Question

高一数学题目 解三角形

看图第一题，或者做完也可以，没时间就看第一题可以了

Answer 1

说明：^2——表示平方
①α∈(π/2,π)，sinα=√3/3
(1) cosα=-√(1-sin^2α)
=-√[1-(√3/3)^2]
=-√(1-1/3)
=-√6/3
sin(π/6-α)
=sin(π/6)cosα-cos(π/6)sinα
=(1/2)(-√6/3)-(√3/2)(√3/3)
=-√6/6-3/6
=-(√6+3)/6
(2) sin2α=2sinαcosα
=2(√3/3)(-√6/3)
=-2√2/3
∵α∈(π/2,π)，sinα=√3/3<√2/2=sin(3π/4)
∴3π/4<α<π
3π/2<2α<2π
cos2α=√[1-sin^2(2α)]
=√[1-(-2√2/3)^2]
=√(1-8/9)
=1/3
cos(5π/3+2α)
=cos(2π-π/3+2α)
=cos(-π/3+2α)
=cos2αcos(π/3)+sin2αsin(π/3)
=(1/3)(1/2)+(-2√2/3)(√3/2)
=1/6-2√6/6
=(1-2√6)/6

Answer 2

如图