Re: [中學] 三角函數化簡
※ 引述《Lieng1207 (Liengs)》之銘言:
: 題目是要化簡
(sinθ-1/sinθ)^2 + (cosθ-1/cosθ)^2 - (tanθ-1/tanθ)^2
= sin^2 - 2 + csc^2 + cos^2 - 2 + sec^2 - tan^2 + 2 - cot^2
= - 2 + [sin^2 + cos^2] + [csc^2 - cot^2] + [sec^2 - tan^2]
= -2 + 1 + 1 + 1
= 1
: 但因為這題是在民國97年的龍騰高中數學第三冊的1-1習題第6題
: 該章節內只有提到sin、cos、tan這三個函數的定義
: 以及
: tanθ=sinθ/cosθ(商數關係)
: (sinθ)^2+(cosθ)^2=1(正餘弦平方關係式)
: sin(90°-θ)=cosθ,cos(90°-θ)=sinθ (餘角關係式)
: 這三個性質
: 想知道這一題能否僅用該章節的內容來解呢?
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抱歉,看錯了,已更正。
※ 編輯: Honor1984 (111.249.189.84), 05/15/2017 01:10:37
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※ 編輯: Honor1984 (111.249.189.84), 05/15/2017 01:42:02
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