Proofs of $cos(x+y) = cos xcos y - sin x sin y$How can I understand and prove the “sum and difference formulas” in trigonometry?Different definitions of trigonometric functionsHow do I prove that $cosleft(2xright)=1-2sin^2left(xright)$?How do we find specific values of sin and cos given the series definitionHow to show that $cos(x+y)=cos(x)cos(y)−sin(x)sin(y)$ by using power series?Show that the equation $cos(sin x)=sin(cos x)$ has no real solutions.Solve for $sin^2(x) = 3cos^2(x)$Convergence of $sum_n |fraccos(3^n)n|$Proof: $sinleft(fracx-x_02right)sinleft(fracx-x_12right) = frac12cos fracx_1-x_02 - frac12cos(x-fracx_1+x_02)$How does one show sin(x) is bounded using the power series?Using the IVP definition of $cos$ and $sin$, how can we show that $cos^2(x)+sin^2(x) = 1$ without any “magic”?Transition from Introductory Proofs/Logic Course to the Proofs in Rudin's Principles of AnalysisHow can we show $cos^6x+sin^6x=1-3sin^2x cos^2x$?Is there a natural way to prove trig identities also hold for complex numbers?General solution of $(sqrt3 - 1)sintheta + (sqrt3 + 1)costheta =2 $
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Proofs of $cos(x+y) = cos xcos y - sin x sin y$
How can I understand and prove the “sum and difference formulas” in trigonometry?Different definitions of trigonometric functionsHow do I prove that $cosleft(2xright)=1-2sin^2left(xright)$?How do we find specific values of sin and cos given the series definitionHow to show that $cos(x+y)=cos(x)cos(y)−sin(x)sin(y)$ by using power series?Show that the equation $cos(sin x)=sin(cos x)$ has no real solutions.Solve for $sin^2(x) = 3cos^2(x)$Convergence of $sum_n |fraccos(3^n)n|$Proof: $sinleft(fracx-x_02right)sinleft(fracx-x_12right) = frac12cos fracx_1-x_02 - frac12cos(x-fracx_1+x_02)$How does one show sin(x) is bounded using the power series?Using the IVP definition of $cos$ and $sin$, how can we show that $cos^2(x)+sin^2(x) = 1$ without any “magic”?Transition from Introductory Proofs/Logic Course to the Proofs in Rudin's Principles of AnalysisHow can we show $cos^6x+sin^6x=1-3sin^2x cos^2x$?Is there a natural way to prove trig identities also hold for complex numbers?General solution of $(sqrt3 - 1)sintheta + (sqrt3 + 1)costheta =2 $
$begingroup$
Define $sin x $ and $cos x$ via their infinite series:
$$
sin x = sum_n (-1)^nfracx^2n+1(2n+1)!, qquad
cos x = sum_n (-1)^n fracx^2n(2n)!.
$$
Is there a short, clever proof that $cos(x+y) = cos x cos y - sin x sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = cos y$, $f'(0) = - sin y$. Does anyone know other (preferably slick!) proofs?
real-analysis trigonometry
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
$endgroup$
|
show 2 more comments
$begingroup$
Define $sin x $ and $cos x$ via their infinite series:
$$
sin x = sum_n (-1)^nfracx^2n+1(2n+1)!, qquad
cos x = sum_n (-1)^n fracx^2n(2n)!.
$$
Is there a short, clever proof that $cos(x+y) = cos x cos y - sin x sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = cos y$, $f'(0) = - sin y$. Does anyone know other (preferably slick!) proofs?
real-analysis trigonometry
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
$endgroup$
$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
2
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39
|
show 2 more comments
$begingroup$
Define $sin x $ and $cos x$ via their infinite series:
$$
sin x = sum_n (-1)^nfracx^2n+1(2n+1)!, qquad
cos x = sum_n (-1)^n fracx^2n(2n)!.
