Proving the angle sum and difference identities for sine and cosine without involving the functions'...
$begingroup$
For well known identities
$$
sin(alpha pm beta) = sinalphacosbeta pm cosalphasinbeta
$$
$$
cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta
$$
is it possible to provide a proof which does not involve geometric meaning of sine and cosine functions (that is use of Ptolemy’s theorem)?
trigonometry
$endgroup$
add a comment |
$begingroup$
For well known identities
$$
sin(alpha pm beta) = sinalphacosbeta pm cosalphasinbeta
$$
$$
cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta
$$
is it possible to provide a proof which does not involve geometric meaning of sine and cosine functions (that is use of Ptolemy’s theorem)?
trigonometry
$endgroup$
1
$begingroup$
Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago
add a comment |
$begingroup$
For well known identities
$$
sin(alpha pm beta) = sinalphacosbeta pm cosalphasinbeta
$$
$$
cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta
$$
is it possible to provide a proof which does not involve geometric meaning of sine and cosine functions (that is use of Ptolemy’s theorem)?
trigonometry
$endgroup$
For well known identities
$$
sin(alpha pm beta) = sinalphacosbeta pm cosalphasinbeta
$$
$$
cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta
$$
is it possible to provide a proof which does not involve geometric meaning of sine and cosine functions (that is use of Ptolemy’s theorem)?
trigonometry
trigonometry
edited 6 hours ago
Blue
48.5k870154
48.5k870154
asked 7 hours ago
scrutariscrutari
1507
1507
1
$begingroup$
Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago
add a comment |
1
$begingroup$
Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago
1
1
$begingroup$
Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago
$begingroup$
Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.
Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=cos a +isin a,$$ and we can compute
begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\
cos (a+b)+isin (a+b)&=(cos a +isin a)(cos b +isin b)\
&=(cos acos b-sin a sin b)+i(sin a cos b+cos asin b).
end{align}
Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.
$endgroup$
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
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@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
|
show 1 more comment
$begingroup$
Just to share my idea. I will not say that my proof doesn't involve geometry.
Let $A=(cosalpha,sinalpha)$ and $B=(cosbeta,sinbeta)$. Then
$$AB^2=(cosalpha-cosbeta)^2+(cosalpha-sinbeta)^2=2-cosalphacosbeta-sinalphasinbeta$$
On the other hand,
$$AB^2=OA^2+OB^2-2(OA)(OB)cosangle AOB=1^2+1^2-2(1)(1)cos(alpha-beta)$$
This proves that $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$. The proof holds for arbitrary $alpha$ and $beta$.
$cos(alpha+beta)=cosalphacos(-beta)+sinalphasin(-beta)=cosalphacosbeta-sinalphasinbeta$
$sin(alpha+beta)=cosleft(frac{pi}{2}-alpha-betaright)=cosleft(frac{pi}{2}-alpharight)cosbeta+sinleft(frac{pi}{2}-alpharight)sinbeta=sinalphacosbeta+cosalphasinbeta$
$sin(alpha-beta)=sinalphacos(-beta)+cosalphasin(-beta)=sinalphacosbeta-cosalphasinbeta$
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$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
add a comment |
$begingroup$
Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.
Suppose you know only the following:
$$sin'=cos, quad cos'=-sin, quad sin 0 = 0, quad cos 0 = 1$$
As a warmup, prove the Pythagorean identity $sin^2 x + cos^2 x = 1$. (Hint: let $f(x) = sin^2 x + cos^2 x$ and compute $f'$.) In particular, $|sin x|le 1$ and $|cos x| le 1$.
Now fix $a$ and consider the function
$$g(x) = sin(x+a) - sin x cos a - cos x sin a.$$
Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n to infty$ you can conclude $g equiv 0$.
For the identity involving $cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = sin(a-x) - sin acos x + cos a sin x$, or by showing separately that $sin$ is an odd function and $cos$ is an even function.
