Why does sin(x) - sin(y) equal this?












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






real-analysis analysis trigonometry







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Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    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






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.







$endgroup$












  • $begingroup$
    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






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.







$endgroup$




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






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




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New contributor




Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.






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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





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        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.










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                                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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                                Verónica Boquete

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                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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                                Ida-Boy-Ed-Garten

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                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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