In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$












1















If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?




My reference gives the solution $2$, but I have no clue of where to start ?



My Attempt
$$
cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
$$






geometry trigonometry triangle







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    1















    If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?




    My reference gives the solution $2$, but I have no clue of where to start ?



    My Attempt
    $$
    cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
    $$






    geometry trigonometry triangle







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      1












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      1


      1






      If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?




      My reference gives the solution $2$, but I have no clue of where to start ?



      My Attempt
      $$
      cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
      $$






      geometry trigonometry triangle







      share|cite|improve this question
















      If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?




      My reference gives the solution $2$, but I have no clue of where to start ?



      My Attempt
      $$
      cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
      $$






      geometry trigonometry triangle




      geometry trigonometry triangle






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      edited Jan 4 at 15:59









      Michael Rozenberg

      97.4k1589188




      97.4k1589188










      asked Jan 4 at 12:23









      ss1729ss1729

      1,8491723




      1,8491723






















          2 Answers
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          Hint:



          $$b+c=3aimpliessin B+sin C=3sin A$$



          $$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$



          Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$



          As $0<A<pi,sinsindfrac A2ne0$



          $impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$






          share|cite|improve this answer





























            1














            In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
            $$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
            $$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
            $$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
            $$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
            Also, we can use the following way:
            $$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$






            share|cite|improve this answer























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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Hint:



              $$b+c=3aimpliessin B+sin C=3sin A$$



              $$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$



              Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$



              As $0<A<pi,sinsindfrac A2ne0$



              $impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$






              share|cite|improve this answer


























                2














                Hint:



                $$b+c=3aimpliessin B+sin C=3sin A$$



                $$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$



                Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$



                As $0<A<pi,sinsindfrac A2ne0$



                $impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$






                share|cite|improve this answer
























                  2












                  2








                  2






                  Hint:



                  $$b+c=3aimpliessin B+sin C=3sin A$$



                  $$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$



                  Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$



                  As $0<A<pi,sinsindfrac A2ne0$



                  $impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$






                  share|cite|improve this answer












                  Hint:



                  $$b+c=3aimpliessin B+sin C=3sin A$$



                  $$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$



                  Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$



                  As $0<A<pi,sinsindfrac A2ne0$



                  $impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 12:28









                  lab bhattacharjeelab bhattacharjee

                  224k15156274




                  224k15156274























                      1














                      In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
                      $$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
                      $$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
                      $$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
                      $$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
                      Also, we can use the following way:
                      $$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$






                      share|cite|improve this answer




























                        1














                        In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
                        $$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
                        $$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
                        $$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
                        $$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
                        Also, we can use the following way:
                        $$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$






                        share|cite|improve this answer


























                          1












                          1








                          1






                          In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
                          $$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
                          $$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
                          $$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
                          $$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
                          Also, we can use the following way:
                          $$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$






                          share|cite|improve this answer














                          In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
                          $$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
                          $$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
                          $$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
                          $$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
                          Also, we can use the following way:
                          $$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 4 at 15:49

























                          answered Jan 4 at 15:43









                          Michael RozenbergMichael Rozenberg

                          97.4k1589188




                          97.4k1589188






























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