$$
Is there a short, clever proof that $cos(x+y) = cos x cos y - sin x sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = cos y$, $f'(0) = - sin y$. Does anyone know other (preferably slick!) proofs?
real-analysis trigonometry
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
$endgroup$
Define $sin x $ and $cos x$ via their infinite series:
$$
sin x = sum_n (-1)^nfracx^2n+1(2n+1)!, qquad
cos x = sum_n (-1)^n fracx^2n(2n)!.
$$
Is there a short, clever proof that $cos(x+y) = cos x cos y - sin x sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = cos y$, $f'(0) = - sin y$. Does anyone know other (preferably slick!) proofs?
real-analysis trigonometry
real-analysis trigonometry
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
asked Apr 2 '13 at 20:07
Umberto P.Umberto P.
39.9k13267
39.9k13267
$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
2
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39
|
show 2 more comments
$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
2
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39
$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
2
2
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39
|
show 2 more comments
8 Answers
8
active
oldest
votes
$begingroup$
New answer to an old question. This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define
$$f(t)=cos(x+y-t)cos(t)-sin(x+y-t)sin(t).$$
Verify $f'(t)=0$, hence $f$ is constant. Now the desired angle sum identity follows from $f(0)=f(y)$
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
One way is to use the fact that
$$cos(theta) = dfrace^i theta+e^-i theta2$$
$$cos(x+y) = dfrace^i(x+y)+e^-i(x+y)2 = left(dfrace^ix+e^-ix2 right) left(dfrace^iy+e^-iy2right) - left(dfrace^ix-e^-ix2i right) left(dfrace^iy-e^-iy2iright)$$
$endgroup$
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
add a comment |
$begingroup$
This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $fequiv 0$. $blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=asin+bcos $$
P Let $g=f-asin+bcos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $gequiv 0$, so that $f=asin+bcos$. $blacktriangle$.
Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–sin y$, $f''(0)=cos y$. Then it must coincide with the unique solution $$f'(0) sin+f(0)cos$$
$endgroup$
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
add a comment |
$begingroup$
The way I learned it as a kid was geometric, and probably looked like the proof seen here on Wikipedia.
The segment $OP$ has length $1$. We have the $sin(alpha + beta) = PB = PR + RB = cos(alpha) sin(beta) + sin(alpha) cos(beta)$.
Then, to prove the cosine identity we can use that $cos(alpha + beta) = sin(alpha + beta + pi/2)$ and use the sine identity.
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
$endgroup$
add a comment |
$begingroup$
I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^i(x + y) = e^ix cdot e^iy.$$
Using the complex multiplication rule $operatornameRe(ab) = operatornameRe(a)operatornameRe(b) - operatornameIm(a)operatornameIm(b)$, taking the real part gives
$$operatornameRe(e^i(x + y)) = operatornameRe(e^ix)operatornameRe(e^iy) - operatornameIm(e^ix)operatornameIm(e^iy).$$
So by $cos(x) = operatornameRe(e^ix)$ and $sin(x) = operatornameIm(e^ix)$
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y).$$
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
$endgroup$
add a comment |
$begingroup$
Let $vecu,vecvinmathbbR^2$ unitary vectors such that
$$
vecu=big(cos(x),sin(x)big)quad mbox and quad vecv=big(cos(-y),sin(-y)big)
$$
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $vecu$ and $vecv$ respectively. Then
beginalign
cosbig( x+ybig) = & cosbig( x-(-y)big)\
= & fracvecubulletvecvcdot \
= & vecubulletvecv\
= & cos(x)cdotcos(-y)+sin(x)cdotsin(-y)\
= & cos(x)cdotcos(y)-sin(x)cdotsin(y)\
endalign
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
$endgroup$
add a comment |
$begingroup$
For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $sin$ and $cos$ have been defined by their Taylor series.
