An interesting approach to Trigonometric Identities , the derivative of Trigonometric function and Euler Formula

Part 1 Trigonometric Identities

When I was a high school student, I wouldn't say I liked the approach to Trigonometric Identities told in school

After I learned complex number, I observed a new approach.

Observe that for $a,ib$$a,ib$, $i$$i$ act on $a,ib$$a,ib$ means rotate 90 degrees counterclockwise.

So according to distributive law, $i$$i$ act on $a+bi$$a+bi$ means rotate 90 degrees counterclockwise.

So we can write for $i^k(\cos (x)+i\sin (x))=\cos (x+\frac{k\pi }{2})+i\sin (x+\frac{k\pi }{2})$$i^k(\cos(x)+i\sin(x))=\cos(x+\frac{k\pi}{2})+i\sin(x+\frac{k\pi}{2})$

Part 2 uses i to get the derivative of the Trigonometric function.

Observe that $f(x)=\cos (x)+i\sin (x)$$f(x)=\cos(x)+i\sin(x)$ is isomorphic to $S^1$$S^1$

And the derivative is on the tangent line, which is orthogonal to $f(x)$$f(x)$

So we get the $f^{\prime }(x)=rif(x)$$f'(x)=ri f(x)$, and $|f^{\prime }(x)|\cdot 2\pi =2\pi R$$|f'(x)|\cdot2\pi=2\pi R$

Thus $r=1$$r=1$

Thus $f^{\prime }(x)=if(x)=-\sin (x)+i\cos (x)$$f'(x)=i f(x)=-\sin(x)+i\cos(x)$

We get the derivative by the complex number(without limit)

Part 3

Then we can solve the ODE

$f^{\prime }(x)=if(x)$$f’(x)=if(x)$

We can get $e^{ix}=\cos (x)+i\sin (x)$$e^{ix}=\cos(x)+i\sin(x)$

Let $x=\pi$$x=\pi$, we get $e^{i\pi }+1=0$$e^{i\pi}+1=0$