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I thought I had defined what cos(ai) and sin(ai) was earlier today when I did the following:

$e^{(vi)} = \cos(v) + i\sin(v)$
If we let $v = ai$, where a is real, we get:
$e^{(aii)} = \cos(ai) + i\sin(ai) = e^{(-a)}$
Since $e^{(-a)}$ is a real number, the $i\sin(ai)$ must be 0, and therefor $\cos(ai)$ must be $e^{(-a)}$.
So I have concluded that $\cos(ai) = e^{(-a)}$ and $\sin(ai) = 0$

I am pretty sure this is wrong since I have seen different answers online, but I would like to know what I did wrong.

Thanks

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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, on this site we use MathJaX to format our maths. [Here](http://meta.math.stackexchange.com/q/5020/145141) you can find a basic tutorial. – 2017-01-17
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    Mh, you didn't consider that $\cos ai$ and $\sin ai$ might be complex numbers... (which they are) – 2017-01-17

5 Answers 5

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It's worth knowing these identities:

$$\cos it = \cosh t$$ $$\sin it = i \sinh t$$

I presume you are familiar with the hyperbolic functions $\cosh$ and $\sinh$. These are usually defined as $\cosh t \equiv \frac12(e^t+e^{-t})$, and $\sinh t \equiv \frac12(e^t-e^{-t})$.

These functions have properties which parallel those of the circular functions.

So, it is indeed true that $\cos it$ is a real number when $t$ is real.

Addendum: An easy way to remember this is that it is completely analogous to how the sign is handled with circular functions: $\cos(-t)=\cos t$ and $\sin(-t)=-\sin t$.

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    :D reminds me of when I told my friends that exponential functions and trig functions were related via differential equations, which is one way of looking at it. – 2017-01-17
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    So $\sin(ia)$ is pure imaginary when $a$ is real, and $i\sin(ia)$ is real (but not zero) when $a$ is real and not zero. – 2017-01-17
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    @DavidK yes, that is correct. – 2017-01-17
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This is wrong. In fact, both cosine and sine could be complex, but the imaginary bits cancel out.

You can find what it comes out to on Wikipedia.

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    But what did I do wrong? – 2017-01-17
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    @user3183005 you assumed that $\cos(ai)$ and $\sin(ai)$ were real, but that is not the case :-) – 2017-01-17
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    Oh, it all makes sense now :P I have never thought of something like that before. Thanks. – 2017-01-17
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    @user3183005 :-) your welcome then. And welcome to the site. – 2017-01-17
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If $z$ is a complex number and $z=u+iv$, it does not follow that $u=Re(z)$ and $v= Im(z)$ !

Example: $2+2i$ can be written as

$$(1+i)+i(1-i).$$

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    I'm pretty sure that this is exactly the definition of $Re(z)$ and $Im(z)$ when $u,v\in \mathbb{R}$. – 2017-01-17
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    This only serves to be correct if $u,v$ are real, which is not stated, and writing it like this is rather confusing.... – 2017-01-17
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    Yes, when $u,v \in \mathbb R$ ! But in $e^{-a}=e^{ai^2} = \cos(ai) + i\sin(ai) $, we do not have that $\cos(ai), \sin(ai) \in \mathbb R$. So, why the downvote ? – 2017-01-17
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    I think this answer is only confusing, as you are mixing terminology without stating what things are. – 2017-01-17
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    It is usually taken that when **constants** are written in the form $u+vi$, then $u,v$ are implicitly assumed to be real, but that is not the case with functions. That is why I downvoted. – 2017-01-17
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    The OP assumed that $cos(ai)$ and $sin(ai)$ were real, which is not the case. That is what I wrote, so why the downvote ? – 2017-01-17
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    Because it is always assumed that if you write $z=a+bi$, then $a,b$ are real, which is in contradiction to your answer and you do nothing to clarify this. – 2017-01-17
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    What ??? If I wrote $2i=i+i$, then it is always assumed that $ i \in \mathbb R $ ???? I can't believe.... – 2017-01-17
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    I do agree with what you are saying, but I'm merely pointing out that you should explicitly state the complex nature of $u,v$. (And its not like you should worry about my down vote...) – 2017-01-17
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    I am not worried about your down vote. But your downvote is not just justified ! – 2017-01-17
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    In all my experience, it is taken that if $a,b$ are not stated to be complex, are constants, *and* $z=a+bi$, then $a=Re(z)$ and $b=Im(z)$. I have never seen this to be not the case **for constants** unless **explicitly stated**, which is why I disagree with this answer. – 2017-01-17
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It is an interesting exercise! The problem you're having though, is that you don't know what $\sin(ai)$ and $\cos(ai)$ are. So instead, you'll may want to try and verify that their outcomes are not complex.

Hint: Try writing the taylor expansion of $e^{-a},\cos{(ai)}$ and $\sin{(ai)}$

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The thing is $\sin{(ai)}$ is complex and not real, so you cannot say $\cos{(ai)}$ and $\sin{(ai)}$ both equal to zero.

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    :-) much better. – 2017-01-17
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    Agreed. Do remember to use MathJax though. (It's just Latex language) – 2017-01-17