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)