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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  • $begingroup$
    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




111




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


















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









$endgroup$





















    3












    $begingroup$

    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











    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        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























                3












                $begingroup$

                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











                $endgroup$


















                  3












                  $begingroup$

                  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











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    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 1 hour ago









                    MarianDMarianD

                    2,0631617




                    2,0631617























                        1












                        $begingroup$

                        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









                        $endgroup$


















                          1












                          $begingroup$

                          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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            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

                            585




                            585






















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                                Tantalo (mitologio)

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                                Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
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                                Erzsébet Schaár

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                                Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
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