arctan(y/3) - arctan(π/3)=ln(x)
y(x)=?
2007-05-20 14:20:04 · 2 個解答 · 發問者 ? 2 in 教育與參考 ➔ 高等教育 (大學或以上)
tan^(– 1) (y/3) – tan^(– 1) (π/3) = ln x
y(x) = ?
Method 1:
tan^(– 1) (y/3) – tan^(– 1) (π/3) = ln x
tan^(– 1) (y/3) = tan^(– 1) (π/3) + ln x
tan tan^(– 1) (y/3) = tan [tan^(– 1) (π/3) + ln x]
y/3 = [tan tan^(– 1) (π/3) + tan ln x]/[1 – [tan tan^(– 1) (π/3)]tan ln x]
y/3 = (π/3 + tan ln x)/[1 – (π/3) tan ln x]
y/3 = (π + 3 tan ln x)/(3 – π tan ln x)
y = (3π + 9 tan ln x)/(3 – π tan ln x)
y(x) = (3π + 9 tan ln x)/(3 – π tan ln x)
Method 2:
tan^(– 1) (y/3) – tan^(– 1) (π/3) = ln x
tan [tan^(– 1) (y/3) – tan^(– 1) (π/3)] = tan ln x
[tan tan^(– 1) (y/3) – tan tan^(– 1) (π/3)]/[1 + [tan tan^(– 1) (y/3)] [tan tan^(– 1) (π/3)]] = tan ln x
(y/3 – π/3)/[1 + (y/3)(π/3)] = tan ln x
(3y – 3π)/(πy + 9) = tan ln x
3y – 3π = πy tan ln x + 9 tan ln x
3y – πy tan ln x = 3π + 9 tan ln x
y(3 – π tan ln x) = 3π + 9 tan ln x
y = (3π + 9 tan ln x)/(3 – π tan ln x)
y(x) = (3π + 9 tan ln x)/(3 – π tan ln x)
2007-05-23 06:39:08 · answer #1 · answered by ? 7 · 0⤊ 0⤋
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2007-05-23 13:11:20 · answer #2 · answered by ? 4 · 0⤊ 0⤋