請問一下這題要怎麼算
拜託大大解釋詳細一點
sin(2arc cot 4)
(數學)反三角函數問題
2009-06-29 11:55 pm
回答 (3)
2009-06-30 12:04 am
✔ 最佳答案
arccot(4)是一個角度,畫一直角三角形,選一銳角x, 對邊為1, 鄰邊為4=>斜邊為17
原題= sin(2x)= 2sinx*cosx
= 2* (1/√17)*(4/√17)
= 8/17
2009-07-01 8:10 pm
藍閃蝶大大算到tan(x/2) = 1/4後
可直接用sinx=2tan(x/2)/[1+tan^2 (x/2)]=(2/4)/[17/16]=8/17
可直接用sinx=2tan(x/2)/[1+tan^2 (x/2)]=(2/4)/[17/16]=8/17
2009-06-30 10:34 am
（不繪圖的方法）
設 x = 2arc cot 4。
arc cot 4 = x/2
cot(x/2) = 4
tan(x/2) = 1/4
tan(x) = 2(1/4) / [1-(1/4)^2] = 8/15
sec^2 (x) = 1 + (8/15)^2 = 289/225
cos^2 (x) = 225/289
sin^2 (x) = 1 - 225/289 = 64/289
sin(2arc cot 4) = √(64/289) = 8/17
設 x = 2arc cot 4。
arc cot 4 = x/2
cot(x/2) = 4
tan(x/2) = 1/4
tan(x) = 2(1/4) / [1-(1/4)^2] = 8/15
sec^2 (x) = 1 + (8/15)^2 = 289/225
cos^2 (x) = 225/289
sin^2 (x) = 1 - 225/289 = 64/289
sin(2arc cot 4) = √(64/289) = 8/17
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