Some variations:
If you know about real analytic functions, and you know that $sin x, cos x,sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.
If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.
$endgroup$
$begingroup$
I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.
Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=cos a +isin a,$$ and we can compute
begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\
cos (a+b)+isin (a+b)&=(cos a +isin a)(cos b +isin b)\
&=(cos acos b-sin a sin b)+i(sin a cos b+cos asin b).
end{align}
Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.
$endgroup$
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
|
show 1 more comment
$begingroup$
The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.
Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=cos a +isin a,$$ and we can compute
begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\
cos (a+b)+isin (a+b)&=(cos a +isin a)(cos b +isin b)\
&=(cos acos b-sin a sin b)+i(sin a cos b+cos asin b).
end{align}
Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.
$endgroup$
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
|
show 1 more comment
$begingroup$
The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.
Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=cos a +isin a,$$ and we can compute
begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\
cos (a+b)+isin (a+b)&=(cos a +isin a)(cos b +isin b)\
&=(cos acos b-sin a sin b)+i(sin a cos b+cos asin b).
end{align}
Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.
$endgroup$
The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.
Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=cos a +isin a,$$ and we can compute
begin{align}
e^{i(a+b)}&=e^{ia}e^{ib}\
cos (a+b)+isin (a+b)&=(cos a +isin a)(cos b +isin b)\
&=(cos acos b-sin a sin b)+i(sin a cos b+cos asin b).
end{align}
Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.
edited 4 hours ago
YawarRaza7349
884
884
answered 6 hours ago
Mark BennetMark Bennet
81.3k983180
81.3k983180
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
|
show 1 more comment
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
could you please explain the third equality?
$endgroup$
– scrutari
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
@scrutari That comes from expanding the product of brackets and using $i^2=-1$. There are two terms in each bracket, therefore four in the product, and these are gathered together in pairs. I've probably put the second pair in a confusing order.
$endgroup$
– Mark Bennet
6 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
$begingroup$
Apologies for the confusion, I meant transition from $cos(a + b) + i sin(a + b)$ to $(cos a + i sin a)(cos b + i sin b)$.
$endgroup$
– scrutari
5 hours ago
1
1
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
@scrutari: That's saying $e^{i(a+b)}=e^{ia}e^{ib}$
$endgroup$
– J. W. Tanner
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
$begingroup$
Right, of course :) thank you!
$endgroup$
– scrutari
5 hours ago
|
show 1 more comment
$begingroup$
Just to share my idea. I will not say that my proof doesn't involve geometry.
Let $A=(cosalpha,sinalpha)$ and $B=(cosbeta,sinbeta)$. Then
$$AB^2=(cosalpha-cosbeta)^2+(cosalpha-sinbeta)^2=2-cosalphacosbeta-sinalphasinbeta$$
On the other hand,
$$AB^2=OA^2+OB^2-2(OA)(OB)cosangle AOB=1^2+1^2-2(1)(1)cos(alpha-beta)$$
This proves that $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$. The proof holds for arbitrary $alpha$ and $beta$.
$cos(alpha+beta)=cosalphacos(-beta)+sinalphasin(-beta)=cosalphacosbeta-sinalphasinbeta$
$sin(alpha+beta)=cosleft(frac{pi}{2}-alpha-betaright)=cosleft(frac{pi}{2}-alpharight)cosbeta+sinleft(frac{pi}{2}-alpharight)sinbeta=sinalphacosbeta+cosalphasinbeta$
$sin(alpha-beta)=sinalphacos(-beta)+cosalphasin(-beta)=sinalphacosbeta-cosalphasinbeta$
$endgroup$
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
add a comment |
$begingroup$
Just to share my idea. I will not say that my proof doesn't involve geometry.
