Proving that $cosleft(ax-frac{bpi}{2}right)=sin(ax)$ for all $a$, $b$












0












$begingroup$



Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.




I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?






trigonometry







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
    $endgroup$
    – Anurag A
    Jan 7 at 6:06








  • 3




    $begingroup$
    Or $a=0$ definitely wrong
    $endgroup$
    – Sorfosh
    Jan 7 at 6:09






  • 2




    $begingroup$
    Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
    $endgroup$
    – Shubham Johri
    Jan 7 at 6:14


















0












$begingroup$



Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.




I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?






trigonometry







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
    $endgroup$
    – Anurag A
    Jan 7 at 6:06








  • 3




    $begingroup$
    Or $a=0$ definitely wrong
    $endgroup$
    – Sorfosh
    Jan 7 at 6:09






  • 2




    $begingroup$
    Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
    $endgroup$
    – Shubham Johri
    Jan 7 at 6:14
















0












0








0





$begingroup$



Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.




I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?






trigonometry







share|cite|improve this question











$endgroup$





Prove that $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$
for all $a$, $b$.




I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?






trigonometry




trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 6:32









Blue

47.8k870152




47.8k870152










asked Jan 7 at 6:04









Anson PangAnson Pang

6215




6215








  • 4




    $begingroup$
    This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
    $endgroup$
    – Anurag A
    Jan 7 at 6:06








  • 3




    $begingroup$
    Or $a=0$ definitely wrong
    $endgroup$
    – Sorfosh
    Jan 7 at 6:09






  • 2




    $begingroup$
    Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
    $endgroup$
    – Shubham Johri
    Jan 7 at 6:14
















  • 4




    $begingroup$
    This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
    $endgroup$
    – Anurag A
    Jan 7 at 6:06








  • 3




    $begingroup$
    Or $a=0$ definitely wrong
    $endgroup$
    – Sorfosh
    Jan 7 at 6:09






  • 2




    $begingroup$
    Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
    $endgroup$
    – Shubham Johri
    Jan 7 at 6:14










4




4




$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06






$begingroup$
This is not true, just take $b=0$, then $cos (ax)$ is not equal to $sin (ax)$ for all $a$.
$endgroup$
– Anurag A
Jan 7 at 6:06






3




3




$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09




$begingroup$
Or $a=0$ definitely wrong
$endgroup$
– Sorfosh
Jan 7 at 6:09




2




2




$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14






$begingroup$
Is it to prove $$cos(ax-fracpi2)=sin(ax)$$
$endgroup$
– Shubham Johri
Jan 7 at 6:14












3 Answers
3






active

oldest

votes


















6












$begingroup$

The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
    $endgroup$
    – John Omielan
    Jan 7 at 6:14












  • $begingroup$
    @JohnOmielan Very true. Thanks for pointing out.
    $endgroup$
    – Kavi Rama Murthy
    Jan 7 at 6:15



















0












$begingroup$

If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$



$ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer



$iff2ax-b=4mpm(1-2ax)$



$-implies2ax-b=4m-(1-2x)iff b=1-4m$



$+implies2ax-b=4m+1-2axiff4ax-b=4m+1$



which is dependent on $x$ hence can not be held for all $a,b$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If
    $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
    $$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.



    So, consider small values of $x$ and use series expansion to get
    $$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
    b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
    b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
    b}{2}right)right)+Oleft(x^4right)$$



    This would imply
    $$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.






    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









      6












      $begingroup$

      The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
        $endgroup$
        – John Omielan
        Jan 7 at 6:14












      • $begingroup$
        @JohnOmielan Very true. Thanks for pointing out.
        $endgroup$
        – Kavi Rama Murthy
        Jan 7 at 6:15
















      6












      $begingroup$

      The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
        $endgroup$
        – John Omielan
        Jan 7 at 6:14












      • $begingroup$
        @JohnOmielan Very true. Thanks for pointing out.
        $endgroup$
        – Kavi Rama Murthy
        Jan 7 at 6:15














      6












      6








      6





      $begingroup$

      The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].






      share|cite|improve this answer











      $endgroup$



      The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $cos(A+B)=cos (A) cos(B)-sin (A) sin (B)$].







