# Differentiating Sin and Cos

1. Aug 1, 2008

### mathfied

Hi. I believe this may have been addressed previously but I wanted to make sure since I dont think it was completed.

Hi I know that differentiating sin = cos , and differentiating cos = -sin. Time to prove it.

Q3:
Prove that: $\frac{d}{{dz}}\sin z = \cos z$

We know the McLaurin form for sin is:
$\sum\limits_{n = 0}^\infty {( - 1)^n } \frac{{z^{2n+1} }}{{(2n+1)!}} = z - z^3/3! + z^5/5! - z^7/7! +...$

differentiating sin term by term i get:
$1 -3z^2/3! + 5z^4/5! - 7z^6/7! +...$ =
$1 -z^2/2! + z^4/4! - z^6/6! +...$ =
cos z

would this be considered a "sufficient proof" . Even though I have only proved the first few terms, Im guessing we can just assume the rest is correct by the theory of induction.????

2. Aug 1, 2008

### t!m

Why not just differentiate the general term in the sum, rather than each individually?

3. Aug 1, 2008

### HallsofIvy

Staff Emeritus
Can you prove it by induction? Then you should do so rather than just saying it can be done.

But I don't see why you only do a few terms. The McLaurin series for sin(x) is, as you say,
$\sum\limits_{n = 0}^\infty {( - 1)^n } \frac{{z^{2n+1} }}{{(2n+1)!}}$
The general term is
[tex](-1)^n \frac{2^{2n+1}}{(2n+1)!}[/itex]

Differentiating that gives
[tex](-1)^n(2n+1)\frac{z^{2n}}{(2n+1)!}= (-1)^n \frac{z^{2n}}{(2n)!}[/itex]

That's all you need. (Except, of course, for some "technicalities" like showing that term by term differentiation of thes series does give the derivatative of the function.)

This is, of course, assuming that you have defined sin(x) and cos(x) by these series. If you defined them in some other way (using the unit circle is common) then you would need to know the derivatives in order to find the McLaurin series. And if you do define sin(x) and cos(x) in terms of series, you would still need to prove a number of other properties from that. Have you thought about how you might prove that sin(x) and cos(x) are periodic with period $2\pi$? Or that sin2(x)+ cos2(x)= 1?

Actually, those aren't that difficult. From the derivatives, as done here, you can show that sin(x) satisfies y"= -y, with y(0)= 0, y'(0)= 1, while cos(x) satisfies y"= -y, with y(0)= 1, y'(0)= 0, and then use properties of differential equations.