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Is there are function to represent $x\mapsto\frac{\cos x}{x}$ as there is for $\sin x$, namely $\frac{\sin x}{x}=\text{sinc }x$?

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    At $0$, it is not indeterminate! – 2017-02-09
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    If there was one, it would rather be $$\cos x:=2\frac{1-\cos x}{x^2}.$$ – 2017-02-09
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    @YvesDaoust How so? $1-\cos x=2\sin^2(x/2)$ so that your $\cos x:=4\frac{\sin^2 x}{x^2}=4\text{ sinc$^2$ }x$, which doesn't make sense to me. – 2017-02-09
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    There is $\frac{1-\cos(x)}{x}$, this is common in literature. – 2017-02-09
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    @user3482534: the second $c$ of $\text{cosc}$ somehow disappeared. Mystery... Now, you probably discovered why there is no need for a $\text{cosc}$. – 2017-02-09
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    Even $\frac{1-\cos x}x$ can be written as $\frac{x\operatorname{sinc}^2x}{1+\cos x}$. – 2018-11-01

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The motivation for functions such as $\text{sinc}x,\,\text{sinch}x,\,\text{tanc}x,\,\text{tanch}x$ is to consider the behaviour of a ratio with limit $1$ as $x\to 0$. There is no such motivation for $\frac{\cos x}{x}$, since $\cos x =1\ne 0$. However, you can write this ratio as $-\text{Ci}' \left( x\right)$.