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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.
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$endgroup$












  • $begingroup$
    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 3 hours ago









Ryan DuranRyan Duran

111




111




New contributor




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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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