Sunday, February 13, 2011

The slope of the sine curve

Just to finish with the subject from last time (here), let's show that the slope of the sine curve at x is cos(x) and for the cosine curve at x it is -sin(x).

We start from x, and then move a little bit h. Using the rule for sum of sines (here):


sin(x + h) = sin(x) cos(h) + cos(x) sin(h)

f(x + h) - f(x) =
= sin(x) cos(h) + cos(x) sin(h) - sin(x)

d/dx sin(x) = lim(h -> 0) [f(x + h) - f(x)]/h
= 1/h [sin(x) cos(h) + cos(x) sin(h) - sin(x)]
= -(1/h) sin(x) (1 - cos(h)) + (1/h) cos(x) sin(h)


The first term is -sin(x) times (1/h)(1 - cos(h)); last time we showed that (1/h)(1 - cos(h)) equals zero in the limit as h -> 0.

The second term is cos(x) times (1/h) sin(h); we showed that (1/h) sin(h) approaches 1 as h -> 0. Thus,


d/dx sin(x) = cos(x)


Since sine and cosine are periodic with cosine "ahead"




sin(x + π/2) =  cos(x)
cos(x + π/2) = -sin(x)


If


y = cos(x) =
= sin(x + π/2)


Let


u = x + π/2


Then


d/dx cos(x) = 
= d/dx sin(x + π/2)
= d/du sin(u) d/dx (x + π/2)
= d/du sin(u)
= cos(u)
= cos(x + π/2)
= -sin(x)

d/dx cos(x) = -sin(x)

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