Proving $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$ (and trig identities in general)












-2












$begingroup$


The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$



I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?



Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
    $endgroup$
    – David H
    May 8 '14 at 22:33










  • $begingroup$
    tan^2(x)*sin^2(x)
    $endgroup$
    – Ila Isabelle
    May 8 '14 at 22:35
















-2












$begingroup$


The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$



I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?



Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
    $endgroup$
    – David H
    May 8 '14 at 22:33










  • $begingroup$
    tan^2(x)*sin^2(x)
    $endgroup$
    – Ila Isabelle
    May 8 '14 at 22:35














-2












-2








-2





$begingroup$


The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$



I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?



Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$



I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?



Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 22:51









Blue

49.1k870156




49.1k870156










asked May 8 '14 at 22:27









Ila IsabelleIla Isabelle

50311




50311












  • $begingroup$
    Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
    $endgroup$
    – David H
    May 8 '14 at 22:33










  • $begingroup$
    tan^2(x)*sin^2(x)
    $endgroup$
    – Ila Isabelle
    May 8 '14 at 22:35


















  • $begingroup$
    Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
    $endgroup$
    – David H
    May 8 '14 at 22:33










  • $begingroup$
    tan^2(x)*sin^2(x)
    $endgroup$
    – Ila Isabelle
    May 8 '14 at 22:35
















$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33




$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33












$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35




$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35










4 Answers
4






active

oldest

votes


















2












$begingroup$

General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
        $



        Typically,
        the only identity that you need is
        $s^2+c^2 = 1$,
        writing $s$ for $sin$
        and $c$ for $cos$.



        Writing
        $t$ for $tan$,
        the difference of the sides is



        $begin{array}\
        t^2-s^2-t^2s^2
        &=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
        &=dfrac{s^2-s^2c^2-s^4}{c^2}\
        &=dfrac{s^2(1-c^2-s^2)}{c^2}\
        &=dfrac{s^2(1-(c^2+s^2))}{c^2}\
        &=0
        qquadtext{since }c^2+s^2=1\
        end{array}
        $



        The only place where
        there might be a problem
        is when
        $c = 0$;
        there
        $t^2 = infty$
        (abusing notation a little)
        and $s^2=1$
        so the identity still holds,
        in a limitly way.






        share|cite|improve this answer









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f787112%2fproving-tan2x-sin2x-tan2x-sin2x-and-trig-identities-in-gener%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
          $$
          tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
          $$
          Multiplying top and bottom by $cos^2 x$, you get
          $$
          sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
          $$






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
            $$
            tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
            $$
            Multiplying top and bottom by $cos^2 x$, you get
            $$
            sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
            $$






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
              $$
              tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
              $$
              Multiplying top and bottom by $cos^2 x$, you get
              $$
              sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
              $$






              share|cite|improve this answer









              $endgroup$



              General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
              $$
              tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
              $$
              Multiplying top and bottom by $cos^2 x$, you get
              $$
              sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
              $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered May 8 '14 at 22:36









              John HughesJohn Hughes

              64.6k24191




              64.6k24191























                  2












                  $begingroup$

                  $x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    $x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.






                      share|cite|improve this answer









                      $endgroup$



                      $x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered May 8 '14 at 22:41









                      bob.sacamentobob.sacamento

                      2,4311919




                      2,4311919























                          2












                          $begingroup$

                          $$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$






                          share|cite|improve this answer











                          $endgroup$


















                            2












                            $begingroup$

                            $$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$






                            share|cite|improve this answer











                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              $$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$






                              share|cite|improve this answer











                              $endgroup$



                              $$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 5 at 7:29









                              KM101

                              6,0901525




                              6,0901525










                              answered Jan 4 at 22:47









                              M. C.M. C.

                              449




                              449























                                  1












                                  $begingroup$

                                  $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
                                  $



                                  Typically,
                                  the only identity that you need is
                                  $s^2+c^2 = 1$,
                                  writing $s$ for $sin$
                                  and $c$ for $cos$.



