# Prove that the derivative of sine is cosine

In an informal exam tonight, my professor asked me to demonstrate that for $$f(x)=\sin(x),\, f'(x)=\cos(x)$$ using the definition of the derivative, $$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$.

Well, plugging things in, we get $$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin(x)}{h}$$ $$=\lim_{h\rightarrow0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}$$$$=\lim_{h\rightarrow0}\frac{\sin(x)\left(\cos(h)-1\right)+\cos(x)\sin(h)}{h}$$

And here I managed to stump him. In order to prove that this equals $$\cos(x)$$, we need to demonstrate that $$\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}=0$$ and that $$\lim_{h\rightarrow0}\frac{\sin(h)}{h}=1$$. You can't simply plug in $$h=0$$ because that would lead to an indeterminate form. L'Hôpital's Rule does indeed get you the required limits, but that requires the very derivative we're trying to prove.

Is there a way, using the definition of the derivative, to actually prove that the derivative of sine is cosine, without a proof by construction or a proof by contradiction?

• In case anyone's wondering, I managed to satisfy my professor by using Taylor Series, glossing over the fact that those equalities can only be proven to hold using the derivatives we're trying to find... – DonielF Jan 28 '20 at 2:54
• These two identities are often proved using a geometric argument and the squeeze theorem. – Michael Burr Jan 28 '20 at 2:55
• The $sinh/h=1$ can be shown through inscribed and circumscribed triangles as well as an accompanied circle sector all spanned by angle $h$. This is a non calculus approach and I am sure it can be found on this site somewhere. I am glad though that you realize that both Taylor series and Hospital Rule is NOT the correct way to prove these limits – imranfat Jan 28 '20 at 2:56
• I mean, if you don't want to use Taylor series, then what definition of $\sin(x)$ and $\cos(x)$ are you using. I personally treat those as the definition but there are other equivalent definitions – QC_QAOA Jan 28 '20 at 2:56
• The mentioned proof using a geometric argument and the squeeze theorem is here. – dxdydz Jan 28 '20 at 2:57

Without a proper definition, you can't prove anything. For instance, the limit

$$\lim_{t\to0}\frac{\sin t}t$$ has no meaning as long as the function $$\sin t$$ has not been defined, or at least some properties given. And it is not so easy to have a definition that avoids a circular argument.

The analytical definition relies on the complex exponential, via Euler's formula $$\sin t=\Im e^{it}$$ where the exponential is the entire function $$e^z=\sum_{n=0}^\infty\dfrac{z^n}{n!}$$ or equivalently a solution of the ODE $$f'(z)=f(z).$$

This obviously implies the derivative of the sine "by definition".

A slightly more geometric approach is by analytical geometry, from the equation of the unit circle, $$x^2+y^2=1,$$ giving by differentiation, $$y+x\frac{dx}{dy}=0.$$

Now if we accept the formula for the element of arc, $$ds^2=dx^2+dy^2,$$ we have

$$s=\int_0^s ds=\int_0^y\frac{dy}{\sqrt{1-y^2}}=f(y)$$ which defines a functional relation between $$s$$ and $$y$$. And by the derivative of the inverse function, let $$g(s):=f^{(-1)}(s)$$, we obtain

$$g'(s)=\sqrt{1-g^2(s)}.$$

A purely geometric argument is more difficult, as it requires to prove facts about lengths/areas of triangles and circular segments, and relate the axioms of geometry to calculus.