Question

tan2θ=﹣2√2,θ∈﹙π/4,π/2﹚,求[2cos^2(θ/2)-sinθ-1]/[√3sin﹙π／3＋θ﹚sin﹙π／3－θ﹚]

Answer 1

∵tan2θ=2tanθ/(1－tan²θ)=﹣2√2 ∴tanθ/(1－tan²θ)=√2

∴√2tan²θ＋tanθ－√2＝0 ∴(tanθ－√2)(√2tanθ－1)＝0
∴tanθ＝√2或tanθ＝√2/2 ∵θ∈﹙π/4,π/2﹚∴tanθ＞1 ∴tanθ＝√2
∵sec²θ＝1＋tan²θ＝3 ∴cos²θ＝1/3 sin²θ＝2/3
∵θ∈﹙π/4,π/2﹚ ∴cosθ＝√3/3 sinθ＝√6/3
[2cos²(θ/2)-sinθ-1]/[√3sin﹙π／3＋θ﹚sin﹙π／3－θ﹚]＝(cosθ－sinθ)/[√3/2(cos2θ－cos2π／3)
＝(√3/3－√6/3 )/[√3/2(2×1/3－1＋1/2)＝[√3/3(1－√2)]/(√3/12)＝4(1－√2)