Apply $fracd^2dx^2$ to each side (viewing $y$ as some constant), and you see that each side a solution to $fracd^2dx^2f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
$endgroup$
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
add a comment |
$begingroup$
$$beginarray rcl
cos(x + y) + i sin(x + y)& = & e^i(x + y) \
&=& e^ixe^iy \
&=& (cos(x) + isin(x))(cos(y) + isin(y)) \
&=& (cos(x)cos(y) - sin(x)sin(y)) + i(sin(x)cos(y) + sin(y)cos(x)) \
endarray$$
Equating real and imaginary parts you get
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$$
$$sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
$endgroup$
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
New answer to an old question. This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define
$$f(t)=cos(x+y-t)cos(t)-sin(x+y-t)sin(t).$$
Verify $f'(t)=0$, hence $f$ is constant. Now the desired angle sum identity follows from $f(0)=f(y)$
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
New answer to an old question. This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define
$$f(t)=cos(x+y-t)cos(t)-sin(x+y-t)sin(t).$$
Verify $f'(t)=0$, hence $f$ is constant. Now the desired angle sum identity follows from $f(0)=f(y)$
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
New answer to an old question. This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define
$$f(t)=cos(x+y-t)cos(t)-sin(x+y-t)sin(t).$$
Verify $f'(t)=0$, hence $f$ is constant. Now the desired angle sum identity follows from $f(0)=f(y)$
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
New answer to an old question. This one, maybe the slickiest of them all, is due to Erhard Schmidt.
Define
$$f(t)=cos(x+y-t)cos(t)-sin(x+y-t)sin(t).$$
Verify $f'(t)=0$, hence $f$ is constant. Now the desired angle sum identity follows from $f(0)=f(y)$
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
answered Nov 13 '18 at 12:03
Michael HoppeMichael Hoppe
11.2k31837
11.2k31837
add a comment |
add a comment |
$begingroup$
One way is to use the fact that
$$cos(theta) = dfrace^i theta+e^-i theta2$$
$$cos(x+y) = dfrace^i(x+y)+e^-i(x+y)2 = left(dfrace^ix+e^-ix2 right) left(dfrace^iy+e^-iy2right) - left(dfrace^ix-e^-ix2i right) left(dfrace^iy-e^-iy2iright)$$
$endgroup$
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
add a comment |
$begingroup$
One way is to use the fact that
$$cos(theta) = dfrace^i theta+e^-i theta2$$
$$cos(x+y) = dfrace^i(x+y)+e^-i(x+y)2 = left(dfrace^ix+e^-ix2 right) left(dfrace^iy+e^-iy2right) - left(dfrace^ix-e^-ix2i right) left(dfrace^iy-e^-iy2iright)$$
$endgroup$
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
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– Umberto P.
Apr 3 '13 at 1:49
add a comment |
$begingroup$
One way is to use the fact that
$$cos(theta) = dfrace^i theta+e^-i theta2$$
$$cos(x+y) = dfrace^i(x+y)+e^-i(x+y)2 = left(dfrace^ix+e^-ix2 right) left(dfrace^iy+e^-iy2right) - left(dfrace^ix-e^-ix2i right) left(dfrace^iy-e^-iy2iright)$$
$endgroup$
One way is to use the fact that
$$cos(theta) = dfrace^i theta+e^-i theta2$$
$$cos(x+y) = dfrace^i(x+y)+e^-i(x+y)2 = left(dfrace^ix+e^-ix2 right) left(dfrace^iy+e^-iy2right) - left(dfrace^ix-e^-ix2i right) left(dfrace^iy-e^-iy2iright)$$
answered Apr 2 '13 at 20:14
user17762
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
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– Umberto P.
Apr 2 '13 at 20:20
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I don't believe it can. Are you looking for, say, an elementary geometry proof?
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– Joe Z.
Apr 2 '13 at 20:21
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look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
add a comment |
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
Can $e^imath x$ be defined without series? The proofs that $e^imath(x+y) = e^imath xe^imath y$ seem to involve product series or uniqueness of ODE, which I'm seeing if I can avoid.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:20
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
I don't believe it can. Are you looking for, say, an elementary geometry proof?
