How to multiply sin by a constant in Euler's formula

Question

Euler's formula is $${\rm e}^{ix} = \cos x + {\rm i}\, \sin x $$

My question is, what if I have

$$ \cos x + {\rm i}\, R \sin x$$

How would the LHS of Euler's formula then look like?

Answer 1

The straightforward relation is

$$\cos{x} + Ri\sin{x}= \dfrac{e^{ix}+e^{-ix}}{2} + e^{\ln{R}} \cdot \dfrac{e^{ix}-e^{-ix}}{2}$$

$$= \dfrac{\left(1+e^{\ln{R}}\right)e^{ix} +\left(1-e^{\ln{R}}\right)e^{-ix}}{2}$$

$$= \dfrac{1+R}{2}e^{ix}+\dfrac{1-R}{2}e^{-ix}$$

Answer 2

For $z=a+bi$, its polar representation is $$z=\sqrt{a^2+b^2}e^{i*atan2(b/a)}$$

So, $$\cos x+iR\sin x=\sqrt{R^2+1}e^{i*atan2(R\sin x/\cos x)}$$

For some range of $x$, this equals a more beautiful expression $$\sqrt{R^2+1}e^{i*atan(R\tan x)}$$

Answer 3

First look at Euler's formula using the identities $$ \left. \begin{aligned} \cosh({\rm i}\,x) & = \cos(x) \\ \sinh({\rm i}\,x) & = {\rm i} \sin(x) \end{aligned} \right\} {\rm e}^{{\rm i} x} = \cosh({\rm i}\,x) + \sinh({\rm i}\,x) $$

and with $a = {\rm i}\, x$

$$ {\rm e}^a = \cosh(a) + \sinh(a) =\left( \tfrac{1}{2} {\rm e}^{a} + \tfrac{1}{2} {\rm e}^{-a}\right) + \left( \tfrac{1}{2} {\rm e}^{a}-\tfrac{1}{2} {\rm e}^{-a} \right) = {\rm e}^{a}\;\;\checkmark $$

Now do the same with your formula

$$ \cosh(a) +R\, \sinh(a) =\left( \tfrac{1}{2} {\rm e}^{a} + \tfrac{1}{2} {\rm e}^{-a}\right) +R \left( \tfrac{1}{2} {\rm e}^{a}-\tfrac{1}{2} {\rm e}^{-a} \right) = \left( \tfrac{1+R}{2} \right){\rm e}^{a} + \left( \tfrac{1-R}{2} \right) {\rm e}^{-a}$$

So you have

$$ \boxed{ \cosh({\rm i}x) +R\, \sinh({\rm i} x) = \left( \tfrac{1+R}{2} \right){\rm e}^{{\rm i} x} + \left( \tfrac{1-R}{2} \right) {\rm e}^{-{\rm i} x} } $$

Answer 4

So you have $cos( x) +iR\cdot sin( x)$

I saw that someone else turned it into polar exponential, so I thought I'd do it another way.

Here I have This formula $ \begin{array}{l} a\cdot sin( \theta ) +b\cdot cos( \theta ) =\sqrt{a^{2} +b^{2}} cos\left( \theta -arctan\left(\frac{a}{b}\right)\right)\\ a=Ri,\ b=1: \end{array}$

So now you get $\boxed{\sqrt{1-R^{2}} cos( \theta -arctan( iR))}$ This is not polar exponential$-$this is a different form not of rectangular or polar. I just thought that I would share this anyways.