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Lesson14: Derivatives of Trigonometric Functions
1. Sections 3.4
Derivatives of Trigonometric Functions
Math S-1ab
Calculus I and II
October 26, 2007
Announcements
Midterm is done. Average=84.2, Median=85
2. Two important trigonometric limits
Theorem
The following two limits hold:
sin θ
lim =1
θ→0 θ
cos θ − 1
lim =0
θ
θ→0
3. Proof of the Sine Limit
Proof.
Notice
sin θ ≤ θ ≤ tan θ
Divide by sin θ:
θ 1
1≤ ≤
sin θ cos θ
Take reciprocals:
sin θ θ tan θ
θ sin θ
1≥ ≥ cos θ
θ
cos θ 1
As θ → 0, the left and right
sides tend to 1. So, then,
must the middle
expression.
4. Now
1 − cos2 θ
1 − cos θ 1 − cos θ 1 + cos θ
·
= =
θ θ 1 + cos θ θ(1 + cos θ)
2
sin θ sin θ θ
·
= =
θ(1 + cos θ) θ 1 + cos θ
So
1 − cos θ sin θ θ
·
lim = lim lim
θ θ→0 θ θ→0 1 + cos θ
θ→0
= 1 · 0 = 0.
5. Derivatives of Sine and Cosine
Theorem
d
sin x = cos x.
dx
Proof.
From the definition:
6. Derivatives of Sine and Cosine
Theorem
d
sin x = cos x.
dx
Proof.
From the definition:
sin(x + h) − sin x
d
sin x = lim
dx h
h→0
7. Derivatives of Sine and Cosine
Theorem
d
sin x = cos x.
dx
Proof.
From the definition:
sin(x + h) − sin x
d
sin x = lim
dx h
h→0
(sin x cos h + cos x sin h) − sin x
= lim
h
h→0
8. Derivatives of Sine and Cosine
Theorem
d
sin x = cos x.
dx
Proof.
From the definition:
sin(x + h) − sin x
d
sin x = lim
dx h
h→0
(sin x cos h + cos x sin h) − sin x
= lim
h
h→0
cos h − 1 sin h
= sin x · lim + cos x · lim
h h→0 h
h→0
9. Derivatives of Sine and Cosine
Theorem
d
sin x = cos x.
dx
Proof.
From the definition:
sin(x + h) − sin x
d
sin x = lim
dx h
h→0
(sin x cos h + cos x sin h) − sin x
= lim
h
h→0
cos h − 1 sin h
= sin x · lim + cos x · lim
h h→0 h
h→0
= sin x · 0 + cos x · 1 = cos x