$endgroup$
– Joe Z.
Apr 2 '13 at 20:21
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
look at Mertens' theorem en.wikipedia.org/wiki/Cauchy_product and you need only case when both series are absolutely convergent, which should be easier to prove. If I can remember right you will need that absolutely convergent series can be summed in any order.
$endgroup$
– tom
Apr 2 '13 at 21:52
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
$begingroup$
@tom: I know the product series proof as stated in the question. I'm looking for novel proofs.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:49
add a comment |
$begingroup$
This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $fequiv 0$. $blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=asin+bcos $$
P Let $g=f-asin+bcos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $gequiv 0$, so that $f=asin+bcos$. $blacktriangle$.
Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–sin y$, $f''(0)=cos y$. Then it must coincide with the unique solution $$f'(0) sin+f(0)cos$$
$endgroup$
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
add a comment |
$begingroup$
This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $fequiv 0$. $blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=asin+bcos $$
P Let $g=f-asin+bcos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $gequiv 0$, so that $f=asin+bcos$. $blacktriangle$.
Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–sin y$, $f''(0)=cos y$. Then it must coincide with the unique solution $$f'(0) sin+f(0)cos$$
$endgroup$
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
add a comment |
$begingroup$
This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $fequiv 0$. $blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=asin+bcos $$
P Let $g=f-asin+bcos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $gequiv 0$, so that $f=asin+bcos$. $blacktriangle$.
Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–sin y$, $f''(0)=cos y$. Then it must coincide with the unique solution $$f'(0) sin+f(0)cos$$
$endgroup$
This was addressed in the question already. I leave it so that the method is fully explained.
LEMMA Let $f$ be a function with second derivative everywhere such that $f''+f=0$ and $f'(0)=0$; $f(0)=0$. Then $f$ is identically zero everywhere.
P We have that $$f''+f=0$$ Then $$f'f''+ff'=0$$ or $$(f')^2+f^2=C$$
But the initial conditions force $f'^2+f^2=0$ everywhere, which means $fequiv 0$. $blacktriangle$.
PROP Let $f$ be a function with second derivative everywhere such that $f''+f=0$, and $f'(0)=a$, $f(0)=b$. Then $$f=asin+bcos $$
P Let $g=f-asin+bcos$. Then $g''+g=0$ and $g'(0)=0$, $g(0)=0$. The lemma implies $gequiv 0$, so that $f=asin+bcos$. $blacktriangle$.
Differentiate with respect to one variable and use the uniqueness of the solution of a second degree ODE with initial conditions.
That is, your cosine on the left vetifies $$f''+f=0$$ and $f'(0)=–sin y$, $f''(0)=cos y$. Then it must coincide with the unique solution $$f'(0) sin+f(0)cos$$
edited Apr 3 '13 at 17:12
community wiki
4 revs
Peter Tamaroff
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
add a comment |
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
This solution was addressed in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:55
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
$begingroup$
@UmbertoP. I just realized.
$endgroup$
– Pedro Tamaroff♦
Apr 3 '13 at 17:07
add a comment |
$begingroup$
The way I learned it as a kid was geometric, and probably looked like the proof seen here on Wikipedia.
The segment $OP$ has length $1$. We have the $sin(alpha + beta) = PB = PR + RB = cos(alpha) sin(beta) + sin(alpha) cos(beta)$.
Then, to prove the cosine identity we can use that $cos(alpha + beta) = sin(alpha + beta + pi/2)$ and use the sine identity.
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
$endgroup$
add a comment |
$begingroup$
The way I learned it as a kid was geometric, and probably looked like the proof seen here on Wikipedia.
The segment $OP$ has length $1$. We have the $sin(alpha + beta) = PB = PR + RB = cos(alpha) sin(beta) + sin(alpha) cos(beta)$.
