Download 19 Derivative of sine and cosine

19
19.1
Derivative of sine and cosine
Two trigonometric limits
The rules for finding the derivative of the functions sin x and cos x depend on
two limits (that are used elsewhere in calculus as well):
Two trigonometric limits.
sin θ
= 1,
θ→0 θ
(a) lim
(b) lim
θ→0
cos θ − 1
= 0.
θ
The verification we give of the first formula is based on the pictured wedge of
the unit circle:
The segment tagged with sin θ has this length by the definition of sin θ as the
y-coordinate of the point P on the unit circle corresponding to the angle θ.
The line segment tagged with tan θ has this length since looking at the large
triangle, we have tan θ = o/a = o. The arc tagged with θ has this length by the
definition of radian measure of an angle. The diagram reveals the inequalities
sin θ < θ < tan θ.
The first inequality implies (sin θ)/θ < 1; the second says θ < (sin θ)/(cos θ),
implying that cos θ < (sin θ)/θ. Therefore,
cos θ <
sin θ
< 1.
θ
As θ goes to 0, both ends go to 1 forcing the middle expression to go to 1 as
well (by the squeeze theorem). This establishes (a).
1
19
DERIVATIVE OF SINE AND COSINE
2
For the second formula, we use a method that is similar to our rationalization
method, as well as the main trigonometric identity, and finally the first formula:
lim
θ→0
cos θ − 1
cos θ − 1 cos θ + 1
= lim
·
θ→0
θ
θ
cos θ + 1
cos2 θ − 1
= lim
θ→0 θ(cos θ + 1)
− sin2 θ
θ→0 θ(cos θ + 1)
sin θ
sin θ
= − lim
·
θ→0 θ
cos θ + 1
sin θ
sin θ
= − lim
· lim
θ→0 θ
θ→0 cos θ + 1
= −1 · 0 = 0.
= lim
This completes the verification of the two trigonometric limit formulas.
19.2
Statement
Derivative of sine and cosine.
(a)
d
[sin x] = cos x,
dx
(b)
d
[cos x] = − sin x.
dx
We verify only the first of these derivative formulas. With f (x) = sin x, the
formula says f 0 (x) = cos x:
f (x + h) − f (x)
h→0
h
sin(x + h) − sin x
= lim
h→0
h
(sin x cos h + cos x sin h) − sin x
= lim
h→0
h
cos h − 1
sin h
= lim sin x ·
+ cos x ·
h→0
h
h
cos h − 1
sin h
= sin x · lim
+ cos x · lim
h→0
h→0 h
h
= sin x · 0 + cos x · 1
f 0 (x) = lim
= cos x.
((4) of 4.3)
((b) and (a))
19
DERIVATIVE OF SINE AND COSINE
3
The formula says that f (x) = sin x has general slope function f 0 (x) = cos x, so
the height of the graph of the cosine function at x should be the slope of the
graph of the sine function at x. The following figures show this relationship for
x a multiple of π/2.
19.3
19.3.1
Examples
Example
Find the derivative of f (x) = 3 cos x + 5 sin x.
Solution We use the rules of this section after first applying the sum rule and
the constant multiple rule:
d
[3 cos x + 5 sin x]
dx
d
d
= 3 [cos x] + 5 [sin x]
dx
dx
= 3(− sin x) + 5(cos x)
f 0 (x) =
= −3 sin x + 5 cos x.
19.3.2 Example
Find all points on the graph of f (x) = sin x at which the
tangent line has slope 1/2.
Solution The general slope function for this function is its derivative, which
is f 0 (x) = cos x. We get the x-coordinates of the desired points by solving
19
DERIVATIVE OF SINE AND COSINE
4
f 0 (x) = 1/2, that is, cos x = 1/2. Picturing the unit circle and using the 30-6090 triangle we find that x = π/3 (60◦ ) in the first quadrant yields a cosine of
1/2, as does x = −π/3 in the fourth quadrant. A multiple of 2π added to these
angles produces the same cosine.
Therefore, the x-coordinates of the desired points are
π/3 + 2πn,
−π/3 + 2πn
(n any integer).
√
The y-coordinate corresponding to x = π/3 is f (π/3) = sin(π/3) = 3/2 and
this is also the y-coordinate corresponding to x = π/3+2πn
√ for any n. Similarly,
the y-coordinate corresponding to x = −π/3+2πn is − 3/2. The desired points
are
√
√
(n any integer).
(π/3 + 2πn, 3/2), (−π/3 + 2πn, − 3/2)
19 – Exercises
19 – 1
Find all points on the graph of f (x) = cos x at which the tangent line has slope
√
3/2.