Proving sine of sum identity for all angles
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Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?
$$sin(a+b) = sin(a) cos(b) + cos(a) sin(b)$$
Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.
The other derivation I know is using Euler's formula, namely this one.
There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?
trigonometry
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
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add a comment |
$begingroup$
Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?
$$sin(a+b) = sin(a) cos(b) + cos(a) sin(b)$$
Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.
The other derivation I know is using Euler's formula, namely this one.
There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?
trigonometry
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
$endgroup$
$begingroup$
What is your definition of sine and cosine?
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– David H
Jan 17 '15 at 23:52
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The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
$endgroup$
– Orest Bucicovschi
Jan 17 '15 at 23:53
$begingroup$
I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
$endgroup$
– servabat
Jan 17 '15 at 23:59
$begingroup$
How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
$endgroup$
– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
$endgroup$
– David K
Jan 18 '15 at 14:06
add a comment |
$begingroup$
Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?
$$sin(a+b) = sin(a) cos(b) + cos(a) sin(b)$$
Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.
The other derivation I know is using Euler's formula, namely this one.
There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?
trigonometry
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
$endgroup$
Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?
$$sin(a+b) = sin(a) cos(b) + cos(a) sin(b)$$
Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.
The other derivation I know is using Euler's formula, namely this one.
There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?
trigonometry
trigonometry
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
edited Apr 13 '17 at 12:21
Community♦
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
asked Jan 17 '15 at 23:43
user4205580user4205580
4611131
4611131
$begingroup$
What is your definition of sine and cosine?
$endgroup$
– David H
Jan 17 '15 at 23:52
$begingroup$
The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
$endgroup$
– Orest Bucicovschi
Jan 17 '15 at 23:53
$begingroup$
I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
$endgroup$
– servabat
Jan 17 '15 at 23:59
$begingroup$
How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
$endgroup$
– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
$endgroup$
– David K
Jan 18 '15 at 14:06
add a comment |
$begingroup$
What is your definition of sine and cosine?
$endgroup$
– David H
Jan 17 '15 at 23:52
$begingroup$
The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
$endgroup$
– Orest Bucicovschi
Jan 17 '15 at 23:53
$begingroup$
I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
$endgroup$
– servabat
Jan 17 '15 at 23:59
$begingroup$
How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
$endgroup$
– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
$endgroup$
– David K
Jan 18 '15 at 14:06
$begingroup$
What is your definition of sine and cosine?
$endgroup$
– David H
Jan 17 '15 at 23:52
$begingroup$
What is your definition of sine and cosine?
$endgroup$
– David H
Jan 17 '15 at 23:52
$begingroup$
The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
$endgroup$
– Orest Bucicovschi
Jan 17 '15 at 23:53
$begingroup$
The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
$endgroup$
– Orest Bucicovschi
Jan 17 '15 at 23:53
$begingroup$
I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
$endgroup$
– servabat
Jan 17 '15 at 23:59
$begingroup$
I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
$endgroup$
– servabat
Jan 17 '15 at 23:59
$begingroup$
How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
$endgroup$
– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
$endgroup$
– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
$endgroup$
– David K
Jan 18 '15 at 14:06
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
$endgroup$
– David K
Jan 18 '15 at 14:06
add a comment |
3 Answers
3
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here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(cos t, sin t)$ is $2(1-cos t)$ using the distance formula. now reinterpret
$$text{ length of chord making an angle $t$ at the center is } 2 - 2cos t $$
now compute the length squared between $cos t, sin t), (cos s, sin s)$ in two different ways:
(i) distance formula gives you $2 - cos t cos s - sin t sin s$
(ii) chord making an angle $t - s$ is $2 - cos(t-s)$
equating the two gives you $$cos (t-s) = cos t cos s + sin t sin s tag 1$$
now use the fact $cos pi/2$ to derive $cos (pi/2 - s) = sin s$ by putting $t = pi/2$ in $(1)$
put $t=0,$ to derive $cos$ is an even function. put $t = -pi/2,$ to show $sin$ is an odd function. after all these you derive
$$sin(t-s) = sin t cos t - cos t sin s $$ and two for the sums.
