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1 代入法例 1 已知 tgα .ctgβ =5,求 sin(α +β) .csc(α -β)值 .解 ∵ tgα .ctgβ =5,∴ sin(α +β) csc(α -β) =sin(α +β)sin(α -β)=sinαcosβ +cosαsinβsinαcosβ - cosαsinβ=tgαctgβ +1tgαctgβ - 1=5+15- 1=32 .2 配凑法例 2 已知 π2
1 Substituting into law 1 Knowing that tgα.ctgβ = 5, find sin(α +β) .csc(α -β) values. Solution ∵ tgα.ctgβ =5, ∴ sin(α +β) csc(α -β) = Sin(α +β)sin(α -β)=sinαcosβ +cosαsinβsinαcosβ - cosαsinβ=tgαctgβ +1tgαctgβ - 1=5+15- 1=32 .2 Puncture 2 Known π2