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In the notation $\sin(x)^2 $

Does this equal $\sin(x^2) $ or $(\sin(x))^2 $ ?

I'm sorry this is such a simple question but Google is unhelpful. There are plenty of sources illustrating $\sin^2(x) $ = $(\sin(x))^2 $ but nothing about $\sin(x)^2 $

Edit

According to the answers in my math book(Calculus of a Single Variable 10e, Ron Larson & Bruce Edwards), it seems as if it equates $\sin(x)^2$ as $\sin(x^2)$ This obviously isn't proof of the notation but it would make some sense when considering $\sin(x+y)^2$ or $\sin(2x)^2$ as opposed to a single variable. I find this notation very confusing and better stated explicitly such as: $\sin((x+y)^2)$ or $\sin((2x)^2)$.

I'm still not happy with this answer and would appreciate if anyone could reference evidence to one side or the other.

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    $\begingroup$ It can mean either, depending on the author. This is precisely why the notation $\sin^2x$ exists. $\endgroup$ – Omnomnomnom Oct 4 '15 at 23:53
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    $\begingroup$ @Omnomnomnom: I would agree there is no universally accepted order of operations, but I suspect most mathematicians would read $\sin(x)^2=(\sin(x))^2$ but $\sin x^2 = \sin(x^2)$. Given that most of them read $\sin^2 x = (\sin(x))^2$ but $\sin^{-1} x \not = \frac{1}{\sin(x)}$, their opinions can be expected to be hopelessly inconsistent. $\endgroup$ – Henry Oct 5 '15 at 0:02
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    $\begingroup$ It is ambiguous, but it "looks" like $(\sin{x})^2$ to me. Or at least I would assume it was until proven otherwise: $\sin^2{x}$ is many times more common than $\sin{(x^2)}$. $\endgroup$ – Chappers Oct 5 '15 at 0:02
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    $\begingroup$ I like the top answer here: math.stackexchange.com/questions/932903/… $\endgroup$ – zahbaz Oct 5 '15 at 0:11
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    $\begingroup$ I would never write $\sin(x)^2$, precisely because of the annoying ambiguity. $(\sin x)^2$ and $\sin(x^2)$ are clear, and it has become universal to write $\sin^2 x$ for the former, even though it makes sense that that notation ought to mean $\sin\sin x$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 5 '15 at 1:11
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$$\sin^2(x)=(\sin x)^2=\sin (x)^2\neq \sin(x^2)$$

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  • $\begingroup$ This doesn't answer the question. I already know this. I'm asking about $\sin (x)^2$ not $\sin^2(x)$. $\endgroup$ – user3304179 Oct 5 '15 at 0:00
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    $\begingroup$ I included that as well. $\endgroup$ – yankeefan11 Oct 5 '15 at 0:07
  • $\begingroup$ You didn't include it in your 2nd edit which was when I posted my comment. $\endgroup$ – user3304179 Oct 5 '15 at 0:35

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