Then, to prove the cosine identity we can use that $cos(alpha + beta) = sin(alpha + beta + pi/2)$ and use the sine identity.
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
$endgroup$
add a comment |
$begingroup$
The way I learned it as a kid was geometric, and probably looked like the proof seen here on Wikipedia.
The segment $OP$ has length $1$. We have the $sin(alpha + beta) = PB = PR + RB = cos(alpha) sin(beta) + sin(alpha) cos(beta)$.
Then, to prove the cosine identity we can use that $cos(alpha + beta) = sin(alpha + beta + pi/2)$ and use the sine identity.
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
$endgroup$
The way I learned it as a kid was geometric, and probably looked like the proof seen here on Wikipedia.
The segment $OP$ has length $1$. We have the $sin(alpha + beta) = PB = PR + RB = cos(alpha) sin(beta) + sin(alpha) cos(beta)$.
Then, to prove the cosine identity we can use that $cos(alpha + beta) = sin(alpha + beta + pi/2)$ and use the sine identity.
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
answered Apr 2 '13 at 20:22
Christopher A. WongChristopher A. Wong
16.6k33159
16.6k33159
add a comment |
add a comment |
$begingroup$
I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^i(x + y) = e^ix cdot e^iy.$$
Using the complex multiplication rule $operatornameRe(ab) = operatornameRe(a)operatornameRe(b) - operatornameIm(a)operatornameIm(b)$, taking the real part gives
$$operatornameRe(e^i(x + y)) = operatornameRe(e^ix)operatornameRe(e^iy) - operatornameIm(e^ix)operatornameIm(e^iy).$$
So by $cos(x) = operatornameRe(e^ix)$ and $sin(x) = operatornameIm(e^ix)$
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y).$$
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
$endgroup$
add a comment |
$begingroup$
I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^i(x + y) = e^ix cdot e^iy.$$
Using the complex multiplication rule $operatornameRe(ab) = operatornameRe(a)operatornameRe(b) - operatornameIm(a)operatornameIm(b)$, taking the real part gives
$$operatornameRe(e^i(x + y)) = operatornameRe(e^ix)operatornameRe(e^iy) - operatornameIm(e^ix)operatornameIm(e^iy).$$
So by $cos(x) = operatornameRe(e^ix)$ and $sin(x) = operatornameIm(e^ix)$
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y).$$
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
$endgroup$
add a comment |
$begingroup$
I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^i(x + y) = e^ix cdot e^iy.$$
Using the complex multiplication rule $operatornameRe(ab) = operatornameRe(a)operatornameRe(b) - operatornameIm(a)operatornameIm(b)$, taking the real part gives
$$operatornameRe(e^i(x + y)) = operatornameRe(e^ix)operatornameRe(e^iy) - operatornameIm(e^ix)operatornameIm(e^iy).$$
So by $cos(x) = operatornameRe(e^ix)$ and $sin(x) = operatornameIm(e^ix)$
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y).$$
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
$endgroup$
I always had a hard time to memorize that formula.
