homework and exercises - Why aren't trigonometric functions dimensionless regardless of the argument?

Consider this equation :-

$$y = a\sin kt$$ where $a$ is amplitude, $y$ is displacement, $t$ is time and $k$ is some dimensionless constant.

My instructor said this equation is dimensionally incorrect because the dimension of $[kt] = [\text{T}^1]$ and since $\text{angles}$ are dimensionless, we can conclude that it is dimensionally incorrect.

I don't understand why it is so. Why do we need to check the dimension homogeneity of the term inside the $\sin$ to conclude whether the equation is dimensionally correct or not?

Why isn't the whole sine function is dimensionless $(\sin kt = \text{[T}^0]) $ regardless of the dimension of the argument inside as the range of sine function is $[-1, 1]$.

Answer

One definition of the sinus function (in fact, the one probably used by your calculator) is $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots $$ This definition is dimensionally inconsistent unless $x$ is dimensionless. The proof that this gives you the familiar trigonometric sine if $x$ is the ratio of two carefully chosen lengths takes you through a surprisingly interesting swath of mathematics.