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
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add a comment |
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You can prove $$e^{ix}=cos(x)+isin(x)$$without using trig sum identities. e.g. let:$$y=cos(x)+isin(x)tag{1}$$$$therefore frac{dy}{dx}=-sin(x)+icos(x)=iy$$$$therefore intfrac{1}{y}dy=int idx$$$$therefore ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$ln(y)=ix$$$$therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
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add a comment |
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Euler’s Formula...
$$ sin(x) = mbox{Im}(e^{ix}) = frac{e^{ix} - e^{-ix}}{2i} $$
$$ sin(a+b) = mbox{Im}(e^{(a+b)i}) = mbox{Im}(e^{ai} * e^{bi}) $$
$$ = frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = frac{((cos(a)+isin(a))(cos(b)+isin(b)) - ((cos(-a)+isin(-a))(cos(-b)+isin(-b))}{2i} $$
After some thorough simplifying...
$$ = frac{(2isin(a)cos(b) + 2icos(a)sin(b))}{2i} $$
$$ sin(a+b) = sin(a)cos(b) + cos(a)sin(b) $$
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(cos t, sin t)$ is $2(1-cos t)$ using the distance formula. now reinterpret
$$text{ length of chord making an angle $t$ at the center is } 2 - 2cos t $$
now compute the length squared between $cos t, sin t), (cos s, sin s)$ in two different ways:
(i) distance formula gives you $2 - cos t cos s - sin t sin s$
(ii) chord making an angle $t - s$ is $2 - cos(t-s)$
equating the two gives you $$cos (t-s) = cos t cos s + sin t sin s tag 1$$
now use the fact $cos pi/2$ to derive $cos (pi/2 - s) = sin s$ by putting $t = pi/2$ in $(1)$
put $t=0,$ to derive $cos$ is an even function. put $t = -pi/2,$ to show $sin$ is an odd function. after all these you derive
$$sin(t-s) = sin t cos t - cos t sin s $$ and two for the sums.
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
$endgroup$
add a comment |
$begingroup$
here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(cos t, sin t)$ is $2(1-cos t)$ using the distance formula. now reinterpret
$$text{ length of chord making an angle $t$ at the center is } 2 - 2cos t $$
now compute the length squared between $cos t, sin t), (cos s, sin s)$ in two different ways:
(i) distance formula gives you $2 - cos t cos s - sin t sin s$
(ii) chord making an angle $t - s$ is $2 - cos(t-s)$
equating the two gives you $$cos (t-s) = cos t cos s + sin t sin s tag 1$$
now use the fact $cos pi/2$ to derive $cos (pi/2 - s) = sin s$ by putting $t = pi/2$ in $(1)$
put $t=0,$ to derive $cos$ is an even function. put $t = -pi/2,$ to show $sin$ is an odd function. after all these you derive
$$sin(t-s) = sin t cos t - cos t sin s $$ and two for the sums.
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
$endgroup$
add a comment |
$begingroup$
here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(cos t, sin t)$ is $2(1-cos t)$ using the distance formula. now reinterpret
$$text{ length of chord making an angle $t$ at the center is } 2 - 2cos t $$
now compute the length squared between $cos t, sin t), (cos s, sin s)$ in two different ways:
(i) distance formula gives you $2 - cos t cos s - sin t sin s$
(ii) chord making an angle $t - s$ is $2 - cos(t-s)$
equating the two gives you $$cos (t-s) = cos t cos s + sin t sin s tag 1$$
now use the fact $cos pi/2$ to derive $cos (pi/2 - s) = sin s$ by putting $t = pi/2$ in $(1)$
put $t=0,$ to derive $cos$ is an even function. put $t = -pi/2,$ to show $sin$ is an odd function. after all these you derive
$$sin(t-s) = sin t cos t - cos t sin s $$ and two for the sums.
