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Approximation using Differentials - Homework
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Derivatives of sin(θ) and cos(θ) Using Differentials - a knol by Stephen Kent Stephenson
3/19/11 9:57 AM
There are no pending text suggestions from your readers
Derivatives of sin(!) and cos(!) Using Differentials
and Defining Angles in Radians Simplifies Trig Derivatives
Consider these figures
(where s and !s are
arc lengths):
Large !"
Small !"
As apparent in the figures, if !" # 0, then dx # !x, dy = !y, and ds # !s.
Segment BG is perpendicular to segment AB, by construction.
!BHG ~ !ACB by Angle-Angle, so we can write these proportions:
dy / ds = r cos(") / r and |dx| / ds = r sin(") / r ; or
dy = ds cos(") and dx = $ ds sin(").
Since y = r sin(") and x = r cos("),
then dy = d(r sin(")) = r d(sin(")) = ds cos("),
and dx = d(r cos(")) = r d(cos(")) = $ ds sin(").
So d(sin(")) = (ds / r) cos(") and d(cos(")) = $ (ds / r) sin(").
We know s = ("° / 360°) 2%r = (% r / 180°) "°, so ds = (% r / 180°) d("°).
Then d(sin("°)) = (% / 180°) cos("°) d("°),
and d(cos("°)) = $ (% / 180°) sin("°) d("°).
But if s / r = (% / 180°) "° is defined as " radians, then s = r " and ds = r d".
So d(sin(")) = cos(") d" and d(cos(")) = $ sin(") d";
or d(sin("))/d" = cos(") and d(cos("))/d" = $ sin(").
Hope you find this useful, -SKStephenson, [email protected],
© 2009-2011, All Rights Reserved.
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