to prove that the first derivative of sine is cosine?

Question

Im this far

2cos(x+x0/2)sin(x-x0/2) all over (x-x0)

Answer 1

I'll use h instead of x0. The difference quotient is

[ sin(x+h) - sin(x)]/h =
[ sin(x)cos(h) + cos(x)sin(h) - sin(x)]/h =
[sin(x)cos(h) - sin(x)]/h + [cos(x)sin(h)]/h =
[sin(x) (cos(h) - 1)/h] + [cos(x) (sin(h)/h)]

if you've previously proved that
lim(h→0) [cos(h)-1]/h = 0 and lim(h→0) [sin(h)]/h = 1, then the last expression above simplifies (as h→0) to cos(x). If not, you have to prove those first.

Answer 2

f `(x) = lim h->0 (sin(x + h) - sin x) / h -- h not = 0
= lim h->0 2.cos [(1/2).(2x+h)].(sin h/2 / h)
= lim h->0 cos(x + h/2). (sin h2 /h/2)
= limh->0 cos(x + h/2).limh->0(sin h/2 /h/2)
= cos x . 1
= cos x

Answer 3

y= sin x
y + dy = sin(x +dx)
dy = sin(x+dx) -sinu
dy/dx = [sin(x +dx) -sin x]/dx <-- Eq 1
Use identity sinA-sinB = 2cos .5(A+B)sin.5(A-B) and transform Eq 1 to:
dy/dx = [2cos(x+.5dx)sin(.5dx)]/dx
dy/dx = cos(x+.5dx)*sin(.5dx)/.5dx
Now since
lim cos(x +.5dx) = cos x
dx --> 0, and
lim [sin(.5dx)]/.5dx =1
dx --> 0,
we have d/dx(sinx) = cos x