Question

Why is d/dx[sin(x)] = cos(x)

Answer 1

Definition of a derivative is lim h --> 0 (f(x + h) - f(x))/h ....

Answer 2

i am sure there is a "proof" in your text
it requires knowing how to compute the difference quotient using the "addition angle formula" for sine, and also knowing two limits
\[\lim_{h\to 0}\frac{\sin(h)}{h}=1\] and also
\[\lim_{h\to 0}\frac{\cos(h)-1}{h}=1\]

Answer 3

if i recall correctly it's a lengthy proof

Answer 4

I need to use this I think... Lim h --> 0 (sin(x + h) - sin(x))/h

Answer 5

\[\sin(x+h)-\sin(x)=\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)\] is step one

Answer 6

There in comes the addition formula satellite

Answer 7

break in to two parts
\[\sin(x)\cos(h)-\sin(x)+\cos(x)\sin(h)=\sin(x)\left(\cos(h)-1\right) +\cos(x)\sin(h)\]

Answer 8

That 2nd limit (the one with cosine) is actually 0, not 1.

Answer 9

then divide both parts by \(h\) and get
\[\sin(x)\times \frac{(\cos(h)-1)}{h}\] and
\[\cos(x)\times\frac{\sin(h)}{h}\]

Answer 10

what @blockcolder said

Answer 11

Oh that simplifies things

Answer 12

so first term goes bye bye and second one gives \(\cos(x)\)

Answer 13

not really a very convincing argument however, there are better ones later

Answer 14

Yeah, in order to prove a basic identity you have to already know a much harder one (angle addition identity).. Thank you here's your medal...