sinh\sinh and cosh\cosh

So I recently found out that

sin(a+bi)=isinh(bai)\sin {(a + bi)} = i\sinh {(b - ai)}

cos(a+bi)=cosh(bai)\cos {(a + bi)} = \cosh {(b - ai)}

and that

sinh(x)=exex2\sinh {(x)} = \frac {e^x - e^{-x}}{2}

cosh(x)=ex+ex2\cosh {(x)} = \frac {e^x + e^{-x}}{2}


ab+ba=a2+b2ab\frac {a}{b} + \frac {b}{a} = \frac {a^2 + b^2}{ab}

abba=a2b2ab\frac {a}{b} - \frac {b}{a} = \frac {a^2 - b^2}{ab}


(ab)x=(ba)x(\frac {a}{b})^{-x} = (\frac {b}{a})^x

Re-arranging the equations for sinh(x)\sinh{(x)} and cosh(x)\cosh {(x)} gives us

sinh(x)=e2x12ex\sinh {(x)} = \frac {e^{2x} - 1}{2e^x}

cosh(x)=e2x+12ex\cosh {(x)} = \frac {e^{2x} + 1}{2e^x}

Substituting these into sin(a+bi)\sin {(a + bi)} and cos(a+bi)\cos {(a + bi)} when x=baix = b - ai gives us

sin(a+bi)=ie2(bai)12ebai\sin {(a + bi)} = i\frac {e^{2(b - ai)} - 1}{2e^{b - ai}}

cos(a+bi)=e2(bai)+12ebai\cos {(a + bi)} = \frac {e^{2(b - ai)} + 1}{2e^{b - ai}}

I'm going to try and expand and substitute by using

ea+bi=ea(cos(b)+isin(b))e^{a + bi} = e^a (\cos {(b)} + i\sin {(b)})

Let cosθ+isinθ=\cos {\theta} + i\sin {\theta} = ciscis θ\theta. So

ea+bi=eacis(b)e^{a + bi} = e^a \cdot cis (b)

Next we substitute this new equation in to get

sin(a+bi)=ie2bcis(2a)12ebcis(a)\sin {(a + bi)} = i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)}

cos(a+bi)=e2bcis(2a)+12ebcis(a)\cos {(a + bi)} = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)}

So what does cis(a+bi)cis(a + bi) equal

Substituting in the two equations gives us

cis(a+bi)=e2bcis(2a)+12ebcis(a)+i(ie2bcis(2a)12ebcis(a))cis(a + bi) = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)} + i(i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)})

That looks unfriendly, let's simplify it

cis(a+bi)=1ebcis(a)cis(a + bi) = \frac {1}{e^{b} \cdot cis(-a)}

(Go through the steps, it works)

This can be simplified even further

cis(a+bi)=cis(a)ebcis(a + bi) = \frac {cis(a)}{e^b}

P.s. (cosθ+isinθ)n=cosθn+isinθn(\cos {\theta} + i\sin {\theta})^n = \cos {\theta n} + i\sin{\theta n}

Note by Jack Rawlin
6 years, 2 months ago

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1 vote

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Although trigonometric functions can operate complex numbers, I think there is no complex angle but only real angles. Just like a doctor who can save lives, not all lives being saved are human being. Do you agree with me?

Lu Chee Ket - 6 years ago

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cis(a+bi)=ei(a+bi)=eiab=eiaeb=cis(a)eb\begin{aligned} \text{cis}(a+bi) &= e^{i(a+bi)}\\ &= e^{ia-b}\\ &= \dfrac{e^{ia}}{e^b}\\ &= \dfrac{\text{cis}(a)}{e^b} \end{aligned}

Daniel Liu - 6 years, 2 months ago

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I did consider that but I wanted to make it as complex as possible so it would be fun to work out.

Jack Rawlin - 6 years, 2 months ago

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Lol..Nice note!!

B.S.Bharath Sai Guhan - 6 years, 2 months ago

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Very nice note. Learned a lot!

Trevor Arashiro - 6 years, 2 months ago

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