Is $\sin^2(x)$ the same as $\sin x^2$?

Question

I'm working with derivatives and need to know if $\sin^2(x)$ the same as $\sin(x^2)$?

I almost don't want to ask because my last question was closed. It was a valid question and so is this one. I've been trying to find the answer on my own but the answers I've discovered are conflicting. Any clarification here is appreciated.

Also,remember that for functions,we generally write $(f(x))^n$ as $f^n(x)$ for all values of n apart from $-1$.$f^-1 indicates the inverse function. — rah4927, Mar 27 '14 at 05:12
@rah4927 that is incorrect. Iterations of functions are also written as $f^n(x)$. — JP McCarthy, Jan 28 '15 at 15:18
@JpMcCarthy true.I wasn't thinking about iterations when I wrote that.Thanks for bringing this up. — rah4927, Feb 06 '15 at 10:17

score 14 · Answer 1 · answered Mar 27 '14 at 03:34

Here is a graphical verification that these two functions are not equal. All the other answers are fine, but it is worth clarifying some other ambiguous notation in math.

Trig functions are notorious for the confusion they create when involving exponents. The conventions listed below are not necessarily optimal but being aware of them is helpful for the future:

$\sin^2(x) = (\sin(x))^2$ "The square of the sine of $x$."
$\sin(x^2)$ "The sine of $x^2$."
$\sin^{-1}(x) = \arcsin(x)$ "The inverse sine of x. That is, if $y=\sin^{-1}(x)$, then $\sin(y) = x$."
$\sin(x^{-1}) = \sin\left(\dfrac{1}{x}\right)$ "The sine of the reciprocal of $x$ (where $x\neq0$)."
${(\sin(x))}^{-1} = \dfrac{1}{\sin(x)} = \csc(x)$ "The reciprocal of the sine of $x$. More commonly denoted the cosecant of $x$."

Other functions seem to adopt similar notation:

$\ln^2(x) = (\ln(x))^2$ "The square of the natural logarithm of $x$."
$\ln(x^2)$ "The natural logarithm of $x^2$."

But it only goes so far. I hope no one ever uses $-1$ as an exponent of the natural logarithm to indicate "the inverse natural logarithm of $x$"...because you would just denote that $e^x$. That is,

$\ln^{-1}(x) = \dfrac{1}{\ln(x)} \neq e^x$ "The reciprocal of the natural logarithm of $x$ (where $x\neq 1$)."

The Golden Rule

Whatever notation you are using: if you have any hint of ambiguity, throw in another pair of parentheses in the appropriate places!

I always thought of $sin^2(x)$ as $sin(sin(x))$ but that does make much sense as the outer sin would not receive an angle but the length of the opposite side on the unit circle. — jbuddy_13, Sep 09 '23 at 20:32

score 4 · Accepted Answer · answered Mar 27 '14 at 01:52

4

No. $\sin^2(x)=(\sin(x))^2$. $\sin(x^2)$, on the other hand, is just $\sin(x^2)$.

answered Mar 27 '14 at 01:52

Shahar

3,302

score 4 · Answer 3 · answered Mar 27 '14 at 01:53

4

No. $\sin(x^2)=\sin(x*x)$ while $\sin^2(x)=\sin(x)*\sin(x)$.

answered Mar 27 '14 at 01:53

Cbjork

696

score 2 · Answer 4 · answered Mar 27 '14 at 01:52

2

No, it's defined to be $sin^2(x)=(sin(x))^2$

answered Mar 27 '14 at 01:52

Stella Biderman

31,155

score 0 · Answer 5 · answered Jan 28 '15 at 15:13

0

No $\sin^2(x)$ means $(\sin x)^2$. This is not the same as $\sin(x^2)$.

answered Jan 28 '15 at 15:13

SQL Injection

123

1

Five other answers posted 10 months ago already said this. – Jonas Meyer Jan 28 '15 at 15:19
@JonasMeyer (i was only joking my dear friend, peace) okay sorry i will delete my answer, you are very right and i'm plain wrong. sorry for that, i'm too stupid y'know, peace – SQL Injection Jan 28 '15 at 15:28

score 0 · Answer 6 · answered Mar 27 '14 at 02:21

0

No, $\sin^2x =(\sin x)^2=(\sin x)\cdot (\sin x)$ and $\sin x^2=\sin (x\cdot x)$ (in this case, the argument $x$ is in square).

answered Mar 27 '14 at 02:21

ZHN

1,419

Three other answers posted half an hour earlier already said this. – Jonas Meyer Jan 28 '15 at 15:21

Is $\sin^2(x)$ the same as $\sin x^2$?

6 Answers6

The Golden Rule

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