
Re: [其他] 三角函數求解

: 第六題解不出來
提供另一個解出tanθ的想法
1 + tanθ
----------- = 3 - 2√2
1 - tanθ
令 tanα = 1
則
1 + tanαtanθ
tan(α-θ) = ----------------
tanα - tanθ
1 + tanθ
= -----------
1 - tanθ
令 A = α-θ
則 θ = α- A, tanA = 3 - 2√2
1 + tanαtanA
tanθ = ---------------
tanα - tanA
1 + 1*(3 - 2√2)
= ------------------
1 - (3 - 2√2)
4 - 2√2
= ----------
-2 + 2√2
1
= -----
√2
sinθ = 1/√3 cosθ = √2/√3
sinθ + cosθ = (1 + √2)/√3
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