Derivatives of Trigonometric Functions
[Pages:7]Derivatives of Trigonometric Functions
The basic trigonometric limit:
Theorem:
lim sin x x0 x
=1=
lim
x0
x sin
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.
Note:
Cosine behaves even better near 0, where
lim cos
x0
x =1.
ex. Show that
lim cos x -1 = 0
x 0
x
lim
x 0
cos x -1 x
=
lim cos x x0 x
-1 cos x +1 cos x +1
=
lim
x0
cos2 x -1 x(cos x +1)
=
lim
x0
- sin 2 x x(cos x +1)
=
lim
x 0
-
sin x
x
lim x0
sin x cos x + 1
=
-
lim
x0
sin x
x
lim x0
sin x cos x + 1
=
-
(1) 1
0 +
1
=
0
ex.
Evaluate
lim
x0
sin 2x 5x
lim
x0
sin 2x 5x
=
1 5
lim
x0
sin 2x x
2 2
=
2 5
lim
x0
sin 2x 2x
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 = 2x. Note that as x approaches 0, so does . Hence,
2 5
lim
x 0
sin 2x 2x
=
2 5
lim
0
sin
=
2 5
1
=
2 5
ex.
Evaluate
lim
x0
sin sin
4x 3x
lim
x0
sin 4x sin 3x
=
lim
x0
sin 4x sin 3x
x x
=
lim
x0
sin 4x x
x sin 3x
=
lim
x0
sin 4x x
lim x0
x sin 3x
Repeat the same trick as in the previous example, let = 4x and = 3x. Both and approach 0 when x does. Then apply the theorem twice.
=
lim
x0
sin 4x x
4 4
lim x0
x sin 3x
3 3
=
4 3
lim
x0
sin 4x 4x
lim x0
3x sin 3x
=
4 3
lim
0
sin
lim 0
sin
=
4 3
1
1
=
4 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
x0 nx
n
lim
x0
mx sin nx
=
m n
lim
x0
sin mx sin nx
=
m n
(Trivially, we also have:
lim
x0
mx nx
=
m n
.)
ex.
Evaluate
lim
x0
tan 7x 2x
lim
x0
tan 7x 2x
=
lim
x0
1 2x
sin 7x cos 7x
=
1 2
lim
x0
sin 7x x
1 cos
x
=
1 2
lim
x0
sin 7x x
lim x0
1 cos
x
= 1 7 1 = 7 211 2
Recall that since cos x is continuous everywhere, the direct substitution property applies, therefore,
lim
x 0
1 cos
x
=
1 lim cos
x
=
1 cos 0
=
1 1
=1
x0
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 = lim sin(x + h) - sin x = lim [sin x cos h + sin h cos x] - sin x
dx
h0
h
h0
h
= lim sin x cos h - sin x + lim sin h cos x = lim sin x(cos h -1) + lim sin h cos x
h0
h
h0
h
h0
h
h0 h
= lim sin x lim cos h -1 + lim sin h lim cos x = sin x (0) + (1) cos x = cos x
h0
h0
h
h h0
h0
Therefore,
d dx
sin
x
=
cos x
ex. Find the derivative of csc x.
d dx
csc
x
=
d dx
1 sin
x
=
sin
x
d 1 -1 d dx dx
(sin x)2
sin
x
=
sin
x
0 -1 cos sin 2 x
x
=
- cos x sin 2 x
=
-
1 sin
x
cos x sin x
=
-
csc
x cot
x
Therefore,
d dx
csc x
=
- csc
x cot
x
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 = sec2 x dx
4.
d sec x = sec x tan x dx
5.
d cot x = - csc2 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 )
=
1
sin t - cos
t
s(t )
=
(1 -
cos
t)(cos t) - (sin t)(0 (1 - cos t)2
-
(- sin
t ))
= cos t - cos2 t - sin 2 t = cos t - (cos2 t + sin 2 t)
(1 - cos t)2
(1 - cos t)2
= cos t - 1 = - (1 - cos t) = -1 = 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 = /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 = /2 is v(/2) = -2 sin (/2) + cos (/2) = -2 + 0 = 2 (m/sec) Its acceleration at the same time is a(/2) = -2 cos (/2) - sin (/2) = 0 - 1 = -1 (m/sec2)
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