
Kathy M. answered 07/24/17
Tutor
5
(13)
High School Math Teacher 9+years
Recall:
{1} cos ( A + B ) = cos(A)cos(B) - sin(A)sin(B)
{2} limh->0 [ (cos(h) - 1 )/h ] = 0
{3} limh->0 [sin(h) / h ] = 1
limh->0 [ ( cos(x+h) - cos(x) ) / h ] **use {1}
limh->0 [ ( cos(x)cos(h) - sin(x)sin(h) - cosx ) / h ]
limh->0 [ (cos(x)cos(h) - cosx - sin(x)sin(h) ) / h ]
limh->0 [ ( cos(x) (cos(h) - 1 ) - sin(x)sin(h) ) / h ]
limh->0 [ ( cos(x) (cos(h) -1) )/h - ( sin(x)sin(h) ) /h ]
limh->0 [ cos(x) · (cos(h) - 1)/h ] - limh->0 [ sin(x) · sin(h)/h ] ** use {2} and {3}
cos(x) · (0) - sin(x) · (1)
- sin(x)