Mixing
what is the difference between $tan^2 (x)$ and $tan(x^2)$
$begingroup$
I know $tan^2(x) neq tan(x^2)$ but I can't find an intuitive way to understand $tan^2(x)$. When I do it in my mind I think: $tan^2(x) = left(frac{text{opp}}{text{adj}}right)^2$ which I think isn't right, that looks more like $tan(x^2)$.
Thanks for any tips, this has been rolling in my head for a while.
algebra-precalculus trigonometry
edited Jan 17 at 22:02
AEngineer
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
$endgroup$
|
show 2 more comments
$begingroup$
I know $tan^2(x) neq tan(x^2)$ but I can't find an intuitive way to understand $tan^2(x)$. When I do it in my mind I think: $tan^2(x) = left(frac{text{opp}}{text{adj}}right)^2$ which I think isn't right, that looks more like $tan(x^2)$.
Thanks for any tips, this has been rolling in my head for a while.
algebra-precalculus trigonometry
edited Jan 17 at 22:02
AEngineer
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
$endgroup$
3
$begingroup$
It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
$endgroup$
– Arturo Magidin
Jan 17 at 21:31
1
$begingroup$
It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
$endgroup$
– Dominik Kutek
Jan 17 at 21:32
1
$begingroup$
The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
$endgroup$
– Rob Arthan
Jan 17 at 21:36
1
$begingroup$
@ArturoMagidin You should post your comment as an official answer.
$endgroup$
– Dog_69
Jan 17 at 21:36
1
$begingroup$
Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
$endgroup$
– Rob Arthan
Jan 17 at 21:38
|
show 2 more comments
$begingroup$
I know $tan^2(x) neq tan(x^2)$ but I can't find an intuitive way to understand $tan^2(x)$. When I do it in my mind I think: $tan^2(x) = left(frac{text{opp}}{text{adj}}right)^2$ which I think isn't right, that looks more like $tan(x^2)$.
Thanks for any tips, this has been rolling in my head for a while.
algebra-precalculus trigonometry
edited Jan 17 at 22:02
AEngineer
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
$endgroup$
I know $tan^2(x) neq tan(x^2)$ but I can't find an intuitive way to understand $tan^2(x)$. When I do it in my mind I think: $tan^2(x) = left(frac{text{opp}}{text{adj}}right)^2$ which I think isn't right, that looks more like $tan(x^2)$.
Thanks for any tips, this has been rolling in my head for a while.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 17 at 22:02
AEngineer
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
edited Jan 17 at 22:02
AEngineer
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
edited Jan 17 at 22:02
AEngineer
1,5441317
edited Jan 17 at 22:02
AEngineer
1,5441317
edited Jan 17 at 22:02
AEngineer
1,5441317
1,5441317
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
asked Jan 17 at 21:28
Daniel WilkDaniel Wilk
82
82
3
$begingroup$
It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
$endgroup$
– Arturo Magidin
Jan 17 at 21:31
1
$begingroup$
It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
$endgroup$
– Dominik Kutek
Jan 17 at 21:32
1
$begingroup$
The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
$endgroup$
– Rob Arthan
Jan 17 at 21:36
1
$begingroup$
@ArturoMagidin You should post your comment as an official answer.
$endgroup$
– Dog_69
Jan 17 at 21:36
1
$begingroup$
Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
$endgroup$
– Rob Arthan
Jan 17 at 21:38
|
show 2 more comments
3
$begingroup$
It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
$endgroup$
– Arturo Magidin
Jan 17 at 21:31
1
$begingroup$
It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
$endgroup$
– Dominik Kutek
Jan 17 at 21:32
1
$begingroup$
The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
$endgroup$
– Rob Arthan
Jan 17 at 21:36
1
$begingroup$
@ArturoMagidin You should post your comment as an official answer.
$endgroup$
– Dog_69
Jan 17 at 21:36
1
$begingroup$
Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
$endgroup$
– Rob Arthan
Jan 17 at 21:38
3
3
$begingroup$
It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
$endgroup$
– Arturo Magidin
Jan 17 at 21:31
$begingroup$
It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
$endgroup$
– Arturo Magidin
Jan 17 at 21:31
1
1
$begingroup$
It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
$endgroup$
– Dominik Kutek
Jan 17 at 21:32
$begingroup$
It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
$endgroup$
– Dominik Kutek
Jan 17 at 21:32
1
1
$begingroup$
The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
$endgroup$
– Rob Arthan
Jan 17 at 21:36
$begingroup$
The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
$endgroup$
– Rob Arthan
Jan 17 at 21:36
1
1
$begingroup$
@ArturoMagidin You should post your comment as an official answer.
