Download 1. Trigonometric Derivatives In this lecture we

1. Trigonometric Derivatives
In this lecture we review trigonometric functions and derive these differentiation formulas.
Theorem 1.
d
(sin x)
dx
d
(cos x)
dx
= − sin x ,
d
(tan x)
dx
d
(sec x)
dx
= cos x ,
= sec2 x ,
= sec x tan x .
1
2. Sines and Cosines
b
(cos t, sin t)
1
We start by recalling basic definitions of trigonometric functions. While
in analysis we define trigonometric functions as limits of polynomials. For
3
x5
− . . . . Here we use the geometric definitions
example, sin(x) = x − x6 + 120
based on the unit circle.
1
−π
−π
2
−1
1
−π
−π
2
−1
sin t
π
2
π
3π
2
2π
π
3π
2
2π
cos t
π
2
Take a real t-axis and wrap it around the unit circle such that the point
t = 0 on the t-axis corresponds to the point (1, 0) on the unit circle. Then
we define cos t and sin t to be the coordinates of the point on the unit circle
determined by the angle t.
Note that f (t) = sin t and of f (t) = cos t are defined for all t, not just
numbers in [0, 2π).
Also, since the coordinates of points on the unit circle vary between −1
and 1, we always have
−1 ≤ sin t ≤ 1,
and
−1 ≤ cos t ≤ 1.
1
t
1
−1
−1
2
Moreover, note that since the angles t and t + 2π determine the same
point on the unit circle, the sine and cosine functions are periodic:
sin(t + 2π) = sin t,
cos(t + 2π) = cos t,
However since the angles t and π + t point to opposite points in the unit
circle,
sin(t + π) = − sin t,
and
cos(t + π) = − cos t,
Finally, since the point (cos t, sin t) lies on the unit circle,
cos2 t + sin2 t = 1 .
3. Tangent function and other trigonometric functions
We define the tangent function to be slope of the line joining (0, 0) and
point (cos t, sin t). Thus
sin t
.
tan t =
cos t
Note that t 6= π2 + kπ for any integer k. Thus at t =
of f (t) = tan t has vertical asymptotes.
π
2
+ kπ, the graph
Other functions
cot t =
1
tan t
, sec t =
1
cos t
, csc t =
1
sin t
.
4. The slope of the sine function at zero
The slope of the tangent line to y = sin x at (0, 0) is, by definition of the
0
slope limh→0 sin h−sin
= limh→0 sinh h . From the graph of sine it seems that
h
this slope is 1. This actually is true and this is an important geometric
limit:
limh→0 sinh h = 1 .
which will be used in the derivation of derivatives of the sine and cosine
functions.
We prove limh→0+
sin h
h
= 1 above using elementary geometry. See how
4
3
2
1
1 2 3 4 5 6
−1
−2
−3
−4
−5
Hence, by multiplying the inequality by 2 and dividing by h,
tan h
sin h
<1<
.
h
h
The right inequality gives:
1<
sin h
cos h
h
.
Several branches of the
graph of tan t.
y=x
1
3
or
cos h <
sin h
.
h
Hence,
sin h
< 1.
h
As h → 0+ , cos h → 1, thus also By the Squeeze Theorem
cos h <
Finally, note that limh→0−
limk→0+ sink k = 1.
sin h
h
= limk→0+
sin(−k)
(−k)
sin h
h
→ 1.
= limk→0+
− sin k
(−k)
=
Combining the left and right limits we obtain: limh→0 sinh h = 1.
1
5. The slope of the cosine function at zero
y = cos x
−1
From the graph of y = cos x, we note the tangent line at the point (0, 1)
is horizontal.
That is
limh→0 cos hh−1 = 0 .
We prove this identity using the previous limit statement, the trigonometric identity: sin2 t +h cos2 t = 1, and
i properties of limits. See how...
limh→0
cos h−1
h
= limh→0
h
= limh→0
− sin h
h
·
cos h−1
h
− sin h
(cos h+1)
i
·
cos h+1
cos h+1
= limh→0
= limh→0
− sin h
h
(cos2 h−12 )
h(cosh +1)
· limh→0
− sin h
(cos h+1)
= limh→0
= −1 ·
0
1+1
− sin2 h
h(cos h+1)
= 0.
6. The derivative of sine is cosine: sin’=cos’
We derive this key formula
d
dx
sin x = cos x ,
which needs to memorized.
In the derivation we use the definition of the derivative of sine and then
we use the trigonometric limits that we just established above. See how?
sin x · cos h + sin h · cos x − sin x
h→0
h
= lim
sin x · (cos h − 1) + sin h · cos x
h→0
h
cos h − 1
sin h
= lim sin x ·
+ cos x ·
h→0
h
h
= lim
1
4
sin h
cos h − 1
+ cos x · lim
h→0 h
h→0
h
= sin x · lim
= sin x · 0 + cos x · 1 = cos x.
Remark. Note that in this derivation we used the sine addition formula
and the information that we had about the derivative at x = 0 of sin x and
cos x (these where the two trigonometric limit statements) to derive the
general formula for the derivative of sin x.
7. cos’=-sin’
Similarly, we show that
d
dx
cos x = − sin x .
This formula needs to be memorized. We leave the details out.
8. Other derivatives
In order to derive these formulas:
d
dx
tan x = sec2 x , and
d
dx
sec x = sec x tan x ,
we use the Quotient Law and
d
dx
sin x = cos x ,
d
dx
cos x = − sin x.