                                  Writing
                                  $t$ for $tan$,
                                  the difference of the sides is



                                  $begin{array}\
                                  t^2-s^2-t^2s^2
                                  &=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
                                  &=dfrac{s^2-s^2c^2-s^4}{c^2}\
                                  &=dfrac{s^2(1-c^2-s^2)}{c^2}\
                                  &=dfrac{s^2(1-(c^2+s^2))}{c^2}\
                                  &=0
                                  qquadtext{since }c^2+s^2=1\
                                  end{array}
                                  $



                                  The only place where
                                  there might be a problem
                                  is when
                                  $c = 0$;
                                  there
                                  $t^2 = infty$
                                  (abusing notation a little)
                                  and $s^2=1$
                                  so the identity still holds,
                                  in a limitly way.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
                                    $



                                    Typically,
                                    the only identity that you need is
                                    $s^2+c^2 = 1$,
                                    writing $s$ for $sin$
                                    and $c$ for $cos$.



                                    Writing
                                    $t$ for $tan$,
                                    the difference of the sides is



                                    $begin{array}\
                                    t^2-s^2-t^2s^2
                                    &=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
                                    &=dfrac{s^2-s^2c^2-s^4}{c^2}\
                                    &=dfrac{s^2(1-c^2-s^2)}{c^2}\
                                    &=dfrac{s^2(1-(c^2+s^2))}{c^2}\
                                    &=0
                                    qquadtext{since }c^2+s^2=1\
                                    end{array}
                                    $



                                    The only place where
                                    there might be a problem
                                    is when
                                    $c = 0$;
                                    there
                                    $t^2 = infty$
                                    (abusing notation a little)
                                    and $s^2=1$
                                    so the identity still holds,
                                    in a limitly way.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
                                      $



                                      Typically,
                                      the only identity that you need is
                                      $s^2+c^2 = 1$,
                                      writing $s$ for $sin$
                                      and $c$ for $cos$.



                                      Writing
                                      $t$ for $tan$,
                                      the difference of the sides is



                                      $begin{array}\
                                      t^2-s^2-t^2s^2
                                      &=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
                                      &=dfrac{s^2-s^2c^2-s^4}{c^2}\
                                      &=dfrac{s^2(1-c^2-s^2)}{c^2}\
                                      &=dfrac{s^2(1-(c^2+s^2))}{c^2}\
                                      &=0
                                      qquadtext{since }c^2+s^2=1\
                                      end{array}
                                      $



                                      The only place where
                                      there might be a problem
                                      is when
                                      $c = 0$;
                                      there
                                      $t^2 = infty$
                                      (abusing notation a little)
                                      and $s^2=1$
                                      so the identity still holds,
                                      in a limitly way.






                                      share|cite|improve this answer









                                      $endgroup$



                                      $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
                                      $



                                      Typically,
                                      the only identity that you need is
                                      $s^2+c^2 = 1$,
                                      writing $s$ for $sin$
                                      and $c$ for $cos$.



                                      Writing
                                      $t$ for $tan$,
                                      the difference of the sides is



                                      $begin{array}\
                                      t^2-s^2-t^2s^2
                                      &=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
                                      &=dfrac{s^2-s^2c^2-s^4}{c^2}\
                                      &=dfrac{s^2(1-c^2-s^2)}{c^2}\
                                      &=dfrac{s^2(1-(c^2+s^2))}{c^2}\
                                      &=0
                                      qquadtext{since }c^2+s^2=1\
                                      end{array}
                                      $



                                      The only place where
                                      there might be a problem
                                      is when
                                      $c = 0$;
                                      there
                                      $t^2 = infty$
                                      (abusing notation a little)
                                      and $s^2=1$
                                      so the identity still holds,
                                      in a limitly way.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 5 at 7:25









                                      marty cohenmarty cohen

                                      74.4k549129




                                      74.4k549129






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f787112%2fproving-tan2x-sin2x-tan2x-sin2x-and-trig-identities-in-gener%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Bressuire

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                          Read more

                                          Cabo Verde

                                          Image
                                            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                                          Read more

                                          Gyllenstierna

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                          Read more