Download Derivatives of Trigonometric Functions

Derivatives of Trigonometric Functions
The basic trigonometric limit:
sin x
x
lim
= 1 = lim
Theorem:
x→0
x →0 sin x
x
(x in radians)
Note: In calculus, unless otherwise noted, all angles are measured in
radians, and not in degrees.
This theorem is sometimes referred to as the small-angle approximation
because it really says that, for very small angles x, sin x ≈ x.
cos x = 1 .
Note: Cosine behaves even better near 0, where lim
x →0
ex. Show that lim
x →0
cos x − 1
=0
x
− sin 2 x
cos 2 x − 1
cos x − 1
cos x − 1 cos x + 1
= lim
= lim
lim
= lim
⋅
x →0
x →0
x
x
cos x + 1 x→0 x(cos x + 1) x→0 x(cos x + 1)
= lim
x→0
− sin x
sin x
sin x
sin x
 0 
⋅ lim
= − lim
⋅ lim
= − (1)
= 0
x →0 cos x + 1
x →0
x
x x→0 cos x + 1
1 +1
ex. Evaluate lim
x→0
lim
x→0
sin 2 x
5x
sin 2 x 1
sin 2 x 2 2
sin 2 x
= lim
⋅ = lim
5x
5 x →0 x
2 5 x →0 2 x
The idea above is to match the angle in the sine function with the
denominator. We’ll then apply the basic trigonometric limit. To do so, first
we substitute θ = 2x. Note that as x approaches 0, so does θ. Hence,
2
sin 2 x 2
sin θ 2
2
lim
= lim
= ⋅1 =
5 x →0 2 x
5 θ →0 θ
5
5
ex. Evaluate lim
x→0
lim
x→0
sin 4 x
sin 3 x
sin 4 x
sin 4 x
sin 4 x x
sin 4 x
x
x
= lim
⋅ = lim
⋅
= lim
⋅ lim
x →0 sin 3 x
sin 3 x x→0 sin 3 x x x→0 x
sin 3 x x→0 x
Repeat the same trick as in the previous example, let θ = 4x and λ = 3x.
Both θ and λ approach 0 when x does. Then apply the theorem twice.
= lim
x →0
=
sin 4 x 4
x
3 4
sin 4 x
3x
⋅ ⋅ lim
⋅ = lim
⋅ lim
x
4 x→0 sin 3 x 3 3 x→0 4 x x→0 sin 3 x
sin θ
4
λ
4
4
lim
⋅ lim
= ⋅1⋅1 =
λ →0 sin λ
3 θ →0 θ
3
3
In fact, after doing a few examples like those, we can see a (very nice)
pattern. To sum it up:
Suppose m and n are nonzero real numbers, then
lim
sin mx m
=
nx
n
lim
mx
m
=
sin nx n
lim
sin mx m
=
sin nx
n
x→0
x →0
x →0
(Trivially, we also have:
lim
x →0
mx m
= .)
nx
n
ex. Evaluate lim
x→0
lim
x→0
=
tan 7 x
2x
tan 7 x
1 sin 7 x 1
sin 7 x
1
1
sin 7 x
1
= lim
⋅
= lim
⋅
= lim
⋅ lim
x →0 2 x cos 7 x
x →0 cos x
2x
2 x →0 x
cos x 2 x→0 x
1 7 1 7
⋅ ⋅ =
2 1 1 2
Recall that since cos x is continuous everywhere, the direct
substitution property applies, therefore,
1
1
1
1
=
=
= =1
x → 0 cos x
lim cos x cos 0 1
lim
x →0
Now, the main topic -Derivatives of Trigonometric Functions
ex. What is the derivative of sin x?
Start with the limit definition of derivative:
d
sin( x + h) − sin x
[sin x cos h + sin h cos x] − sin x
= lim
sin x = lim
h →0
h →0
dx
h
h
= lim
h →0
sin x cos h − sin x
sin h cos x
sin x(cos h − 1)
sin h
+ lim
= lim
+ lim
⋅ cos x
h →0
h →0
h →0
h
h
h
h
= lim sin x ⋅ lim
h →0
Therefore,
h →0
cos h − 1
sin h
+ lim
⋅ lim cos x = sin x ⋅ (0) + (1) cos x = cos x
h →0
h
h h→0
d
sin x = cos x
dx
ex. Find the derivative of csc x.

d  d
sin x 1 − 1 sin x 
d 1
d
 = sin x ⋅ 0 − 1 ⋅ cos x
 dx   dx
csc x =
=
2
(sin x)
sin 2 x
dx sin x
dx
=
− cos x
1 cos x
=−
⋅
= − csc x cot x
2
sin x
sin x sin x
Therefore,
d
csc x = − csc x cot x
dx
The complete list of derivatives of trigonometric functions:
1.
d
sin x = cos x
dx
2.
d
cos x = − sin x
dx
3.
d
tan x = sec 2 x
dx
4.
d
sec x = sec x tan x
dx
5.
d
cot x = − csc 2 x
dx
6.
d
csc x = − csc x cot x
dx
ex. Differentiate f (x) = sec x + 5 csc x
f ′(x) = sec x tan x + 5( −csc x cot x) = sec x tan x − 5 csc x cot x
ex. Differentiate f (x) = x2 cos x − 2x sin x − 3 cos x
f ′(x) = [x2(−sin x) + (2x) cos x] − 2[x (cos x) + (1)sin x] − 3(−sin x)
= − x2 sin x + 2x cos x − 2x cos x − 2sin x + 3sin x
= − x2 sin x + sin x
ex. Differentiate s (t ) =
s ′(t ) =
sin t
1 − cos t
(1 − cos t )(cos t ) − (sin t )(0 − (− sin t ))
(1 − cos t ) 2
cos t − cos 2 t − sin 2 t cos t − (cos 2 t + sin 2 t )
=
=
(1 − cos t ) 2
(1 − cos t ) 2
=
cos t − 1
1
− (1 − cos t )
−1
=
=
=
(1 − cos t ) 2 (1 − cos t ) 2 1 − cos t cos t − 1
ex. Simple Harmonic Motion Suppose the oscillating motion (in meters)
of a weight attached to a spring is described by the displacement function
s(t) = 2 cos t + sin t
Find its velocity and acceleration functions, and its speed and acceleration at
t = π/2 sec.
Velocity: v(t) = s′(t) = −2 sin t + cos t
Acceleration: a(t) = v′(t) = −2 cos t − sin t
Its speed when t = π/2 is
│v(π/2)│ = │−2 sin (π/2) + cos (π/2) │ = │−2 + 0│ = 2 (m/sec)
Its acceleration at the same time is
a(π/2) = −2 cos (π/2) − sin (π/2) = 0 − 1 = −1 (m/sec2)