1面积=根3/2求sinA+sinB

Answer 1

已知函数f(x)=2cosx/2(根号3cosx/2-sinx/2)在三角形ABC中c=1f(c)=根3+1面积=根3/2求sinA+sinB

f(c)=2cosc/2(√3cosc/2-sinc/2)
=2√3(cosc/2)²-2(sinc/2)(cosc/2)
=2√3*(cosc+1)/2-sinc
=√3cosc+√3-sinc
=√3-(sinc-√3cosc)
=√3-2sin(c-π/3)
=√3+1

√3-2sin(c-π/3)=√3+1
-2sin(c-π/3)=1

sin(c-π/3)=-1/2
c-π/3=-π/6
c=π/6

S=(ab*sinC)/2=ab/4=√3/2
ab=2√3
cosC=(a²+b²-c²)/2ab
√3/2=(a²+b²-1)/2*2√3
a²+b²=7
a²+2ab+b²=7+2√3
(a+b)²=7+2√3
a+b=√(7+2√3)

sinA+sinB=(asinC)/c+(bsinC)/c=(a+b)/2=√(7+2√3)/2