Question

已知三角形ABC中，A=π/4,CB=2,设角ACB=θ,θ属于（π/2,3π/4)

1,.用θ表示CA
2.求f(θ)=向量CA×CB的单调递增区间

Answer 1

1，
由正弦定理，CB/sinA=CA/sinB
CA=CBsinB/sinA
其中B=π-A-θ
sinB=sin(A＋θ)=sin(θ＋π/4)
所以
CA=2 sin(θ＋π/4)/sin(π/4)
=2√2 sin(θ＋π/4)

2，
f(θ)=CA·CB·cosθ
= 2√2 sin(θ＋π/4)·2·cosθ
= 4√2 sin(θ＋π/4)·cosθ
求导
f'(θ)= 4√2 cos(θ＋π/4)·cosθ - 4√2 sin(θ＋π/4)·sinθ
= 4√2 cos(2θ＋π/4)·cosθ

Answer 2

1. 用正弦定理：CA/sin(3π/4-θ)=CB/sin(π/4), 所以：CA=2√2 sin(3π/4 -θ),
2. 应该是点乘而不是叉乘吧，点乘的话：f(θ) =|CA||CB|cos∠ACB=2√2 sin(3π/4 -θ)×2cosθ
=2(sin2θ+cos2θ)+2=2√2 sin(π/4 +2θ)+2, 所以f(θ)在θ∈（5π/8, 3π/4）上递增

Answer 3

用正弦定理。第2题用降幂公式化简。