Why does sin(x) - sin(y) equal this? The Next CEO of Stack OverflowProve that...

Can Sri Krishna be called 'a person'?

Are British MPs missing the point, with these 'Indicative Votes'?

Is it OK to decorate a log book cover?

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

Prodigo = pro + ago?

Can I hook these wires up to find the connection to a dead outlet?

What did the word "leisure" mean in late 18th Century usage?

Incomplete cube

Finitely generated matrix groups whose eigenvalues are all algebraic

Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?

What is the difference between 'contrib' and 'non-free' packages repositories?

Why can't we say "I have been having a dog"?

Read/write a pipe-delimited file line by line with some simple text manipulation

Find a path from s to t using as few red nodes as possible

Is there a rule of thumb for determining the amount one should accept for a settlement offer?

Could a dragon use its wings to swim?

How to show a landlord what we have in savings?

How dangerous is XSS

Creating a script with console commands

Could you use a laser beam as a modulated carrier wave for radio signal?

How to pronounce fünf in 45

Is it correct to say moon starry nights?

What is a typical Mizrachi Seder like?

How should I connect my cat5 cable to connectors having an orange-green line?



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 $cos(2x)$












2












$begingroup$


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










share|cite|improve this question







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
















2












$begingroup$


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










share|cite|improve this question







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














2












2








2





$begingroup$


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










share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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




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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















5












$begingroup$

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}






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      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 $y$.






      share|cite|improve this answer









      $endgroup$














        Your Answer





        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.










        draft saved

        draft discarded


















        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









        5












        $begingroup$

        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}






        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          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}






          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            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}






            share|cite|improve this answer











            $endgroup$



            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}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            MarianDMarianD

            2,0831617




            2,0831617























                4












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    John DoeJohn Doe

                    11.4k11239




                    11.4k11239























                        2












                        $begingroup$

                        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 $y$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          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 $y$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            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 $y$.






                            share|cite|improve this answer









                            $endgroup$



                            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 $y$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            AdmuthAdmuth

                            685




                            685






















                                Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                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.




                                draft saved


                                draft discarded














                                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





















































                                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







                                Popular posts from this blog

                                Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                                Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                                Simple Scan not detecting my scanner (Brother DCP-7055W)Brother MFC-L2700DW printer can print, can't...