Why does $sin(x) - sin(y)=2 cos(fracx+y2) sin(fracx-y2)$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do I integrate $frac sin^3xcos^2x$How to evaluate the limit $lim_xto 0 frac1-cos(4x)sin^2(7x)$Prove $(1 + sin x + cos x -cos^2x sin x + sin^2x cos x )/ (cos x sin x) = (1+ sin^3x + cos^3x ) / (cos x sin x)$Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosUnderstanding equality $frac12int_0^2pi fraccos(theta)2 + cos(theta)dtheta = pi - int_0^2pi fracdtheta2 + cos(theta)}$Prove $fraccos(x)1 - tan(x) + fracsin(x){1 - cot(x) = sin(x) + cos(x)$Does $sin^2x-cos^2x$ equal $cos(2x)$Prove $sin(alpha -beta)+sin(alpha-gamma)+sin(beta-gamma)=4cosfracalpha-beta2sinfracalpha-gamma2cosfracbeta-gamma2$Trouble proving the trigonometric identity $frac1-2sin(x)sec(x)=fraccos(3x)1+2sin(x)$
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Why does $sin(x) - sin(y)=2 cos(fracx+y2) sin(fracx-y2)$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How do I integrate $frac sin^3xcos^2x$How to evaluate the limit $lim_xto 0 frac1-cos(4x)sin^2(7x)$Prove $(1 + sin x + cos x -cos^2x sin x + sin^2x cos x )/ (cos x sin x) = (1+ sin^3x + cos^3x ) / (cos x sin x)$Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosUnderstanding equality $frac12int_0^2pi fraccos(theta)2 + cos(theta)dtheta = pi - int_0^2pi fracdtheta2 + cos(theta)}$Prove $fraccos(x)1 - tan(x) + fracsin(x){1 - cot(x) = sin(x) + cos(x)$Does $sin^2x-cos^2x$ equal $cos(2x)$Prove $sin(alpha -beta)+sin(alpha-gamma)+sin(beta-gamma)=4cosfracalpha-beta2sinfracalpha-gamma2cosfracbeta-gamma2$Trouble proving the trigonometric identity $frac1-2sin(x)sec(x)=fraccos(3x)1+2sin(x)$
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
$endgroup$
add a comment |
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
$endgroup$
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
Apr 2 at 3:02
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
Apr 2 at 3:03
add a comment |
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
$endgroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
real-analysis analysis trigonometry
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
edited Apr 2 at 7:20
Asaf Karagila♦
308k33441775
308k33441775
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
asked Apr 2 at 2:54
Ryan DuranRyan Duran
132
132
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
Apr 2 at 3:02
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
Apr 2 at 3:03
add a comment |
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
Apr 2 at 3:02
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
Apr 2 at 3:03
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
Apr 2 at 3:02
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
Apr 2 at 3:02
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
Apr 2 at 3:03
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
Apr 2 at 3:03
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered Apr 2 at 4:08
AdmuthAdmuth
886
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
$endgroup$
add a comment |
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
$endgroup$
add a comment |
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
$endgroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
edited Apr 2 at 4:01
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
answered Apr 2 at 3:41
MarianDMarianD
2,2761618
2,2761618
add a comment |
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
$endgroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
answered Apr 2 at 3:04
John DoeJohn Doe
12.1k11339
12.1k11339
add a comment |
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered Apr 2 at 4:08
AdmuthAdmuth
886
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered Apr 2 at 4:08
AdmuthAdmuth
886
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered Apr 2 at 4:08
AdmuthAdmuth
886
$endgroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered Apr 2 at 4:08
AdmuthAdmuth
886
answered Apr 2 at 4:08
AdmuthAdmuth
886
answered Apr 2 at 4:08
AdmuthAdmuth
886
answered Apr 2 at 4:08
AdmuthAdmuth
886
886
add a comment |
add a comment |
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Required, but never shown
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Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
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– Newman
Apr 2 at 3:02
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After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
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– R_D
Apr 2 at 3:03