Question

证明 |sinA-sinB|=<|A-B|

解法:
f(x)=sinx, f(x)连续可导
[f(A)-f(B)]/(A-B)=f'(c)=cos(c), |cos(c)|=<1
then |[f(A)-f(B)]/(A-B) |=|f(A)-f(B)| / |A-B|=|cos(c)|=<1
so |sinA-sinB|=<|A-B|

Answer 1

sinA=sin(A+B/2)cos(A-B/2)+sin(A-B/2)cos(A+B/2)
sinB=sin(A+B/2)cos(A-B/2)-sin(A-B/2)cos(A+B/2)
sinA-sinB=2sin(A-B/2)cos(A+B/2)
所以 |sinA-sinB|=| 2sin(A-B/2)cos(A+B/2)|=<| 2sin(A-B/2)|=<| 2(A-B/2)|=|A-B|
第一个不等号因为余弦的绝对值永远小于等于1
第二个不等式因为在x>0的情况下x>sinx