推 ytyt5239 :謝謝了，人超好，好心有好報><~ 12/08 16:42
※ 引述《ytyt5239 (ytyt)》之銘言：
: 1.use the mean value theorem to prove the inequality
: ｜sin a - sin b｜ <= ｜a-b｜ for all a and b
: 是要分三個case嗎?? a>b a=b a<b
: 2.prove the identity
: arcsin x-1/x+1 = 2arctan 開根號X -π/2
: 我是先把上是假設一個函式，想讓其函數=0
: 但函式微分後變好複雜，牽扯隱函數微分~誰來幫幫我啊
1. If a = b, sin a = sin b. |sin a - sin b| = |a - b| = 0 and
｜sin a - sin b｜ <= ｜a-b｜
f(x) = sin x is continous everywhere, f'(x) = cos x for all x.
Suppose a≠b.
Let k1 = min(a, b), k2 = max(a,b)
(sin k1 - sin k2)/ (k1-k2) = cos c for some c in (k1, k2) for a≠b.
Obviously, |sin k1 - sin k2| = |sin a - sin b|, |k1-k2|=|a-b|
=> |sin a - sin b| / |a - b| = |cos c| ≦ 1
=> |sin a - sin b| ≦ |a - b| for a≠b
-1
2. d sin x 1
-------- = -----------
d x √(1-x^2)
-1 x-1
Let y = sin ------
x+1
[(x+1) - (x-1)]/(x+1)^2 2
=> y' = ------------------- = ------------------------
√{1-[(x-1)/(x+1)]^2} (x+1)√[(x+1)^2-(x-1)^2]
2
= -----------
2(x+1)√x
1
=> y' = ---------
(x+1)√x
1
=> y = ∫--------- dx + C
(x+1)√x
2
= ∫-------- du + C, u = √x, du = dx/(2√x)
(u^2+1)
-1
= 2tan u + C
-1
= 2tan √x + C
-1 -1
y(0) = sin [(0-1)/(0+1)] = -π/2 = 2tan 0 + C
=> C = -π/2
-1 -1
So, y = sin [(x-1)/(x+1)] = 2tan √x - π/2
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