推 pop10353 :sin2Θ = 2x/(1+x^2) ?? 02/14 01:51
漏了"則有"
→ wayn2008 :cosx-cosy= - 2sin()sin()吧?! 02/14 02:27
THX
※ 編輯: FAlin 來自: 124.11.128.7 (02/14 02:50)
※ 引述《deryann (星辰)》之銘言：
: sinx+siny=1/2，cosx-cosy=1/3 求sin(x-y)?
利用和差化積
sinx+siny = 2 * sin[(x+y)/2] * cos[(x-y)/2] = 1/2
cosx-cosy = -2 * sin[(x+y)/2] * sin[(x-y)/2] = 1/3
把下式除上式：
tan[(x-y)/2] = -2/3
而tanΘ = x 的話，則有sin2Θ = 2x/(1+x^2)
所以 sin(x-y) = 2 * (-2/3) / (1 + 4/9) = -12/13
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◆ From: 124.11.128.7