Formulas for Derivatives: Derivatives of Trig Functions
Goal: Finally prove the derivative formula for sine and cosine. Use them to get derivative
formulas for other trig functions.
We’ve thought all along that the derivative of sine is cosine. Let’s prove it:
d
sin(x + h) − sin(x)
sin(x) = lim
h→0
dx
h
sin(x)cos(h) + cos(x)sin(h) − sin(x)
= lim
h→0
h
sin(x)(cos(h) − 1) + cos(x)sin(h)
= lim
h→0
h
cos(h) − 1 ⎞
sin(h) ⎞
⎛
⎛
= sin(x) ⎜ lim
+ cos(x) ⎜ lim
⎟
⎝ h→0
⎠
⎝ h→0 h ⎟⎠
h
The second line is the angle addition formula for sine. The third line is just using algebra to
rearrange the formula, and the fourth is justified if both limits exist (lim(a + b) = lim(a) + lim(b)
if both pieces exist). The first limit in the last line is zero and the second is one by our work with
special trig limits. So the final result is cos(x).
We could do a similar calculation with cosine and its addition formula, but there is an easier
d
d
way:
cos(x) = sin(x + π / 2) = cos(x + π / 2) = − sin(x) , just using properties of derivatives
dx
dx
of horizontal shifts.
From these, we can easily calculate the derivatives of the other trig functions using the quotient
rule:
d
d sin(x) cos(x)⋅ cos(x) − sin(x)⋅(− sin(x))
1
tan(x) =
=
=
= sec 2 (x)
2
dx
dx cos(x)
cos (x)
cos 2 (x)
d
d
1
cos(x)⋅ 0 − 1⋅(− sin(x)) sin(x)
sec(x) =
=
=
= sec(x)tan(x)
dx
dx cos(x)
cos 2 (x)
cos 2 (x)
The formulas for cot(x) and csc(x) work out similarly. Watch the signs!
To summarize:
f(x)
sin(x)
cos(x)
f′(x)
cos(x)
–sin(x)
tan(x)
sec2(x)
cot(x)
–csc2(x)
sec(x)
sec(x)tan(x)
csc(x)
–csc(x)cot(x)
Notice the relationship between derivatives of functions and derivatives of their cofunctions—
you change the sign and the things in the derivative to their co-things.
Examples:
d
sin(3x) = 3cos(3x)
dx
d
tan 2 (θ ) = 2 tan(θ )sec 2 (θ )
dθ
(don’t forget the chain rule!)
Example:
d
sin 2 (x) = 2sin(x)⋅ cos(x) = sin(2x) , the first equality being from
dx
the chain rule and the second from the double-angle identity. Now remember the double-angle
identity for cosine: cos(2x) = cos2(x) – sin2(x) = 2cos2(x) – 1 = 1 – 2sin2(x). We could rewrite
1− cos(2x)
this as sin 2 (x) =
. We could also differentiate this last formula:
2
d
d ⎛ 1 cos(2x) ⎞
−2sin(2x)
sin 2 (x) = ⎜ −
= 0−
= sin(2x).
⎟
dx
dx ⎝ 2
2 ⎠
2
An interesting relationship:
Problems: 3 – 39 odd, 40 – 44, 47, 49, 50, 51, 52