Determinant of a matrix with trigonometric functions












1












$begingroup$



Prove:



det
$begin{pmatrix}
cos(a-b) & cos(b-c) & cos(c-a) \
cos(a+b) & cos(b+c) & cos(c+a) \
sin(a+b) & sin(b+c) & sin(c+a)
end{pmatrix}=-2sin(a-b)sin(b-c)sin(c-a)
$.




I know that



$detbegin{pmatrix}
cos(a-b) & cos(b-c) & cos(c-a) \
cos(a+b) & cos(b+c) & cos(c+a) \
sin(a+b) & sin(b+c) & sin(c+a)
end{pmatrix}=2cdot detbegin{pmatrix}
sin asin b & sin bsin c & sin csin a \
cos acos b & cos bcos c & cos ccos a \
sin(a+b) & sin(b+c) & sin(c+a)
end{pmatrix}
$



Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.






linear-algebra matrices determinant







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$endgroup$

















    1












    $begingroup$



    Prove:



    det
    $begin{pmatrix}
    cos(a-b) & cos(b-c) & cos(c-a) \
    cos(a+b) & cos(b+c) & cos(c+a) \
    sin(a+b) & sin(b+c) & sin(c+a)
    end{pmatrix}=-2sin(a-b)sin(b-c)sin(c-a)
    $    .




    I know that



    $detbegin{pmatrix}
    cos(a-b) & cos(b-c) & cos(c-a) \
    cos(a+b) & cos(b+c) & cos(c+a) \
    sin(a+b) & sin(b+c) & sin(c+a)
    end{pmatrix}=2cdot detbegin{pmatrix}
    sin asin b & sin bsin c & sin csin a \
    cos acos b & cos bcos c & cos ccos a \
    sin(a+b) & sin(b+c) & sin(c+a)
    end{pmatrix}
    $



    Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.






    linear-algebra matrices determinant







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Prove:



      det
      $begin{pmatrix}
      cos(a-b) & cos(b-c) & cos(c-a) \
      cos(a+b) & cos(b+c) & cos(c+a) \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}=-2sin(a-b)sin(b-c)sin(c-a)
      $      .




      I know that



      $detbegin{pmatrix}
      cos(a-b) & cos(b-c) & cos(c-a) \
      cos(a+b) & cos(b+c) & cos(c+a) \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}=2cdot detbegin{pmatrix}
      sin asin b & sin bsin c & sin csin a \
      cos acos b & cos bcos c & cos ccos a \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}
      $



      Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.






      linear-algebra matrices determinant







      share|cite|improve this question











      $endgroup$





      Prove:



      det
      $begin{pmatrix}
      cos(a-b) & cos(b-c) & cos(c-a) \
      cos(a+b) & cos(b+c) & cos(c+a) \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}=-2sin(a-b)sin(b-c)sin(c-a)
      $      .




      I know that



      $detbegin{pmatrix}
      cos(a-b) & cos(b-c) & cos(c-a) \
      cos(a+b) & cos(b+c) & cos(c+a) \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}=2cdot detbegin{pmatrix}
      sin asin b & sin bsin c & sin csin a \
      cos acos b & cos bcos c & cos ccos a \
      sin(a+b) & sin(b+c) & sin(c+a)
      end{pmatrix}
      $



      Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.






      linear-algebra matrices determinant




      linear-algebra matrices determinant






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      share|cite|improve this question




      share|cite|improve this question








      edited Jan 27 at 11:59









      Michael Rozenberg

      109k1896200




      109k1896200










      asked Jan 27 at 10:54









      Tao XTao X

      865




      865






















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          $begingroup$

          $$det=sum_{cyc}(cos(a-b)cos(b+c)sin(a+c)-cos(a-b)sin(b+c)cos(c+a))=$$
          $$=sum_{cyc}cos(a-b)sin(a+c-b-c)=sum_{cyc}cos(a-b)sin(a-b).$$
          In another hand,
          $$-2sin(c-a)sin(b-c)sin(a-b)=-(cos(2c-a-b)-cos(a-b))sin(a-b)=$$
          $$=cos(a-b)sin(a-b)-frac{1}{2}(sin(2c-2b)+sin(2a-2c))=sum_{cyc}cos(a-b)sin(a-b)$$
          and we are done.






          share|cite|improve this answer









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            $begingroup$

            $$det=sum_{cyc}(cos(a-b)cos(b+c)sin(a+c)-cos(a-b)sin(b+c)cos(c+a))=$$
            $$=sum_{cyc}cos(a-b)sin(a+c-b-c)=sum_{cyc}cos(a-b)sin(a-b).$$
            In another hand,
            $$-2sin(c-a)sin(b-c)sin(a-b)=-(cos(2c-a-b)-cos(a-b))sin(a-b)=$$
            $$=cos(a-b)sin(a-b)-frac{1}{2}(sin(2c-2b)+sin(2a-2c))=sum_{cyc}cos(a-b)sin(a-b)$$
            and we are done.






            share|cite|improve this answer









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              1












              $begingroup$

              $$det=sum_{cyc}(cos(a-b)cos(b+c)sin(a+c)-cos(a-b)sin(b+c)cos(c+a))=$$
              $$=sum_{cyc}cos(a-b)sin(a+c-b-c)=sum_{cyc}cos(a-b)sin(a-b).$$
              In another hand,
              $$-2sin(c-a)sin(b-c)sin(a-b)=-(cos(2c-a-b)-cos(a-b))sin(a-b)=$$
              $$=cos(a-b)sin(a-b)-frac{1}{2}(sin(2c-2b)+sin(2a-2c))=sum_{cyc}cos(a-b)sin(a-b)$$
              and we are done.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $$det=sum_{cyc}(cos(a-b)cos(b+c)sin(a+c)-cos(a-b)sin(b+c)cos(c+a))=$$
                $$=sum_{cyc}cos(a-b)sin(a+c-b-c)=sum_{cyc}cos(a-b)sin(a-b).$$
                In another hand,
                $$-2sin(c-a)sin(b-c)sin(a-b)=-(cos(2c-a-b)-cos(a-b))sin(a-b)=$$
                $$=cos(a-b)sin(a-b)-frac{1}{2}(sin(2c-2b)+sin(2a-2c))=sum_{cyc}cos(a-b)sin(a-b)$$
                and we are done.






                share|cite|improve this answer









                $endgroup$



                $$det=sum_{cyc}(cos(a-b)cos(b+c)sin(a+c)-cos(a-b)sin(b+c)cos(c+a))=$$
                $$=sum_{cyc}cos(a-b)sin(a+c-b-c)=sum_{cyc}cos(a-b)sin(a-b).$$
                In another hand,
                $$-2sin(c-a)sin(b-c)sin(a-b)=-(cos(2c-a-b)-cos(a-b))sin(a-b)=$$
                $$=cos(a-b)sin(a-b)-frac{1}{2}(sin(2c-2b)+sin(2a-2c))=sum_{cyc}cos(a-b)sin(a-b)$$
                and we are done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 27 at 11:58









                Michael RozenbergMichael Rozenberg

                109k1896200




                109k1896200






























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