But actually, that's not really needed, because there it is an easy way to reconstruct it from the from the laws of exponentiation applied to complex exponentiation:
$$e^i(x + y) = e^ix cdot e^iy.$$
Using the complex multiplication rule $operatornameRe(ab) = operatornameRe(a)operatornameRe(b) - operatornameIm(a)operatornameIm(b)$, taking the real part gives
$$operatornameRe(e^i(x + y)) = operatornameRe(e^ix)operatornameRe(e^iy) - operatornameIm(e^ix)operatornameIm(e^iy).$$
So by $cos(x) = operatornameRe(e^ix)$ and $sin(x) = operatornameIm(e^ix)$
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y).$$
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
edited Nov 2 '13 at 11:32
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
answered May 7 '13 at 12:12
azimutazimut
16.5k1052101
16.5k1052101
add a comment |
add a comment |
$begingroup$
Let $vecu,vecvinmathbbR^2$ unitary vectors such that
$$
vecu=big(cos(x),sin(x)big)quad mbox and quad vecv=big(cos(-y),sin(-y)big)
$$
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $vecu$ and $vecv$ respectively. Then
beginalign
cosbig( x+ybig) = & cosbig( x-(-y)big)\
= & fracvecubulletvecvcdot \
= & vecubulletvecv\
= & cos(x)cdotcos(-y)+sin(x)cdotsin(-y)\
= & cos(x)cdotcos(y)-sin(x)cdotsin(y)\
endalign
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
$endgroup$
add a comment |
$begingroup$
Let $vecu,vecvinmathbbR^2$ unitary vectors such that
$$
vecu=big(cos(x),sin(x)big)quad mbox and quad vecv=big(cos(-y),sin(-y)big)
$$
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $vecu$ and $vecv$ respectively. Then
beginalign
cosbig( x+ybig) = & cosbig( x-(-y)big)\
= & fracvecubulletvecvcdot \
= & vecubulletvecv\
= & cos(x)cdotcos(-y)+sin(x)cdotsin(-y)\
= & cos(x)cdotcos(y)-sin(x)cdotsin(y)\
endalign
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
$endgroup$
add a comment |
$begingroup$
Let $vecu,vecvinmathbbR^2$ unitary vectors such that
$$
vecu=big(cos(x),sin(x)big)quad mbox and quad vecv=big(cos(-y),sin(-y)big)
$$
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $vecu$ and $vecv$ respectively. Then
beginalign
cosbig( x+ybig) = & cosbig( x-(-y)big)\
= & fracvecubulletvecvcdot \
= & vecubulletvecv\
= & cos(x)cdotcos(-y)+sin(x)cdotsin(-y)\
= & cos(x)cdotcos(y)-sin(x)cdotsin(y)\
endalign
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
$endgroup$
Let $vecu,vecvinmathbbR^2$ unitary vectors such that
$$
vecu=big(cos(x),sin(x)big)quad mbox and quad vecv=big(cos(-y),sin(-y)big)
$$
Here $x$ and $-y$ are the smallest angle formed between the x-axis and the vectors $vecu$ and $vecv$ respectively. Then
beginalign
cosbig( x+ybig) = & cosbig( x-(-y)big)\
= & fracvecubulletvecvcdot \
= & vecubulletvecv\
= & cos(x)cdotcos(-y)+sin(x)cdotsin(-y)\
= & cos(x)cdotcos(y)-sin(x)cdotsin(y)\
endalign
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
edited Nov 2 '13 at 15:33
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
answered Nov 2 '13 at 11:49
MathOverviewMathOverview
8,95043164
8,95043164
add a comment |
add a comment |
$begingroup$
For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $sin$ and $cos$ have been defined by their Taylor series.
Apply $fracd^2dx^2$ to each side (viewing $y$ as some constant), and you see that each side a solution to $fracd^2dx^2f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
$endgroup$
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
add a comment |
$begingroup$
For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $sin$ and $cos$ have been defined by their Taylor series.
Apply $fracd^2dx^2$ to each side (viewing $y$ as some constant), and you see that each side a solution to $fracd^2dx^2f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
$endgroup$
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
add a comment |
$begingroup$
For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $sin$ and $cos$ have been defined by their Taylor series.
Apply $fracd^2dx^2$ to each side (viewing $y$ as some constant), and you see that each side a solution to $fracd^2dx^2f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
$endgroup$
For variety, here is a different proof. Unfortunately it might be considered circular since it relies on differentiation of trig functions. But maybe you know their derivatives without using this identity, if say $sin$ and $cos$ have been defined by their Taylor series.
Apply $fracd^2dx^2$ to each side (viewing $y$ as some constant), and you see that each side a solution to $fracd^2dx^2f(x)=-f(x)$
Both sides are in agreement at $x=-y$. Also the first derivatives of each side are in agreement at $x=-y$. Therefore they are the same expression.
