Given the above - what about some Trig relationships?



We know that

\(e^{ix} = cos(x) +i\,sin(x)\)

So

\(e^{-ix} = e^{i(-x)} = cos(-x) +i\,sin(-x) = cos(x) -i\,sin(x)\)

Multiply these 2 equations to get:

\(e^{ix}e^{-ix} = e^0 = 1 = (cos(x) +i\,sin(x))(cos(x) - i\,sin(x)) = cos^2(x)+sin^2(x)\)

Giving us:

$$sin^2(x) + cos^2(x) = 1$$


Next,


\(e^{ia} = cos(a)+i\,sin(a)$ and $e^{ib} = cos(b)+i\,sin(b)\)

Multiplying these we get:

\(e^{i(a+b)} = cos(a)cos(b) -sin(a)sin(b) + i(cos(a)sin(b) + sin(a)cos(b))\)

But also, we know:

\(e^{i(a+b)} = cos(a+b) + i\,sin(a+b)\)

Now real and imaginary parts must be equal so:

$$cos(a+b) = cos(a)cos(b) - sin(a)sin(b)$$

and

$$sin(a+b) = cos(a)sin(b) + sin(a)cos(b)$$

Finally, setting $a=b=x$:

$$cos(2x) = cos^2(x)-sin^2(x)$$

and

$$sin(2x) = 2\,sin(x)cos(x)$$


Now, isn't that easier than remembering them?