Use the sine rule to prove trig identity [duplicate]












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  • Proof of the identity: $csin frac{A-B}{2} equiv (a-b) cos frac{C}{2}$

    2 answers




Using the sine rule:
$$ frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)}$$



prove, for triangle ABC:



$$sinleft(frac{B-C}{2}right) = frac{b-c}{a} cosleft(frac A2right)$$



Using the sine rule it's easy to translate the RHS into:



$$sinleft(frac{B-C}{2}right) = frac{sin(B)-sin(C)}{sin(A)} cosleft(frac A2right)$$



Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.



A nudge in the right direction would really be appreciated. Thanks.






trigonometry







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 Dec 9 '18 at 19:47


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.























    1












    $begingroup$



    This question already has an answer here:




    • Proof of the identity: $csin frac{A-B}{2} equiv (a-b) cos frac{C}{2}$

      2 answers




    Using the sine rule:
    $$ frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)}$$



    prove, for triangle ABC:



    $$sinleft(frac{B-C}{2}right) = frac{b-c}{a} cosleft(frac A2right)$$



    Using the sine rule it's easy to translate the RHS into:



    $$sinleft(frac{B-C}{2}right) = frac{sin(B)-sin(C)}{sin(A)} cosleft(frac A2right)$$



    Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.



    A nudge in the right direction would really be appreciated. Thanks.






    trigonometry







    share|cite|improve this question











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    marked as duplicate by Blue trigonometry
    Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.

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     Dec 9 '18 at 19:47


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      1












      1








      1





      $begingroup$



      This question already has an answer here:




      • Proof of the identity: $csin frac{A-B}{2} equiv (a-b) cos frac{C}{2}$

        2 answers




      Using the sine rule:
      $$ frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)}$$



      prove, for triangle ABC:



      $$sinleft(frac{B-C}{2}right) = frac{b-c}{a} cosleft(frac A2right)$$



      Using the sine rule it's easy to translate the RHS into:



      $$sinleft(frac{B-C}{2}right) = frac{sin(B)-sin(C)}{sin(A)} cosleft(frac A2right)$$



      Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.



      A nudge in the right direction would really be appreciated. Thanks.






      trigonometry







      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • Proof of the identity: $csin frac{A-B}{2} equiv (a-b) cos frac{C}{2}$

        2 answers




      Using the sine rule:
      $$ frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)}$$



      prove, for triangle ABC:



      $$sinleft(frac{B-C}{2}right) = frac{b-c}{a} cosleft(frac A2right)$$



      Using the sine rule it's easy to translate the RHS into:



      $$sinleft(frac{B-C}{2}right) = frac{sin(B)-sin(C)}{sin(A)} cosleft(frac A2right)$$



      Yes, I can expand out the LHS, and use the difference of 2 sines in the RHS, but neither makes an obvious equality, especially with terms in a and A in the RHS.



      A nudge in the right direction would really be appreciated. Thanks.





      This question already has an answer here:




      • Proof of the identity: $csin frac{A-B}{2} equiv (a-b) cos frac{C}{2}$

        2 answers







      trigonometry




      trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 19:25









      Larry

      2,39131129




      2,39131129










      asked Dec 9 '18 at 19:01









      GurnemanzGurnemanz

      112




      112




      marked as duplicate by Blue trigonometry
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       Dec 9 '18 at 19:47


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Blue trigonometry
      Users with the  trigonometry badge can single-handedly close trigonometry questions as duplicates and reopen them as needed.

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       Dec 9 '18 at 19:47


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          2 Answers
          2






          active

          oldest

          votes


















          -1












          $begingroup$

          Since $A+B+C=pi$, the difference formula for $sin$ gives begin{align}frac{sin B-sin C}{sin A} cosfrac A2&=frac{2cosfrac{B+C}2sinfrac{B-C}2}{sin(pi-B-C)}cosfrac{pi-B-C}2\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1+cos(pi-B-C)}2}\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1-cos(B+C)}2}\&=frac{sinfrac{B-C}2}{sin(B+C)}sqrt{1-cos^2(B+C)}\&=frac{sinfrac{B-C}2}{sin(B+C)}sin(B+C)=sinfrac{B-C}2end{align} as required.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.



            So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.



            Now substitute this and you will get the desired result.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Are you sure this leads anywhere different from what OP's tried?
              $endgroup$
              – TheSimpliFire
              Dec 9 '18 at 19:20










            • $begingroup$
              Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
              $endgroup$
              – Jimmy
              Dec 9 '18 at 19:22










            • $begingroup$
              I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
              $endgroup$
              – TheSimpliFire
              Dec 9 '18 at 19:25




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            -1












            $begingroup$

            Since $A+B+C=pi$, the difference formula for $sin$ gives begin{align}frac{sin B-sin C}{sin A} cosfrac A2&=frac{2cosfrac{B+C}2sinfrac{B-C}2}{sin(pi-B-C)}cosfrac{pi-B-C}2\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1+cos(pi-B-C)}2}\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1-cos(B+C)}2}\&=frac{sinfrac{B-C}2}{sin(B+C)}sqrt{1-cos^2(B+C)}\&=frac{sinfrac{B-C}2}{sin(B+C)}sin(B+C)=sinfrac{B-C}2end{align} as required.






