3 added 633 characters in body
source | link

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

As a side note, harmonic motion is often started off with nonzero displacement but zero velocity, corresponding to a multiple of $\cos$, whereas $\sin$ corresponds to zero initial displacement but nonzero initial velocity. In some sense, rest is simpler than motion, just as first-order terms in a Taylor series are 'more' important than second-order terms.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

In fact, a lot of mathematical objects that seem tied to the anti-clockwise direction are nowhere near arbitrary but have to do with the way we have chosen to position the coordinate axes on paper. For example signed area is positive for anti-clockwise traversal of a non-self-intersecting polygon in the plane. The relation between the anti-clockwise contour integral around a pole and the residue there as given by the Laurent series is another. Not coincidental at all. (I decided to put this paragraph in because amazingly people don't seem to get that my answer explains all these seemingly arbitrary conventions at one go.)

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

As a side note, harmonic motion is often started off with nonzero displacement but zero velocity, corresponding to a multiple of $\cos$, whereas $\sin$ corresponds to zero initial displacement but nonzero initial velocity. In some sense, rest is simpler than motion, just as first-order terms in a Taylor series are 'more' important than second-order terms.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

As a side note, harmonic motion is often started off with nonzero displacement but zero velocity, corresponding to a multiple of $\cos$, whereas $\sin$ corresponds to zero initial displacement but nonzero initial velocity. In some sense, rest is simpler than motion, just as first-order terms in a Taylor series are 'more' important than second-order terms.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

In fact, a lot of mathematical objects that seem tied to the anti-clockwise direction are nowhere near arbitrary but have to do with the way we have chosen to position the coordinate axes on paper. For example signed area is positive for anti-clockwise traversal of a non-self-intersecting polygon in the plane. The relation between the anti-clockwise contour integral around a pole and the residue there as given by the Laurent series is another. Not coincidental at all. (I decided to put this paragraph in because amazingly people don't seem to get that my answer explains all these seemingly arbitrary conventions at one go.)

2 added 361 characters in body
source | link

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

As a side note, harmonic motion is often started off with nonzero displacement but zero velocity, corresponding to a multiple of $\cos$, whereas $\sin$ corresponds to zero initial displacement but nonzero initial velocity. In some sense, rest is simpler than motion, just as first-order terms in a Taylor series are 'more' important than second-order terms.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

As a side note, harmonic motion is often started off with nonzero displacement but zero velocity, corresponding to a multiple of $\cos$, whereas $\sin$ corresponds to zero initial displacement but nonzero initial velocity. In some sense, rest is simpler than motion, just as first-order terms in a Taylor series are 'more' important than second-order terms.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.

1
source | link

There are many possible reasons why it should be defined that way, but I'm not sure what were the initial motivating reasons. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

Firstly, $(\cos,\sin)$ is a basis for all the real-valued functions satisfying $f''=-f$. It is then natural to choose the basis elements so that $\cos(0) = 1$ and $\cos'(0) = 0$ and $\sin(0) = 0$ and $\sin'(0) = 1$. Another viewpoint is that this differential equation immediately implies that if they have a Taylor series $x \mapsto \sum_{k\in\nn} a_k \frac{x^k}{k!}$, then the coefficients $(a_n)_{n\in\nn}$ would have to satisfy $a_{n+2}=a_n$ for any $n\in\nn$. If we do not want to invoke the Taylor theorem we could also have defined them by the infinite series, prove that it converges, and then prove directly that it satisfies the differential equation. Either way, we obviously want the basis elements to be $(1,0,-1,0,\cdots)$ and $(0,1,0,-1,\cdots)$.

Secondly, $\exp(it)$ goes round the unit circle for $t\in\rr$. We can see this from the properties of $\exp$ that follows from the infinite series definition, which is motivated by the differential equation $f'=f$. The other definitions do not reveal the underlying structure. We can then choose $\cos,\sin$ as the $x,y$ coordinates.

In both reasons above, they do not fix a starting point or direction for the path traced by $(\cos,\sin)$ in the plane. What fixes it is the choice of our coordinate system where the $y$-axis is $90^\circ$ anti-clockwise from the $x$-axis, and we plot a pair with the first element as $x$-coordinate and the second element as $y$-coordinate.