Question

高中三角函数问题求解答必采纳谢谢

高中三角函数问题求解答必采纳谢谢在△ABC中 sinBsin(B+π/3)=3/4
求B
求sinA+sinC的取值范围

Answer 1

3/4 = sin(B)sin(B+π/3) = 1/4 + 1/2 sin(2*B-π/6)
所以 sin(2*B-π/6)=1，由角度范围，2*B-π/6=π/2, B=π/3
所以 0<A<π-π/3=2π/3
sin(A)+sin(C)=sin(A)+sin(π-π/3-A)=sin(A)+sin(π/3+A)
=sin(A) + √3/2 cos(A) + 1/2 sin(A)
=√3 (√3/2 sin(A) +1/ cos(A))
=√3 (cos(π/6) sin(A) +sin(π/6) cos(A))
=√3 sin(A+π/6)
由前面A的范围， 1/2<sin(A)+sin(C)≤√3
最大值当且仅当 A=π/3 时成立，当A接近0或2π/3时，接近下界1/2

Answer 2

参考

Answer 3

Answer 4