$endgroup$
– Dog_69
Jan 17 at 21:36
$begingroup$
@ArturoMagidin You should post your comment as an official answer.
$endgroup$
– Dog_69
Jan 17 at 21:36
1
1
$begingroup$
Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
$endgroup$
– Rob Arthan
Jan 17 at 21:38
$begingroup$
Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
$endgroup$
– Rob Arthan
Jan 17 at 21:38
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The traditional notation for the trigonometrc functions is that $tan^2$ is the function that maps an angle to the square of its tangent. If, for a given $x$, $mathsf{opp}$ (i.e., $sin(x)$) and $mathsf{adj}$ (i.e., $cos(x)$) denote the lengths of the legs of a right-angled triangle with unit hypotenuse that are opposite and adjacent to the angle $x$, then $tan^2(x)$ is indeed equal to $(mathsf{opp}/mathsf{adj})^2$, i.e., $(tan(x))^2$. Your thinking that this is more like $tan(x^2)$ is incorrect: we are holding the right-angle triangle fixed and looking at different functions of the lengths of its edges.
In modern notation, $f^2(x)$ usually means $f(f(x))$. So as $tan(tan(x))$ and $(tan(x))^2$ are certainly not the same function, you need to keep in mind that the convention for trigonometric functions is a special one.
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
Define $f(x):=(tan x)^2,,g(x):=tan(tan x),,h(x):=tan (x^2)$ (i.e. if $y=x^2$ then $h(x)=tan y$). These are all completely different functions.
Either $f$ or $g$ could be denoted $tan^2 x$; if it means $f$ we're composing the squaring function with the tangent function, while if it means $g$ we're composing the tangent function with itself. The reason the latter can be denoted $tan^2 x$ is because functions form an associative algebra under composition, for which composition is like a "multiplication" of functions. And if you "multiply" the tangent function by itself, you're "squaring" it.
However, it's a matter of convention that, unless very clearly stated otherwise, $tan^2 x$ refers to $f$ rather than $g$, if only because $f$ is far more likely to come up in a problem than $g$ is. In particular, the squared opposite-to-adjacent ratio is indeed $f$.
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
$endgroup$
add a comment |
$begingroup$
Here is the difference. Consider these operations:
- Hit spot marked X with an axe with a mighty blow
- Put your hand on spot X.
Suppose you do these operations in succession Does the order matter? In general, functional
composition is not commutative.
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The traditional notation for the trigonometrc functions is that $tan^2$ is the function that maps an angle to the square of its tangent. If, for a given $x$, $mathsf{opp}$ (i.e., $sin(x)$) and $mathsf{adj}$ (i.e., $cos(x)$) denote the lengths of the legs of a right-angled triangle with unit hypotenuse that are opposite and adjacent to the angle $x$, then $tan^2(x)$ is indeed equal to $(mathsf{opp}/mathsf{adj})^2$, i.e., $(tan(x))^2$. Your thinking that this is more like $tan(x^2)$ is incorrect: we are holding the right-angle triangle fixed and looking at different functions of the lengths of its edges.
In modern notation, $f^2(x)$ usually means $f(f(x))$. So as $tan(tan(x))$ and $(tan(x))^2$ are certainly not the same function, you need to keep in mind that the convention for trigonometric functions is a special one.
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
The traditional notation for the trigonometrc functions is that $tan^2$ is the function that maps an angle to the square of its tangent. If, for a given $x$, $mathsf{opp}$ (i.e., $sin(x)$) and $mathsf{adj}$ (i.e., $cos(x)$) denote the lengths of the legs of a right-angled triangle with unit hypotenuse that are opposite and adjacent to the angle $x$, then $tan^2(x)$ is indeed equal to $(mathsf{opp}/mathsf{adj})^2$, i.e., $(tan(x))^2$. Your thinking that this is more like $tan(x^2)$ is incorrect: we are holding the right-angle triangle fixed and looking at different functions of the lengths of its edges.
