# $\sinh$ and $\cosh$

So I recently found out that

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

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

and that

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

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

Since

$\frac {a}{b} + \frac {b}{a} = \frac {a^2 + b^2}{ab}$

$\frac {a}{b} - \frac {b}{a} = \frac {a^2 - b^2}{ab}$

and

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

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

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

$\cosh {(x)} = \frac {e^{2x} + 1}{2e^x}$

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

$\sin {(a + bi)} = i\frac {e^{2(b - ai)} - 1}{2e^{b - ai}}$

$\cos {(a + bi)} = \frac {e^{2(b - ai)} + 1}{2e^{b - ai}}$

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

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

Let $\cos {\theta} + i\sin {\theta} =$ $cis$ $\theta$. So

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

Next we substitute this new equation in to get

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

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

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

Substituting in the two equations gives us

$cis(a + bi) = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)} + i(i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)})$

That looks unfriendly, let's simplify it

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

(Go through the steps, it works)

This can be simplified even further

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

P.s. $(\cos {\theta} + i\sin {\theta})^n = \cos {\theta n} + i\sin{\theta n}$

Note by Jack Rawlin
6 years, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Very nice note. Learned a lot!

- 6 years, 7 months ago

\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}

- 6 years, 7 months ago

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

- 6 years, 7 months ago

Lol..Nice note!!

- 6 years, 7 months ago

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?

- 6 years, 6 months ago