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
answered Apr 2 '13 at 21:39
alex.jordanalex.jordan
39.5k560122
39.5k560122
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
add a comment |
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
$begingroup$
That solution was mentioned in the question.
$endgroup$
– Umberto P.
Apr 3 '13 at 1:48
add a comment |
$begingroup$
$$beginarray rcl
cos(x + y) + i sin(x + y)& = & e^i(x + y) \
&=& e^ixe^iy \
&=& (cos(x) + isin(x))(cos(y) + isin(y)) \
&=& (cos(x)cos(y) - sin(x)sin(y)) + i(sin(x)cos(y) + sin(y)cos(x)) \
endarray$$
Equating real and imaginary parts you get
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$$
$$sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
$endgroup$
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
add a comment |
$begingroup$
$$beginarray rcl
cos(x + y) + i sin(x + y)& = & e^i(x + y) \
&=& e^ixe^iy \
&=& (cos(x) + isin(x))(cos(y) + isin(y)) \
&=& (cos(x)cos(y) - sin(x)sin(y)) + i(sin(x)cos(y) + sin(y)cos(x)) \
endarray$$
Equating real and imaginary parts you get
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$$
$$sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
$endgroup$
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
add a comment |
$begingroup$
$$beginarray rcl
cos(x + y) + i sin(x + y)& = & e^i(x + y) \
&=& e^ixe^iy \
&=& (cos(x) + isin(x))(cos(y) + isin(y)) \
&=& (cos(x)cos(y) - sin(x)sin(y)) + i(sin(x)cos(y) + sin(y)cos(x)) \
endarray$$
Equating real and imaginary parts you get
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$$
$$sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
$endgroup$
$$beginarray rcl
cos(x + y) + i sin(x + y)& = & e^i(x + y) \
&=& e^ixe^iy \
&=& (cos(x) + isin(x))(cos(y) + isin(y)) \
&=& (cos(x)cos(y) - sin(x)sin(y)) + i(sin(x)cos(y) + sin(y)cos(x)) \
endarray$$
Equating real and imaginary parts you get
$$cos(x + y) = cos(x)cos(y) - sin(x)sin(y)$$
$$sin(x + y) = sin(x)cos(y) + sin(y)cos(x)$$
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
answered Mar 13 at 9:38
DanielVDanielV
18.1k42755
18.1k42755
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
add a comment |
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
$begingroup$
Thanks for the answer - this question is nearly 6 years old. At the time I was interested in proving the trig angle-sum formulas using nothing but the series definition of sine and cosine. The issue is that the identity $cos x + i sin x = e^ix$ is outside the scope of the question, and the justification of the further identity $e^w+z = e^w e^z$ is requires tools that essentially prove what was being asked in the first place.
$endgroup$
– Umberto P.
Mar 13 at 18:59
add a comment |
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$begingroup$
There's probably a proof involving complex numbers.
$endgroup$
– Joe Z.
Apr 2 '13 at 20:12
$begingroup$
Not exactly what you want, but still, you might be interested in the geometric proof.
$endgroup$
– dtldarek
Apr 2 '13 at 20:24
$begingroup$
Allow me to clarify. I'm wondering if there is a proof using real analytic methods (e.g. power series). I'm not assuming anything about $sin$ and $cos$ other than what can be derived from their definitions as series.
$endgroup$
– Umberto P.
Apr 2 '13 at 20:25
2
$begingroup$
I happen to have answered a very similar question today. You will find there in particular a proof of the two Addition Laws, rather condensed, using the series definition.
$endgroup$
– André Nicolas
Apr 2 '13 at 20:46
$begingroup$
ProofWiki features a geometric proof. As a PW herald, I'd like to invite any users to add their proofs (as far as they're different from the present quadruple) to it as well!
$endgroup$
– Lord_Farin
Apr 2 '13 at 21:39