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 7 at 6:15

























      answered Jan 7 at 6:11









      Kavi Rama MurthyKavi Rama Murthy

      54.4k32055




      54.4k32055












      • $begingroup$
        If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
        $endgroup$
        – John Omielan
        Jan 7 at 6:14












      • $begingroup$
        @JohnOmielan Very true. Thanks for pointing out.
        $endgroup$
        – Kavi Rama Murthy
        Jan 7 at 6:15


















      • $begingroup$
        If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
        $endgroup$
        – John Omielan
        Jan 7 at 6:14












      • $begingroup$
        @JohnOmielan Very true. Thanks for pointing out.
        $endgroup$
        – Kavi Rama Murthy
        Jan 7 at 6:15
















      $begingroup$
      If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
      $endgroup$
      – John Omielan
      Jan 7 at 6:14






      $begingroup$
      If $b = frac{left(4n+1right)pi}{2}$, then $-frac{bpi}{2} = -frac{left(4n+1right)pi^2}{4}$. I believe you need to remove the $pi$ from your formula.
      $endgroup$
      – John Omielan
      Jan 7 at 6:14














      $begingroup$
      @JohnOmielan Very true. Thanks for pointing out.
      $endgroup$
      – Kavi Rama Murthy
      Jan 7 at 6:15




      $begingroup$
      @JohnOmielan Very true. Thanks for pointing out.
      $endgroup$
      – Kavi Rama Murthy
      Jan 7 at 6:15











      0












      $begingroup$

      If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$



      $ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer



      $iff2ax-b=4mpm(1-2ax)$



      $-implies2ax-b=4m-(1-2x)iff b=1-4m$



      $+implies2ax-b=4m+1-2axiff4ax-b=4m+1$



      which is dependent on $x$ hence can not be held for all $a,b$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$



        $ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer



        $iff2ax-b=4mpm(1-2ax)$



        $-implies2ax-b=4m-(1-2x)iff b=1-4m$



        $+implies2ax-b=4m+1-2axiff4ax-b=4m+1$



        which is dependent on $x$ hence can not be held for all $a,b$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$



          $ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer



          $iff2ax-b=4mpm(1-2ax)$



          $-implies2ax-b=4m-(1-2x)iff b=1-4m$



          $+implies2ax-b=4m+1-2axiff4ax-b=4m+1$



          which is dependent on $x$ hence can not be held for all $a,b$






          share|cite|improve this answer









          $endgroup$



          If $$cosleft(ax-dfrac{bpi}2right)=sin(ax)=cosleft(dfrac{pi}2-axright)$$



          $ax-dfrac{bpi}2=2mpipmleft(dfrac{pi}2-axright)$ where $m$ is any integer



          $iff2ax-b=4mpm(1-2ax)$



          $-implies2ax-b=4m-(1-2x)iff b=1-4m$



          $+implies2ax-b=4m+1-2axiff4ax-b=4m+1$



          which is dependent on $x$ hence can not be held for all $a,b$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 7:32









          lab bhattacharjeelab bhattacharjee

          224k15156274




          224k15156274























              0












              $begingroup$

              If
              $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
              $$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.



              So, consider small values of $x$ and use series expansion to get
              $$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
              b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
              b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
              b}{2}right)right)+Oleft(x^4right)$$



              This would imply
              $$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If
                $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
                $$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.



                So, consider small values of $x$ and use series expansion to get
                $$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
                b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
                b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
                b}{2}right)right)+Oleft(x^4right)$$



                This would imply
                $$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If
                  $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
                  $$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.



                  So, consider small values of $x$ and use series expansion to get
                  $$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
                  b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
                  b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
                  b}{2}right)right)+Oleft(x^4right)$$



                  This would imply
                  $$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.






                  share|cite|improve this answer









                  $endgroup$



                  If
                  $$cosleft(ax-frac{bpi}{2}right)=sin(ax)$$ is true, it means that
                  $$f(x)=cosleft(ax-frac{bpi}{2}right)-sin(ax)$$ is $0$.



                  So, consider small values of $x$ and use series expansion to get
                  $$f(x)=cos left(frac{pi b}{2}right)+x left(a sin left(frac{pi
                  b}{2}right)-aright)-frac{1}{2} x^2 left(a^2 cos left(frac{pi
                  b}{2}right)right)+frac{1}{6} x^3 left(a^3-a^3 sin left(frac{pi
                  b}{2}right)right)+Oleft(x^4right)$$



                  This would imply
                  $$cos left(frac{pi b}{2}right)=0 implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 7:36









                  Claude LeiboviciClaude Leibovici

                  120k1157132




                  120k1157132






























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