In modern notation, $f^2(x)$ usually means $f(f(x))$. So as $tan(tan(x))$ and $(tan(x))^2$ are certainly not the same function, you need to keep in mind that the convention for trigonometric functions is a special one.
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
The traditional notation for the trigonometrc functions is that $tan^2$ is the function that maps an angle to the square of its tangent. If, for a given $x$, $mathsf{opp}$ (i.e., $sin(x)$) and $mathsf{adj}$ (i.e., $cos(x)$) denote the lengths of the legs of a right-angled triangle with unit hypotenuse that are opposite and adjacent to the angle $x$, then $tan^2(x)$ is indeed equal to $(mathsf{opp}/mathsf{adj})^2$, i.e., $(tan(x))^2$. Your thinking that this is more like $tan(x^2)$ is incorrect: we are holding the right-angle triangle fixed and looking at different functions of the lengths of its edges.
In modern notation, $f^2(x)$ usually means $f(f(x))$. So as $tan(tan(x))$ and $(tan(x))^2$ are certainly not the same function, you need to keep in mind that the convention for trigonometric functions is a special one.
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
$endgroup$
The traditional notation for the trigonometrc functions is that $tan^2$ is the function that maps an angle to the square of its tangent. If, for a given $x$, $mathsf{opp}$ (i.e., $sin(x)$) and $mathsf{adj}$ (i.e., $cos(x)$) denote the lengths of the legs of a right-angled triangle with unit hypotenuse that are opposite and adjacent to the angle $x$, then $tan^2(x)$ is indeed equal to $(mathsf{opp}/mathsf{adj})^2$, i.e., $(tan(x))^2$. Your thinking that this is more like $tan(x^2)$ is incorrect: we are holding the right-angle triangle fixed and looking at different functions of the lengths of its edges.
In modern notation, $f^2(x)$ usually means $f(f(x))$. So as $tan(tan(x))$ and $(tan(x))^2$ are certainly not the same function, you need to keep in mind that the convention for trigonometric functions is a special one.
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
answered Jan 17 at 22:13
Rob ArthanRob Arthan
29.3k42966
29.3k42966
add a comment |
add a comment |
$begingroup$
Define $f(x):=(tan x)^2,,g(x):=tan(tan x),,h(x):=tan (x^2)$ (i.e. if $y=x^2$ then $h(x)=tan y$). These are all completely different functions.
Either $f$ or $g$ could be denoted $tan^2 x$; if it means $f$ we're composing the squaring function with the tangent function, while if it means $g$ we're composing the tangent function with itself. The reason the latter can be denoted $tan^2 x$ is because functions form an associative algebra under composition, for which composition is like a "multiplication" of functions. And if you "multiply" the tangent function by itself, you're "squaring" it.
However, it's a matter of convention that, unless very clearly stated otherwise, $tan^2 x$ refers to $f$ rather than $g$, if only because $f$ is far more likely to come up in a problem than $g$ is. In particular, the squared opposite-to-adjacent ratio is indeed $f$.
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
$endgroup$
add a comment |
$begingroup$
Define $f(x):=(tan x)^2,,g(x):=tan(tan x),,h(x):=tan (x^2)$ (i.e. if $y=x^2$ then $h(x)=tan y$). These are all completely different functions.
Either $f$ or $g$ could be denoted $tan^2 x$; if it means $f$ we're composing the squaring function with the tangent function, while if it means $g$ we're composing the tangent function with itself. The reason the latter can be denoted $tan^2 x$ is because functions form an associative algebra under composition, for which composition is like a "multiplication" of functions. And if you "multiply" the tangent function by itself, you're "squaring" it.
However, it's a matter of convention that, unless very clearly stated otherwise, $tan^2 x$ refers to $f$ rather than $g$, if only because $f$ is far more likely to come up in a problem than $g$ is. In particular, the squared opposite-to-adjacent ratio is indeed $f$.
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
$endgroup$
add a comment |
$begingroup$
Define $f(x):=(tan x)^2,,g(x):=tan(tan x),,h(x):=tan (x^2)$ (i.e. if $y=x^2$ then $h(x)=tan y$). These are all completely different functions.
