※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: Prove that for any a, b in R,
: | sin(a＋b) － ( sin(a)＋b cos(a) ) | ≦ b^2/2.
令 f(x) = sin(a+x) - sin(a) - xcos(a) - (x^2)/2
f'(x) = cos(a+x) - cos(a) - x
f"(x) = -sin(a+x) - 1
f"(x)≦0 for all real x => f'(x) is decreasing on R
f'(0) = 0 => f'(x)≧0 when x<0 and f'(x)≦0 when x>0
Hence f(x) ≦ f(0) = 0
i.e. sin(a+x) - sin(a) - xcos(a) ≦ (x^2)/2
類似地, 令 g(x) = -sin(a+x) + sin(a) + xcos(a) - (x^2)/2
相同的作法可得 -sin(a+x) + sin(a) + xcos(a) ≦ (x^2)/2
故 |sin(a＋x) - sin(a) - bcos(x)| ≦ x^2/2
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