PHY1106: Waves and Oscillators: Dr. Pete Vukusic
Differentiating sine / cosine
The sine / cosine functions differentiate using standard sets of rules:
d
(sin t ) = cos t
dt
d
(cos t ) = − sin t
dt
d
( − sin t ) = − cos t
dt
d
( − cos t ) = sin t
dt
Slightly more complicated is sin (ωt) or sin (ωt + φ) etc. In these cases the
expression within the sine or cosine is itself differentiated w.r.t the variable “t”, in
addition to the sin or cos function itself: these are then multiplied together.
d
( A sin ωt ) = ωA cos ωt and
dt
d
( A cos ωt ) = −ωA sin ωt
dt
since
d
(ωt ) = ω
dt
For a further example, if the sin / cos function is now a more complicated function of
t say, notice how we still differentiate it (as well as the sin / cos itself) ; e.g. for sin
(kt2) and cos (kt3) etc.
[
]
d
sin(kt 2 ) = 2kt. cos(kt 2 )
dt
and
[
]
d
cos(kt 3 ) = −3kt 2 . sin kt 3
dt
If the function has two variables in it, (as with some of the functions in our lecture
course, e.g. y = y 0 sin(ωt − kx ) ), then the differentiation works in the same way;
d
( y ) = ωy 0 cos(ωt − kx )
dt
In general:
d
( y ) = −ky 0 cos(ωt − kx )
and
dx
d
df dg
[f(g(x))] =
×
dx
dx dx
This means we differentiate the outside function, leave the argument of the
outside function alone, and then multiply by the derivative of the inside function.
For example;
d
df dg
[cos(3x 2 + 1)] =
×
= − sin(3x 2 + 1) × 6 x = −6 x. sin(3x 2 + 1)
dx
dx dx
Practice Problems:
Complete the following (see PV or your tutor for solutions):
d
[Aω sin(ωt )] =
dt
d
[A cos(ωt − kx )] =
dx
d
[Aω cos(ωt − kx )] =
dt
d  − A cos(ωt − kx ) 
 =
dt 
ω

d
dk
 c. sin(ka / 2 ) 
 =

(a / 2) 