Let $A=(cosalpha,sinalpha)$ and $B=(cosbeta,sinbeta)$. Then
$$AB^2=(cosalpha-cosbeta)^2+(cosalpha-sinbeta)^2=2-cosalphacosbeta-sinalphasinbeta$$
On the other hand,
$$AB^2=OA^2+OB^2-2(OA)(OB)cosangle AOB=1^2+1^2-2(1)(1)cos(alpha-beta)$$
This proves that $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$. The proof holds for arbitrary $alpha$ and $beta$.
$cos(alpha+beta)=cosalphacos(-beta)+sinalphasin(-beta)=cosalphacosbeta-sinalphasinbeta$
$sin(alpha+beta)=cosleft(frac{pi}{2}-alpha-betaright)=cosleft(frac{pi}{2}-alpharight)cosbeta+sinleft(frac{pi}{2}-alpharight)sinbeta=sinalphacosbeta+cosalphasinbeta$
$sin(alpha-beta)=sinalphacos(-beta)+cosalphasin(-beta)=sinalphacosbeta-cosalphasinbeta$
$endgroup$
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
add a comment |
$begingroup$
Just to share my idea. I will not say that my proof doesn't involve geometry.
Let $A=(cosalpha,sinalpha)$ and $B=(cosbeta,sinbeta)$. Then
$$AB^2=(cosalpha-cosbeta)^2+(cosalpha-sinbeta)^2=2-cosalphacosbeta-sinalphasinbeta$$
On the other hand,
$$AB^2=OA^2+OB^2-2(OA)(OB)cosangle AOB=1^2+1^2-2(1)(1)cos(alpha-beta)$$
This proves that $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$. The proof holds for arbitrary $alpha$ and $beta$.
$cos(alpha+beta)=cosalphacos(-beta)+sinalphasin(-beta)=cosalphacosbeta-sinalphasinbeta$
$sin(alpha+beta)=cosleft(frac{pi}{2}-alpha-betaright)=cosleft(frac{pi}{2}-alpharight)cosbeta+sinleft(frac{pi}{2}-alpharight)sinbeta=sinalphacosbeta+cosalphasinbeta$
$sin(alpha-beta)=sinalphacos(-beta)+cosalphasin(-beta)=sinalphacosbeta-cosalphasinbeta$
$endgroup$
Just to share my idea. I will not say that my proof doesn't involve geometry.
Let $A=(cosalpha,sinalpha)$ and $B=(cosbeta,sinbeta)$. Then
$$AB^2=(cosalpha-cosbeta)^2+(cosalpha-sinbeta)^2=2-cosalphacosbeta-sinalphasinbeta$$
On the other hand,
$$AB^2=OA^2+OB^2-2(OA)(OB)cosangle AOB=1^2+1^2-2(1)(1)cos(alpha-beta)$$
This proves that $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$. The proof holds for arbitrary $alpha$ and $beta$.
$cos(alpha+beta)=cosalphacos(-beta)+sinalphasin(-beta)=cosalphacosbeta-sinalphasinbeta$
$sin(alpha+beta)=cosleft(frac{pi}{2}-alpha-betaright)=cosleft(frac{pi}{2}-alpharight)cosbeta+sinleft(frac{pi}{2}-alpharight)sinbeta=sinalphacosbeta+cosalphasinbeta$
$sin(alpha-beta)=sinalphacos(-beta)+cosalphasin(-beta)=sinalphacosbeta-cosalphasinbeta$
answered 6 hours ago
CY AriesCY Aries
16.7k11744
16.7k11744
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
add a comment |
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
$begingroup$
(+1) haven't seen you in a while!
$endgroup$
– TheSimpliFire
2 hours ago
add a comment |
$begingroup$
Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.
Suppose you know only the following:
$$sin'=cos, quad cos'=-sin, quad sin 0 = 0, quad cos 0 = 1$$
As a warmup, prove the Pythagorean identity $sin^2 x + cos^2 x = 1$. (Hint: let $f(x) = sin^2 x + cos^2 x$ and compute $f'$.) In particular, $|sin x|le 1$ and $|cos x| le 1$.