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
$endgroup$
here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(cos t, sin t)$ is $2(1-cos t)$ using the distance formula. now reinterpret
$$text{ length of chord making an angle $t$ at the center is } 2 - 2cos t $$
now compute the length squared between $cos t, sin t), (cos s, sin s)$ in two different ways:
(i) distance formula gives you $2 - cos t cos s - sin t sin s$
(ii) chord making an angle $t - s$ is $2 - cos(t-s)$
equating the two gives you $$cos (t-s) = cos t cos s + sin t sin s tag 1$$
now use the fact $cos pi/2$ to derive $cos (pi/2 - s) = sin s$ by putting $t = pi/2$ in $(1)$
put $t=0,$ to derive $cos$ is an even function. put $t = -pi/2,$ to show $sin$ is an odd function. after all these you derive
$$sin(t-s) = sin t cos t - cos t sin s $$ and two for the sums.
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
edited Jan 18 '15 at 13:43
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
answered Jan 18 '15 at 0:46
abelabel
26.6k12048
26.6k12048
add a comment |
add a comment |
$begingroup$
You can prove $$e^{ix}=cos(x)+isin(x)$$without using trig sum identities. e.g. let:$$y=cos(x)+isin(x)tag{1}$$$$therefore frac{dy}{dx}=-sin(x)+icos(x)=iy$$$$therefore intfrac{1}{y}dy=int idx$$$$therefore ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$ln(y)=ix$$$$therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
$endgroup$
add a comment |
$begingroup$
You can prove $$e^{ix}=cos(x)+isin(x)$$without using trig sum identities. e.g. let:$$y=cos(x)+isin(x)tag{1}$$$$therefore frac{dy}{dx}=-sin(x)+icos(x)=iy$$$$therefore intfrac{1}{y}dy=int idx$$$$therefore ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$ln(y)=ix$$$$therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
$endgroup$
add a comment |
$begingroup$
You can prove $$e^{ix}=cos(x)+isin(x)$$without using trig sum identities. e.g. let:$$y=cos(x)+isin(x)tag{1}$$$$therefore frac{dy}{dx}=-sin(x)+icos(x)=iy$$$$therefore intfrac{1}{y}dy=int idx$$$$therefore ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$ln(y)=ix$$$$therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
$endgroup$
You can prove $$e^{ix}=cos(x)+isin(x)$$without using trig sum identities. e.g. let:$$y=cos(x)+isin(x)tag{1}$$$$therefore frac{dy}{dx}=-sin(x)+icos(x)=iy$$$$therefore intfrac{1}{y}dy=int idx$$$$therefore ln(y)=ix+C$$and we can show $C=0$ because from (1) $y=1$ when $x=0$, therefore:$$ln(y)=ix$$$$therefore y=e^{ix}$$that therefore removes the circular argument you mentioned in yur question.
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
answered Jan 17 '15 at 23:51
MufasaMufasa
5,03311323
5,03311323
add a comment |
add a comment |
$begingroup$
Euler’s Formula...
$$ sin(x) = mbox{Im}(e^{ix}) = frac{e^{ix} - e^{-ix}}{2i} $$
$$ sin(a+b) = mbox{Im}(e^{(a+b)i}) = mbox{Im}(e^{ai} * e^{bi}) $$
$$ = frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = frac{((cos(a)+isin(a))(cos(b)+isin(b)) - ((cos(-a)+isin(-a))(cos(-b)+isin(-b))}{2i} $$
After some thorough simplifying...
$$ = frac{(2isin(a)cos(b) + 2icos(a)sin(b))}{2i} $$
$$ sin(a+b) = sin(a)cos(b) + cos(a)sin(b) $$
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
$endgroup$
add a comment |
$begingroup$
Euler’s Formula...
