The Differentiation of Sine and Cosine Jonathan M. Sterling 14 June 2008 Abstract Where f (θ) = sin θ and g(θ) = cos θ, their derivatives are defined df by dθ = cos θ and dg = − sin θ respectively. The limits lim sinh h and dθ h→0
1−cos h h h→0
lim
were evaluated using the Squeeze Theorem; their values are 1
and 0 respectively.
Contents 1 The differentiation of the Sine function 1.1 Evaluating the limit lim sinh h . . . . . . . . . . . . . . . . . h→0
1−cos h h
1.2
Evaluating the limit lim
. . . . . . . . . . . . . . . .
2
1.3
Final Steps . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
h→0
2 The differentiation of the Cosine function
1
1 2
3
The differentiation of the Sine function
Given the definition of the derivative of f (x) with respect to x, f (x + h) − f (x) df = lim h→0 dx h the function f (θ) = sin(θ) can be differentiated. sin(θ + h) − sin θ df = lim h→0 dθ h sin θ cos h + cos θ sin h − sin θ = lim h→0 h Next, the numerator is factored. · · · = lim
h→0
− sin θ(1 − cos h) + cos θ sin h h
The fraction is now split into two separate limits. − sin θ(1 − cos h) cos θ sin h + lim h→0 h h df 1 − cos h sin h = − sin θ lim + cos θ lim h→0 h→0 dθ h h
. . . = lim
h→0
1
(1)
The limits lim
h→0
1.1
1−cos h h
and lim
h→0
sin h h
must now be evaluated. sin h h→0 h
Evaluating the limit lim
Let h be an arc on the unit circle, where 0 < h < inequality is true: sin h < h < tan h
π . 2
The following
The following is equivalent: sin h h tan h < < sin h sin h sin h which is simplified to 1 h < sin h cos h That inequality can also be expressed as 1<
1<
sin h < cos h h
The Squeeze Theorem[1] holds that f , g and h are functions in an interval around point a, and h(x) ≤ f (x) ≤ g(x). If lim h(x) = lim g(x) = a→0
a→0
L, then lim f (x) = L. a→0
This fits the Squeeze Theorem, because lim 1 = lim cos h = 1: h→0
lim 1 = 1
h→0
lim cos h = cos 0
h→0
h→0
lim cos h = 1
h→0
Given that lim 1 = lim cos h = 1, the following must also be true: h→0
h→0
lim
h→0
1.2
sin h =1 h
Evaluating the limit lim
h→0
The limit lim
h→0
1−cos h h
lim
h→0
1−cos h h
is evaluated as follows:
1 − cos h = lim h→0 h
1 − cos h h
1 + cos h 1 + cos h
= lim
1 − cos2 h h(1 + cos h)
= lim
sin2 h + cos2 h − cos2 h h(1 + cos h)
h→0
h→0
sin2 h h→0 h(1 + cos h) sin h sin h = lim h→0 h 1 + cos h sin h sin h = lim lim h→0 h→0 1 + cos h h = lim
2
Based on 1.1, that is simplified to . . . = (1)
lim
h→0
sin h 1 + cos h
sin 0 = 1 + cos 0 0 = 1+1 lim
h→0
1.3
1 − cos h =0 h
Final Steps
The last step before the limits were evaluated in 1.1 and 1.2 was df 1 − cos h sin h = − sin θ lim + cos θ lim h→0 h→0 dθ h h Given the values of those limits, this is simplified to df = − sin θ(0) + cos θ(1) dθ df = cos θ dθ
2 The differentiation of the Cosine function df Given the definition of dx as the derivative of f (x) in Equation 1, the function g(θ) = cos θ can be differentiated.
cos(θ + h) − cos θ dg = lim h→0 dθ h cos θ cos h − sin θ sin h − cos θ = lim h→0 h − cos θ(1 − cos h) − sin θ sin h = lim h→0 h − cos θ(1 − cos h) sin θ sin h = lim − lim h→0 h→0 h h 1 − cos h sin h = − cos θ lim − sin θ lim h→0 h→0 h h Given the values of the two limits found in 1.1 and 1.2, that can be simplified. . . . = − cos θ(0) − sin θ(1) dg = − sin θ dθ
3
References [1] Joseph M. Ling. Examples on limits of functions: The squeeze theorem. http://math.ucalgary.ca/~ling/calculus/examples/ squeeze.pdf, 2001.
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