Now fix $a$ and consider the function
$$g(x) = sin(x+a) - sin x cos a - cos x sin a.$$
Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n to infty$ you can conclude $g equiv 0$.
For the identity involving $cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = sin(a-x) - sin acos x + cos a sin x$, or by showing separately that $sin$ is an odd function and $cos$ is an even function.
Some variations:
If you know about real analytic functions, and you know that $sin x, cos x,sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.
If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.
$endgroup$
$begingroup$
I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
add a comment |
$begingroup$
Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.
Suppose you know only the following:
$$sin'=cos, quad cos'=-sin, quad sin 0 = 0, quad cos 0 = 1$$
As a warmup, prove the Pythagorean identity $sin^2 x + cos^2 x = 1$. (Hint: let $f(x) = sin^2 x + cos^2 x$ and compute $f'$.) In particular, $|sin x|le 1$ and $|cos x| le 1$.
Now fix $a$ and consider the function
$$g(x) = sin(x+a) - sin x cos a - cos x sin a.$$
Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n to infty$ you can conclude $g equiv 0$.
For the identity involving $cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = sin(a-x) - sin acos x + cos a sin x$, or by showing separately that $sin$ is an odd function and $cos$ is an even function.
Some variations:
If you know about real analytic functions, and you know that $sin x, cos x,sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.
If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.
$endgroup$
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I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
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– scrutari
1 hour ago
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I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
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– Nate Eldredge
31 mins ago
add a comment |
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Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.
Suppose you know only the following:
$$sin'=cos, quad cos'=-sin, quad sin 0 = 0, quad cos 0 = 1$$
As a warmup, prove the Pythagorean identity $sin^2 x + cos^2 x = 1$. (Hint: let $f(x) = sin^2 x + cos^2 x$ and compute $f'$.) In particular, $|sin x|le 1$ and $|cos x| le 1$.
Now fix $a$ and consider the function
$$g(x) = sin(x+a) - sin x cos a - cos x sin a.$$
Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n to infty$ you can conclude $g equiv 0$.
For the identity involving $cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = sin(a-x) - sin acos x + cos a sin x$, or by showing separately that $sin$ is an odd function and $cos$ is an even function.
Some variations:
If you know about real analytic functions, and you know that $sin x, cos x,sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.
If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.
$endgroup$
Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.
Suppose you know only the following:
$$sin'=cos, quad cos'=-sin, quad sin 0 = 0, quad cos 0 = 1$$
As a warmup, prove the Pythagorean identity $sin^2 x + cos^2 x = 1$. (Hint: let $f(x) = sin^2 x + cos^2 x$ and compute $f'$.) In particular, $|sin x|le 1$ and $|cos x| le 1$.
Now fix $a$ and consider the function
$$g(x) = sin(x+a) - sin x cos a - cos x sin a.$$
Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n to infty$ you can conclude $g equiv 0$.
For the identity involving $cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = sin(a-x) - sin acos x + cos a sin x$, or by showing separately that $sin$ is an odd function and $cos$ is an even function.
Some variations:
If you know about real analytic functions, and you know that $sin x, cos x,sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.
If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.
answered 4 hours ago
Nate EldredgeNate Eldredge
63.6k682171
63.6k682171
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I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
add a comment |
$begingroup$
I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
$begingroup$
I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I like this approach, quite intense in using theorems from various parts of Maths, though "No complex numbers" in fact means "but use the rest of Maths instead" :)
$endgroup$
– scrutari
1 hour ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
$begingroup$
I think if you look carefully, other than the variations, that this really doesn't use anything except basic facts about derivatives (product rule, chain rule, etc) and the mean value theorem. (Taylor's theorem is just a repeated application of the mean value theorem.)
$endgroup$
– Nate Eldredge
31 mins ago
add a comment |
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Then do you want to start with alternative definitions of these functions? Any algebraic definitions would reduce the proof to an easy calculation. For example, define them as the real and imaginary parts of $e^{ix}$.
$endgroup$
– Aravind
6 hours ago