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


















3












$begingroup$

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









        3












        $begingroup$

        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












          $begingroup$

          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












            3








            3





            $begingroup$

            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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                                Vallis Paradisi

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                                Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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