# Derivative of sinx

1. Apr 13, 2007

### nathaniell

Hey guys,

I know that the derivative of sinx is cosx but how do we get to that answer

can someone show me the steps followed

2. Apr 13, 2007

### Data

Well, that depends on the definition you use for $\sin x$ and $\cos x$. From the geometric definition that you've probably seen, the proof is not trivial; This website seems to do all of it at some point or another: http://people.hofstra.edu/faculty/Stefan_Waner/trig/trigintro.html [Broken]

Typically, though, the sine and cosine functions are defined either as the solutions to a DE (or a pair of coupled DEs) or as their Taylor expansions. Consider the pair of DEs

$$y^\prime(x) = -z(x), z^\prime(x) = y(x)$$

with initial conditions $z(0) = 0, \ y(0) = 1$. We can define $\sin(x)$ to be the unique solution for $z(x)$ and $\cos(x)$ to be the unique solution for $y(x)$. In that case, the fact that $\frac{d}{dx}\sin x = \cos x$ is by definition!

A couple of other common definitions for sine and cosine (on the reals) follow. Proving the derivative identities using either of these definitions only requires a few basic results:

1) We define $\sin$ and $\cos$ to be the unique solutions $y_1$ and $y_2$ to

$$y'' = -y[/itex] with initial conditions $y(0)=0, \ y'(0) = 1$ and $y(0)=1, \ y'(0) = 0$, respectively. OR 2) For all $x \in \mathbb{R}$, we define [tex]\sin x = \sum_{n=0}^\infty \frac{x^{2n+1}(-1)^n}{(2n+1)!}$$

and

$$\cos x = \sum_{n=0}^\infty \frac{x^{2n}(-1)^n}{(2n)!}$$

(you need to prove convergence for these series, but that's not difficult).

Of course, starting from any of these definitions, it takes some work to prove that they're equivalent to the geometric one.

Last edited by a moderator: May 2, 2017
3. Apr 13, 2007

### Rainbow

Just apply the first principle and you will get it.

4. Apr 13, 2007

### quasar987

Like Rainbow said, you can use the definition of the derivative and the greeks' definitions of cos and sin and get the result simply by applying the trigonometric identity sin(x+h)-sin(x)=2*sin(h/2)*cos(x+h/2). The fact that sin(h/2)/h --> 1 together with the continuity of cos gives the answer.

5. Apr 13, 2007

### symbolipoint

Check the old Howard Anton book on Calculus from the 1970's. That's the best proof for derivative of sin(x) that I ever found. Clever Algebra tricks, but understandable.

6. Apr 13, 2007

### Data

Rather, $\sin(h/2)/h \rightarrow 1/2$. The nontrivial part that I mentioned is mostly deriving a bunch of trig identities, as well as that limit; if you already have those then the proof is straightforward.

The problem is that the way that you'd usually approach the limit

$$\lim_{h \rightarrow 0} \frac{\sin x}{x}$$

is using l'Hopital's rule. But that's circular in this case, because it requires you to already know the derivative of sin. You have to derive that limit by geometric arguments (and the website I linked to does so).

Last edited: Apr 13, 2007
7. Apr 13, 2007

### quasar987

Yes, yes, apologies.

8. Apr 14, 2007

### mathwonk

the usual proof in all books is the one from courant calculus. it derives the limit 1 = sin(x)/x as x-->0, by an area argument and using the squeeze law.

you draw a sector of the unit circle with angle t, then drop a perpendicular from the point on the circle to the x axis, to form a right triangle of hypotenuse 1, and base length cos(t). then one extends that hypotenuse to form a slightly larger similar right triangle of base length 1 along the x axis.

Then one has three nested figures, two right triangles and a circular sector in between, all with acute angle t.

comparing the three areas gives the inequality

sin(t)cos(t) < t < tan(t). dividing by sin(t) gives

cos(t) < t/sin(t) < 1/cos(t).

now as t-->0 we get all 3 limits equal to 1, hence the reciprocal limit sin(t)/t is also 1.