prove the derivative of sin(x)=cos(x)?

Question

i'll give the answer after some people try

Answer 1

d/dx(sinx) = cosx

we know lim{h→0} (sin(x) / x ) = 1


d/dx(sinx) = lim{h→0} (sin(x+h) - sin(x))/h
= lim{h→0} (sin(x)cos(h) + cos(x)sin(h) - sin(x))/h
= lim{h→0} (sin(x)cos(h) /h + cos(x)sin(h) /h - sin(x)/h)
= lim{h→0} (cos(x)sin(h) /h )
= lim{h→0} (cos(x)) * lim{h→0} ( sin(h) /h )
= cos(x) * 1 [lim{h→0} (sin(x) / x ) = 1]
= cos(x)

Answer 2

d/dx sin x

The derivative formula,
= lim (∆x -> 0) [sin (x + ∆x) - sin x]/∆x

Use the expansion sin (a + b) = sin a cos b + cos a sin b
= lim (∆x -> 0) (sin x cos ∆x + cos x sin ∆x - sin x)/∆x

We rearrange terms
= lim (∆x -> 0) (sin x cos ∆x - sin x + cos x sin ∆x)/∆x

We factor out sin x fro the first 2 terms, and distribute the denominator ∆x,
= lim (∆x -> 0) [sin x (cos ∆x - 1)/∆x + cos x sin ∆x/∆x]

Then, distribute the limit sign
= lim (∆x -> 0) sin x (cos ∆x - 1)/∆x + lim (∆x -> 0) cos x sin ∆x/∆x

We remove the sin x and cos x from the limit, since it is unneeded,
= sin x lim (∆x -> 0) (cos ∆x - 1)/∆x + cos x lim (∆x -> 0) sin ∆x/∆x

Since lim (∆x -> 0) (cos ∆x - 1)/∆x = 0, and lim (∆x -> 0) sin ∆x/∆x = 1, then
= sin x (0) + cos x (1)

Thus,
= 0 + cos x

Therefore,
d/dx sin x = cos x. QED

^_^

Answer 3

wite down the tailor expansion of son x.
differentitiate each term
compare the result with the expansion of thecos x

Answer 4

limit (sin(x+t)-sinx)/t = limit 2cos((x+t+x)/2)sin((x+t-x)/2)/t = cos (x+t/2) * sin(t/2) / (t/2) when t -> 0 sint/t -> 1 so the limit = cosx

Answer 5

sin(X)is not equal to cos(x)
sin(x) is only equal to cos(90-x)
if sin(x)=cos(x),,,,,then,
sin(X)-cos(x)=0(this is not true)
virtually sin(X) is equal to cos(90-x)