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Why does sin(x) - sin(y) equal this?
The Next CEO of Stack OverflowProve that $sin(2A)+sin(2B)+sin(2C)=4sin(A)sin(B)sin(C)$ when $A,B,C$ are angles of a triangleWhy $sin(pi)$ sometimes equal to $0$?Understanding expanding trig identitiesWhy does this always equal $1$?When does this equation $cos(alpha + beta) = cos(alpha) + cos(beta)$ hold?Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosSolve Trigonometric Equality, Multiple Angle TrigonometryFinding relationships between angles, a, b and c when $sin a - sin b - sin c = 0$Does $sin^2x-cos^2x$ equal $cos(2x)$
$begingroup$
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
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
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
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
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
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
asked 2 hours ago
Ryan DuranRyan Duran
111
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
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
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
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 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
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 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
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 1 hour ago
AdmuthAdmuth
685
$endgroup$
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
$begingroup$
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 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
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 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
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 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
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 1 hour ago
MarianDMarianD
2,0831617
edited 1 hour ago
answered 1 hour ago
MarianDMarianD
2,0831617
answered 1 hour ago
MarianDMarianD
2,0831617
answered 1 hour ago
MarianDMarianD
2,0831617
2,0831617
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 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
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 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
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 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
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 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
11.4k11239
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 1 hour ago
AdmuthAdmuth
685
$endgroup$
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 1 hour ago
AdmuthAdmuth
685
$endgroup$
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 1 hour ago
AdmuthAdmuth
685
$endgroup$
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 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
685
add a comment |
add a comment |
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago