※ 引述《gto770 (比克)》之銘言:
: 1.年級:高二上
: 2.科目:數學
: 3.章節:第三章 圓與球 第一節 圓的參數式
: 4.題目:X= sin2θ Y=2sin^2θ (0≦θ<π),求圓心以及半徑?
: 5.想法:因為θ介於0到π之間,所以顯然是要使用2θ,
: 於是把Y=2sin^2θ利用二倍角公式轉換成Y=1-cos2θ
: 如此一來就變成 X= sin2θ
: Y=1-cos2θ
: 乍看之下圓心似乎是(0,1),但是半徑卻無法得知?
: 但是我不懂的是,以圓的參數式來看的話 X的值應該是要搭配cosθ
: Y值應該是要配上sinθ(此題剛好相反)
: 在來第二個問題就是說sinθ與cosθ的系數及是圓的半徑值
: 但是這題的系數卻是不一樣的兩個值(分別為1和-1),所以在這邊想不通 = =
: 不知道是哪邊卡住了還是說我的觀念上有些錯誤?
: 小弟不才,請板上各位高手指點迷津
: 第一次PO文,若有違反版規的地方請告知,會自D的,謝謝!
X= sin2θ ..... (1)
Y-1=-cos2θ ......(2)
X^2+(Y-1)^2= 1
我想可能是圓心(0,1) R=1
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X= sinθ=cos(pi/2-θ) ..... (1)
Y-1=-cosθ=-sin(pi/2-θ) ......(2)
X^2+(Y-1)^2= 1
※ 編輯: win1 來自: 210.66.48.95 (05/18 00:14)
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