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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.

Muhatashim
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3 Answers3

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$$\Large \sin(x)^2 \equiv [\sin(x)]^2 \equiv \sin^2(x).$$

Think about it: if $\sin(x)^2$ were equivalent to $\sin(x^2),$ then the exponent $\quad ^2 \quad$ would be inside the brackets.

beep-boop
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  • That x was just a condensed version of the original expression that we were arguing on. The original one was x+a. And what about the difference for Texas and Casio calculators? Does the space really mean that difference? – Muhatashim Sep 05 '14 at 23:59
  • So, is Sin (XX) same as (Sin X)(Sin X)? – Rao A. Sep 06 '14 at 00:04
  • And now, what is it without brackets: $\sin x^2$ ... ?? – GEdgar Sep 06 '14 at 00:22
  • @Muhatashim I don't know about all calculators, but as far as mathematical convention goes, $\sin(x^2)$ is interpreted as in my answer, and spacing makes no difference. – beep-boop Sep 06 '14 at 00:39
  • @RaoA. No. the former is $\sin(x^2)$ (i.e. the sine of the square of $x$) and the latter is is $[\sin(x)]^2$ (i.e. the square of the sine of $x$). The two are not (necessarily the same). – beep-boop Sep 06 '14 at 00:41
  • @GEdgar If there are no brackets, when in doubt, assume that everything after the $\sin$ is to be bracketed. i.e. assume that $\sin x^2 \equiv \sin (x^2)$. – beep-boop Sep 06 '14 at 00:44
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    "if $\sin(x)^2$ were equivalent to $\sin(x^2),$ then the exponent $\quad ^2 \quad$ would be inside the brackets." unless you interpret $\sin(x)^2$ as equivalent to $\sin((x)^2)$ – BartekChom Jan 17 '15 at 20:04
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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 $(\sin{x})^{-1}$.

layman
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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.

UserX
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    I've seen the iterated logarithm $\log(\log x)$ written as $\log^2 x$. So it's not completely universal convention of interpreting $f^2(x)$ as $f(x)^2$ rather than $(f\circ f)(x)$. – Semiclassical Sep 06 '14 at 02:04
  • @Semiclassical True, though I usually see the exponent in brackets when indicating composition, e.g. $f^2(x) = (f(x))^2$ and $f^{(2)}(x) = (f \circ f)(x) = f(f(x))$. It should hopefully be clear from context in most cases though. – Thomas Sep 06 '14 at 02:05
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    Aye, though that overlaps with $f^{(n)}(x)$ as the $n$th derivative, though. Furthermore, if one writes differentiation as an operator $D=\frac{d}{dx}$ then the usual notation is $f^{(n)}(x)=D^n f(x)$. So it all comes down to conventions... @Thomas – Semiclassical Sep 06 '14 at 02:06
  • @Semiclassical Agreed, in case of ambiguity always define your terms and there will be no problem :) – Thomas Sep 06 '14 at 02:10