## 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)`