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Why does sin(x) - sin(y) equal this?
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$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.
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 $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. 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?
$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.
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 $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. 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 $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. 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?
$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.
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 $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. 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?
$sin x - sin y = 2 cos(frac{x+y}{2}) sin(frac{x-y}{2})$.
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 $A=frac{x+y}{2}$ and $B=frac{x-y}{2}$. 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 $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. 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 $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. 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 $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. 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
$begingroup$
Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. 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 |
$begingroup$
The main trick is here:
begin{align}
color{red} {x = {x+yover2} + {x-yover2}}\[1em]
color{blue}{y = {x+yover2} - {x-yover2}}
end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain
begin{align}
sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
end{align}
All the rest is then only a routine calculation:
begin{align}
require{enclose}
&= sin left({x+yover2}right) cosleft( {x-yover2} right) +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
&= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)
\[3em]
&=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
end{align}
answered 1 hour ago
MarianDMarianD
2,0631617
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
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
585
$endgroup$
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
$begingroup$
Following your professor's advice, let $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. 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 $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. 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 $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. 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 $A=frac{x+y}{2}$, $B=frac{x-y}{2}$. 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 |
$begingroup$
The main trick is here:
begin{align}
color{red} {x = {x+yover2} + {x-yover2}}\[1em]
color{blue}{y = {x+yover2} - {x-yover2}}
end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain
begin{align}
sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
end{align}
All the rest is then only a routine calculation:
begin{align}
require{enclose}
&= sin left({x+yover2}right) cosleft( {x-yover2} right) +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
&= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)
\[3em]
&=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
end{align}
answered 1 hour ago
MarianDMarianD
2,0631617
$endgroup$
add a comment |
$begingroup$
The main trick is here:
begin{align}
color{red} {x = {x+yover2} + {x-yover2}}\[1em]
color{blue}{y = {x+yover2} - {x-yover2}}
end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain
begin{align}
sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
end{align}
All the rest is then only a routine calculation:
begin{align}
require{enclose}
&= sin left({x+yover2}right) cosleft( {x-yover2} right) +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
&= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)
\[3em]
&=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
end{align}
answered 1 hour ago
MarianDMarianD
2,0631617
$endgroup$
add a comment |
$begingroup$
The main trick is here:
begin{align}
color{red} {x = {x+yover2} + {x-yover2}}\[1em]
color{blue}{y = {x+yover2} - {x-yover2}}
end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain
begin{align}
sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
end{align}
All the rest is then only a routine calculation:
begin{align}
require{enclose}
&= sin left({x+yover2}right) cosleft( {x-yover2} right) +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
&= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)
\[3em]
&=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
end{align}
answered 1 hour ago
MarianDMarianD
2,0631617
$endgroup$
The main trick is here:
begin{align}
color{red} {x = {x+yover2} + {x-yover2}}\[1em]
color{blue}{y = {x+yover2} - {x-yover2}}
end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $color{red}x$ and $color{blue}y,,$ you will obtain
begin{align}
sin color{red} x - sin color{blue }y = sin left(color{red}{{x+yover2} + {x-yover2} }right) - sin left(color{blue }{{x+yover2} - {x-yover2}} right) \[1em]
end{align}
All the rest is then only a routine calculation:
begin{align}
require{enclose}
&= sin left({x+yover2}right) cosleft( {x-yover2} right) +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-left[sin left({x+yover2}right) cosleft( {x-yover2} right) -
sin left({x-yover2}right) cosleft( {x+yover2} right)right]\[3em]
&= enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)\[1em]
&-enclose{updiagonalstrike}{sin left({x+yover2}right) cosleft( {x-yover2} right)} +
sin left({x-yover2}right) cosleft( {x+yover2} right)
\[3em]
&=2sin left({x-yover2}right) cosleft( {x+yover2} right)\
end{align}
answered 1 hour ago
MarianDMarianD
2,0631617
edited 1 hour ago
answered 1 hour ago
MarianDMarianD
2,0631617
answered 1 hour ago
MarianDMarianD
2,0631617
answered 1 hour ago
MarianDMarianD
2,0631617
2,0631617
add a comment |
add a comment |
$begingroup$
Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
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
585
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
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
585
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
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
585
$endgroup$
Following your notation, let $A=dfrac{x+y}{2}$ and $B=dfrac{x-y}{2}$.
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
585
answered 1 hour ago
AdmuthAdmuth
585
answered 1 hour ago
AdmuthAdmuth
585
answered 1 hour ago
AdmuthAdmuth
585
585
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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Let A and B be as you defined. Then $sin(A+B)=sin(frac{x+y}{2}+frac{x-y}{2})$. Evaluate this and use the given identities.
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– Newman
2 hours ago
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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
2 hours ago