$$ sin(x) = mbox{Im}(e^{ix}) = frac{e^{ix} - e^{-ix}}{2i} $$
$$ sin(a+b) = mbox{Im}(e^{(a+b)i}) = mbox{Im}(e^{ai} * e^{bi}) $$
$$ = frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = frac{((cos(a)+isin(a))(cos(b)+isin(b)) - ((cos(-a)+isin(-a))(cos(-b)+isin(-b))}{2i} $$
After some thorough simplifying...
$$ = frac{(2isin(a)cos(b) + 2icos(a)sin(b))}{2i} $$
$$ sin(a+b) = sin(a)cos(b) + cos(a)sin(b) $$
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
$endgroup$
add a comment |
$begingroup$
Euler’s Formula...
$$ sin(x) = mbox{Im}(e^{ix}) = frac{e^{ix} - e^{-ix}}{2i} $$
$$ sin(a+b) = mbox{Im}(e^{(a+b)i}) = mbox{Im}(e^{ai} * e^{bi}) $$
$$ = frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = frac{((cos(a)+isin(a))(cos(b)+isin(b)) - ((cos(-a)+isin(-a))(cos(-b)+isin(-b))}{2i} $$
After some thorough simplifying...
$$ = frac{(2isin(a)cos(b) + 2icos(a)sin(b))}{2i} $$
$$ sin(a+b) = sin(a)cos(b) + cos(a)sin(b) $$
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
$endgroup$
Euler’s Formula...
$$ sin(x) = mbox{Im}(e^{ix}) = frac{e^{ix} - e^{-ix}}{2i} $$
$$ sin(a+b) = mbox{Im}(e^{(a+b)i}) = mbox{Im}(e^{ai} * e^{bi}) $$
$$ = frac{e^{(a+b)i} - e^{-(a+b)i}}{2i} = frac{(e^{ai} * e^{bi}) - (e^{-ai} * e^{-bi})}{2i} $$
$$ = frac{((cos(a)+isin(a))(cos(b)+isin(b)) - ((cos(-a)+isin(-a))(cos(-b)+isin(-b))}{2i} $$
After some thorough simplifying...
$$ = frac{(2isin(a)cos(b) + 2icos(a)sin(b))}{2i} $$
$$ sin(a+b) = sin(a)cos(b) + cos(a)sin(b) $$
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
edited Feb 11 at 15:25
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
answered Dec 31 '18 at 19:06
Michael LeeMichael Lee
5618
5618
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$begingroup$
What is your definition of sine and cosine?
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– David H
Jan 17 '15 at 23:52
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The answer may depend on your definition of $sin$, $cos$. If you use the definition of $e^{ia}$ to be $cos a + i sin a$, then yes, the argument is circular. What is $e^z$? What is $cos x$, $sin x$? This needs to be sorted out.
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– Orest Bucicovschi
Jan 17 '15 at 23:53
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I would use crossproduct but I guess it's the geometric approach (else this can be of course proven with analytical tools : derivatives and stuffs).
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– servabat
Jan 17 '15 at 23:59
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How many definitions of sine and cosine are there? Let $P=(x,y)$ be a point on cartesian plane, $r$ is the distance from the origin in the straight line. Then sine of the angle between $|OP|$ and positive $X$ is $frac{y}{r}$ and cosine - $frac{x}{r}$.
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– user4205580
Jan 18 '15 at 8:31
$begingroup$
How many definitions? For any really interesting mathematical system, it is very likely that different people have constructed the definitions of that system in different (but mutually consistent) ways. Usually the $theta$ in $sintheta$ and $costheta$ is a number obtained in some way (such as the length of an arc of a unit circle). If it is defined as a particular geometric figure that must include the positive $x$-axis, then you have to define what you mean by "any angle" and "adding two angles".
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– David K
Jan 18 '15 at 14:06