Why does sin(x) - sin(y) equal this?
$begingroup$
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 |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171404%2fwhy-does-sinx-siny-equal-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171404%2fwhy-does-sinx-siny-equal-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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