Does exponentiation distribute over the multiplication of trigoneomtric functions?
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I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?
algebra-precalculus trigonometry
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add a comment |
$begingroup$
I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?
algebra-precalculus trigonometry
$endgroup$
2
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Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10
add a comment |
$begingroup$
I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?
algebra-precalculus trigonometry
$endgroup$
I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 8 at 4:29
naveen dankal
4,60931348
4,60931348
asked Jan 2 at 3:08
StartecStartec
1306
1306
2
$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10
add a comment |
2
$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10
2
2
$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10
$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.
$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:
$$(a cdot b)^c = a^c cdot b^c$$
In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:
$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$
$endgroup$
add a comment |
$begingroup$
What sort of numbers are you dealing with? What sort of function space?
In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:
begin{equation}
(a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
end{equation}
If it's endowed with associativity:
begin{equation}
(a*b)^n = a*b * a*b * a*b cdots * a*b
end{equation}
If it's also endowed with commutativity then:
begin{align}
(a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
&= a^n * b^n = b^n * a^n
end{align}
I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.
I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).
If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.
$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:
$$(a cdot b)^c = a^c cdot b^c$$
In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:
$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$
$endgroup$
add a comment |
$begingroup$
Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.
$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:
$$(a cdot b)^c = a^c cdot b^c$$
In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:
$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$
$endgroup$
add a comment |
$begingroup$
Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.
$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:
$$(a cdot b)^c = a^c cdot b^c$$
In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:
$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$
$endgroup$
Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.
$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:
$$(a cdot b)^c = a^c cdot b^c$$
In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:
$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$
edited Jan 8 at 4:27
user150203
answered Jan 2 at 3:12
Eevee TrainerEevee Trainer
10.7k31843
10.7k31843
add a comment |
add a comment |
$begingroup$
What sort of numbers are you dealing with? What sort of function space?
In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:
begin{equation}
(a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
end{equation}
If it's endowed with associativity:
begin{equation}
(a*b)^n = a*b * a*b * a*b cdots * a*b
end{equation}
If it's also endowed with commutativity then:
begin{align}
(a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
&= a^n * b^n = b^n * a^n
end{align}
I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.
I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).
If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.
$endgroup$
add a comment |
$begingroup$
What sort of numbers are you dealing with? What sort of function space?
In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:
begin{equation}
(a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
end{equation}
If it's endowed with associativity:
begin{equation}
(a*b)^n = a*b * a*b * a*b cdots * a*b
end{equation}
If it's also endowed with commutativity then:
begin{align}
(a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
&= a^n * b^n = b^n * a^n
end{align}
I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.
I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).
If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.
$endgroup$
add a comment |
$begingroup$
What sort of numbers are you dealing with? What sort of function space?
In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:
begin{equation}
(a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
end{equation}
If it's endowed with associativity:
begin{equation}
(a*b)^n = a*b * a*b * a*b cdots * a*b
end{equation}
If it's also endowed with commutativity then:
begin{align}
(a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
&= a^n * b^n = b^n * a^n
end{align}
I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.
I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).
If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.
$endgroup$
What sort of numbers are you dealing with? What sort of function space?
In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:
begin{equation}
(a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
end{equation}
If it's endowed with associativity:
begin{equation}
(a*b)^n = a*b * a*b * a*b cdots * a*b
end{equation}
If it's also endowed with commutativity then:
begin{align}
(a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
&= a^n * b^n = b^n * a^n
end{align}
I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.
I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).
If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.
edited Jan 2 at 7:34
answered Jan 2 at 7:25
user150203
add a comment |
add a comment |
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$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10