Either $f$ or $g$ could be denoted $tan^2 x$; if it means $f$ we're composing the squaring function with the tangent function, while if it means $g$ we're composing the tangent function with itself. The reason the latter can be denoted $tan^2 x$ is because functions form an associative algebra under composition, for which composition is like a "multiplication" of functions. And if you "multiply" the tangent function by itself, you're "squaring" it.
However, it's a matter of convention that, unless very clearly stated otherwise, $tan^2 x$ refers to $f$ rather than $g$, if only because $f$ is far more likely to come up in a problem than $g$ is. In particular, the squared opposite-to-adjacent ratio is indeed $f$.
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
$endgroup$
Define $f(x):=(tan x)^2,,g(x):=tan(tan x),,h(x):=tan (x^2)$ (i.e. if $y=x^2$ then $h(x)=tan y$). These are all completely different functions.
Either $f$ or $g$ could be denoted $tan^2 x$; if it means $f$ we're composing the squaring function with the tangent function, while if it means $g$ we're composing the tangent function with itself. The reason the latter can be denoted $tan^2 x$ is because functions form an associative algebra under composition, for which composition is like a "multiplication" of functions. And if you "multiply" the tangent function by itself, you're "squaring" it.
However, it's a matter of convention that, unless very clearly stated otherwise, $tan^2 x$ refers to $f$ rather than $g$, if only because $f$ is far more likely to come up in a problem than $g$ is. In particular, the squared opposite-to-adjacent ratio is indeed $f$.
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
answered Jan 17 at 22:15
J.G.J.G.
27.9k22843
27.9k22843
add a comment |
add a comment |
$begingroup$
Here is the difference. Consider these operations:
- Hit spot marked X with an axe with a mighty blow
- Put your hand on spot X.
Suppose you do these operations in succession Does the order matter? In general, functional
composition is not commutative.
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
add a comment |
$begingroup$
Here is the difference. Consider these operations:
- Hit spot marked X with an axe with a mighty blow
- Put your hand on spot X.
Suppose you do these operations in succession Does the order matter? In general, functional
composition is not commutative.
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
$endgroup$
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
add a comment |
$begingroup$
Here is the difference. Consider these operations:
- Hit spot marked X with an axe with a mighty blow
- Put your hand on spot X.
Suppose you do these operations in succession Does the order matter? In general, functional
composition is not commutative.
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
$endgroup$
Here is the difference. Consider these operations:
- Hit spot marked X with an axe with a mighty blow
- Put your hand on spot X.
Suppose you do these operations in succession Does the order matter? In general, functional
composition is not commutative.
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
answered Jan 17 at 22:18
ncmathsadistncmathsadist
42.9k260103
42.9k260103
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
add a comment |
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
$begingroup$
I must say that your example is in beautiful harmony with your nickname.
$endgroup$
– Ivan Neretin
Jan 17 at 22:57
add a comment |
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It’s the difference between first squaring and then finding the tangent, and first finding the tangent and then squaring the result. The “in your mind” is correct; it does not look like $tan(x^2)$, because that would require you to square the size of the angle, thus moving to a different triangle before taking the tangent.
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– Arturo Magidin
Jan 17 at 21:31
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It's just composing function in different order, if $g(x) = tan(x)$, and $f(x) = x^2$ then $ f(g(x)) = f(tan(x)) = (tan(x))^2 = tan^2(x)$, and $g(f(x)) = g(x^2) = tan(x^2)$, so in $tan^2(x)$ you firstly compute $tan(x)$ for a fixed x, then raise it to the power of two, and opposite in $tan(x^2)$ you firstly raise x to the power of two, ten plug it into tangent function.
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– Dominik Kutek
Jan 17 at 21:32
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The traditional notation is a bit confusing: $tan^2$ is used to denote the function that takes the tangent of its argument and then squares the result. I.e., $tan^2(x) = (tan(x))^2$. If you think about $(tan(x))^2$, it may be easier to understand.
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– Rob Arthan
Jan 17 at 21:36
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@ArturoMagidin You should post your comment as an official answer.
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– Dog_69
Jan 17 at 21:36
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Any answer should include the point that $tan^2$ and similar notations are special for the trigonometric functions. $f^2(x)$ usually means $f(f(x))$.
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– Rob Arthan
Jan 17 at 21:38