            share|cite|improve this answer









            $endgroup$


















              -1












              $begingroup$

              Since $A+B+C=pi$, the difference formula for $sin$ gives begin{align}frac{sin B-sin C}{sin A} cosfrac A2&=frac{2cosfrac{B+C}2sinfrac{B-C}2}{sin(pi-B-C)}cosfrac{pi-B-C}2\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1+cos(pi-B-C)}2}\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1-cos(B+C)}2}\&=frac{sinfrac{B-C}2}{sin(B+C)}sqrt{1-cos^2(B+C)}\&=frac{sinfrac{B-C}2}{sin(B+C)}sin(B+C)=sinfrac{B-C}2end{align} as required.






              share|cite|improve this answer









              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                Since $A+B+C=pi$, the difference formula for $sin$ gives begin{align}frac{sin B-sin C}{sin A} cosfrac A2&=frac{2cosfrac{B+C}2sinfrac{B-C}2}{sin(pi-B-C)}cosfrac{pi-B-C}2\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1+cos(pi-B-C)}2}\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1-cos(B+C)}2}\&=frac{sinfrac{B-C}2}{sin(B+C)}sqrt{1-cos^2(B+C)}\&=frac{sinfrac{B-C}2}{sin(B+C)}sin(B+C)=sinfrac{B-C}2end{align} as required.






                share|cite|improve this answer









                $endgroup$



                Since $A+B+C=pi$, the difference formula for $sin$ gives begin{align}frac{sin B-sin C}{sin A} cosfrac A2&=frac{2cosfrac{B+C}2sinfrac{B-C}2}{sin(pi-B-C)}cosfrac{pi-B-C}2\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1+cos(pi-B-C)}2}\&=frac{2sqrt{frac{1+cos(B+C)}2}sinfrac{B-C}2}{sin(B+C)}sqrt{frac{1-cos(B+C)}2}\&=frac{sinfrac{B-C}2}{sin(B+C)}sqrt{1-cos^2(B+C)}\&=frac{sinfrac{B-C}2}{sin(B+C)}sin(B+C)=sinfrac{B-C}2end{align} as required.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 19:20









                TheSimpliFireTheSimpliFire

                12.4k62460




                12.4k62460























                    1












                    $begingroup$

                    Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.



                    So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.



                    Now substitute this and you will get the desired result.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Are you sure this leads anywhere different from what OP's tried?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:20










                    • $begingroup$
                      Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                      $endgroup$
                      – Jimmy
                      Dec 9 '18 at 19:22










                    • $begingroup$
                      I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:25


















                    1












                    $begingroup$

                    Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.



                    So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.



                    Now substitute this and you will get the desired result.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Are you sure this leads anywhere different from what OP's tried?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:20










                    • $begingroup$
                      Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                      $endgroup$
                      – Jimmy
                      Dec 9 '18 at 19:22










                    • $begingroup$
                      I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:25
















                    1












                    1








                    1





                    $begingroup$

                    Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.



                    So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.



                    Now substitute this and you will get the desired result.






                    share|cite|improve this answer









                    $endgroup$



                    Just assume that $a/sin(A)=b/sin(B)=c/sin(C)=1/k$.



                    So $sin(A)=ka, sin(B)=kb,sin(C)=kc$.



                    Now substitute this and you will get the desired result.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 9 '18 at 19:08









                    JimmyJimmy

                    29813




                    29813












                    • $begingroup$
                      Are you sure this leads anywhere different from what OP's tried?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:20










                    • $begingroup$
                      Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                      $endgroup$
                      – Jimmy
                      Dec 9 '18 at 19:22










                    • $begingroup$
                      I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:25




















                    • $begingroup$
                      Are you sure this leads anywhere different from what OP's tried?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:20










                    • $begingroup$
                      Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                      $endgroup$
                      – Jimmy
                      Dec 9 '18 at 19:22










                    • $begingroup$
                      I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                      $endgroup$
                      – TheSimpliFire
                      Dec 9 '18 at 19:25


















                    $begingroup$
                    Are you sure this leads anywhere different from what OP's tried?
                    $endgroup$
                    – TheSimpliFire
                    Dec 9 '18 at 19:20




                    $begingroup$
                    Are you sure this leads anywhere different from what OP's tried?
                    $endgroup$
                    – TheSimpliFire
                    Dec 9 '18 at 19:20












                    $begingroup$
                    Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                    $endgroup$
                    – Jimmy
                    Dec 9 '18 at 19:22




                    $begingroup$
                    Well he has given here his method and his last step. Now if we do what I have stated we will get the RHS.
                    $endgroup$
                    – Jimmy
                    Dec 9 '18 at 19:22












                    $begingroup$
                    I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                    $endgroup$
                    – TheSimpliFire
                    Dec 9 '18 at 19:25






                    $begingroup$
                    I might have overlooked this, but how would you handle $sin(B-C/2)$ and $cos(A/2)$ using those substitutions?
                    $endgroup$
                    – TheSimpliFire
                    Dec 9 '18 at 19:25





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