What is the order of operations in trig functions? Is $\sin(x)^2$ the same as $\sin^2(x)$ or $\sin(x^2)$? I thought it would mean the former interpretation, $\sin^2(x)$, rather than the latter, but my teacher and I had a long argument on this and in the end I found that Casio calculators have a space between before the parenthesis, so it would look like $\sin\text{ }(x)^2$ and what the calculator would do is calculate $x^2$, and then take the $\sin$ of that whereas on Texas calculators, there is no space, so it would look like $\sin(x)^2$ and it will calculate $\sin$ of $x$ first, and then take the result and square it.
 A: $$\Large \sin(x)^2 \equiv [\sin(x)]^2 \equiv \sin^2(x).$$
Think about it: if $\sin(x)^2$ were equivalent to $\sin(x^2),$ then the exponent $\quad ^2 \quad$ would be inside the brackets.
A: If you see $(\sin{x})^{2}$, to me this means $(\sin{x})(\sin{x})$.  Mathematicians are lazy and don't like writing parentheses around the $\sin{x}$ every time when squaring it, so they decided to write $\sin^{2}{x}$ when they want to square $\sin{x}$.
So basically, yes, $\sin^{2}{x} = (\sin{x})^{2}$.
In fact, for any whole number $n \geq 2$, when we write $\sin^{n}{x}$, we really mean $(\sin{x})^{n}$.
There is one exception. 
$\sin^{-1}{x} \neq (\sin{x})^{-1}$.
This is because we write $\sin^{-1}{x}$ for the inverse function of $\sin{x}$ (in other words, $\sin^{-1}{x}$ is what we write for $\arcsin{x}$).
However, $(\sin{x})^{-1} = \frac{1}{\sin{x}} = \csc{x}$.  And if you know your trig functions well, you will know $\arcsin{x}$ and $\csc{x}$ are not the same functions.  So with $-1$, $\sin^{-1}{x}$ is a different function from $(\sin{x})^{-1}$.
A: It is not only in trigonometric functions, all functions fall under this notation. $$(f(x))^2 \equiv f(x)^2 \equiv f^2(x)$$
It is commonly used for trigonometric functions and logarithmic ones, but it doesn't mean it is wrong for the rest of functions.
