Proving $sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$












0












$begingroup$


Knowing



$$sin^2theta +cos^2theta equiv 1$$



how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$



Can I substitute the first equation to prove the second one? If so, how can I?



Please help.






trigonometry







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    Please read this tutorial on how to typeset mathematics on this site.
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    Jan 15 at 13:47






  • 6




    $begingroup$
    The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
    $endgroup$
    – Blue
    Jan 15 at 13:47












  • $begingroup$
    See also math.stackexchange.com/questions/175143/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 13:54
















0












$begingroup$


Knowing



$$sin^2theta +cos^2theta equiv 1$$



how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$



Can I substitute the first equation to prove the second one? If so, how can I?



Please help.






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 15 at 13:47






  • 6




    $begingroup$
    The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
    $endgroup$
    – Blue
    Jan 15 at 13:47












  • $begingroup$
    See also math.stackexchange.com/questions/175143/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 13:54














0












0








0





$begingroup$


Knowing



$$sin^2theta +cos^2theta equiv 1$$



how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$



Can I substitute the first equation to prove the second one? If so, how can I?



Please help.






trigonometry







share|cite|improve this question











$endgroup$




Knowing



$$sin^2theta +cos^2theta equiv 1$$



how would I prove:
$$sin^2x cos^2y - cos^2x sin^2y ;equiv; cos^2y - cos^2x$$



Can I substitute the first equation to prove the second one? If so, how can I?



Please 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 15 at 13:46









Blue

49.3k870157




49.3k870157










asked Jan 15 at 13:42









user8469209user8469209

123




123












  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 15 at 13:47






  • 6




    $begingroup$
    The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
    $endgroup$
    – Blue
    Jan 15 at 13:47












  • $begingroup$
    See also math.stackexchange.com/questions/175143/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 13:54


















  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 15 at 13:47






  • 6




    $begingroup$
    The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
    $endgroup$
    – Blue
    Jan 15 at 13:47












  • $begingroup$
    See also math.stackexchange.com/questions/175143/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 13:54
















$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47




$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 15 at 13:47




6




6




$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47






$begingroup$
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side?
$endgroup$
– Blue
Jan 15 at 13:47














$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54




$begingroup$
See also math.stackexchange.com/questions/175143/…
$endgroup$
– lab bhattacharjee
Jan 15 at 13:54










3 Answers
3






active

oldest

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1












$begingroup$

Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Yes you can substitute the first equation into the second one.



    $sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$



    Substitute to get,



    $begin{eqnarray}
    sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
    &=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
    &=& cos^2y - cos^2x
    end{eqnarray}$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $sin^2x cos^2y - cos^2x sin^2y \
      = sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
      = (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
      = cos^2y - cos^2x$






      share|cite|improve this answer









      $endgroup$














        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

        oldest

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        1












        $begingroup$

        Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.






            share|cite|improve this answer











            $endgroup$



            Notice that the right hand side only has cosines in it. Try replacing all of the $sin^2(x)$ and $sin^2(y)$ with $1-cos^2(x)$ and $1-cos^2(y)$ respectively. If you simplify, you will see the right hand side.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 18 at 17:44

























            answered Jan 15 at 14:40









            CuhrazateeCuhrazatee

            372110




            372110























                0












                $begingroup$

                Yes you can substitute the first equation into the second one.



                $sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$



                Substitute to get,



                $begin{eqnarray}
                sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
                &=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
                &=& cos^2y - cos^2x
                end{eqnarray}$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Yes you can substitute the first equation into the second one.



                  $sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$



                  Substitute to get,



                  $begin{eqnarray}
                  sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
                  &=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
                  &=& cos^2y - cos^2x
                  end{eqnarray}$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Yes you can substitute the first equation into the second one.



                    $sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$



                    Substitute to get,



                    $begin{eqnarray}
                    sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
                    &=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
                    &=& cos^2y - cos^2x
                    end{eqnarray}$






                    share|cite|improve this answer









                    $endgroup$



                    Yes you can substitute the first equation into the second one.



                    $sin^2theta +cos^2theta equiv 1 implies sin^2theta = 1-cos^2theta$



                    Substitute to get,



                    $begin{eqnarray}
                    sin^2x cos^2y - cos^2x sin^2y &=& ((1-cos^2(x)) (cos^2y)) - ((cos^2x) (1-cos^2(y)) \\
                    &=& cos^2y - ((cos^2y)(cos^2x)) - cos^2x + ((cos^2x)(cos^2y)) \\
                    &=& cos^2y - cos^2x
                    end{eqnarray}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 15 at 14:54









                    E.NoleE.Nole

                    301114




                    301114























                        0












                        $begingroup$

                        $sin^2x cos^2y - cos^2x sin^2y \
                        = sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
                        = (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
                        = cos^2y - cos^2x$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $sin^2x cos^2y - cos^2x sin^2y \
                          = sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
                          = (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
                          = cos^2y - cos^2x$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $sin^2x cos^2y - cos^2x sin^2y \
                            = sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
                            = (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
                            = cos^2y - cos^2x$






                            share|cite|improve this answer









                            $endgroup$



                            $sin^2x cos^2y - cos^2x sin^2y \
                            = sin^2x cos^2y + (cos^2x cos^2y - cos^2x cos^2y) - cos^2x sin^2y \
                            = (sin^2x + cos^2x) cos^2y - cos^2x (cos^2y + sin^2y) \
                            = cos^2y - cos^2x$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 15 at 15:31









                            gandalf61gandalf61

                            9,174825




                            9,174825






























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