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












-2














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
























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










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
















-2














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
























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










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














-2












-2








-2







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















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 yesterday









Blue

47.7k870151




47.7k870151










asked May 8 '14 at 22:27









Ila Isabelle

50310




50310












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










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


















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










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
















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




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












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




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










4 Answers
4






active

oldest

votes


















1














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





























    2














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






    share|cite|improve this answer










    New contributor




    M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


























      1














      $x^2-y^2$ 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





























        0














        $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





















          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









          1














          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


























            1














            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
























              1












              1








              1






              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












              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 Hughes

              62.5k24090




              62.5k24090























                  2














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






                  share|cite|improve this answer










                  New contributor




                  M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.























                    2














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






                    share|cite|improve this answer










                    New contributor




                    M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





















                      2












                      2








                      2






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






                      share|cite|improve this answer










                      New contributor




                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









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







                      share|cite|improve this answer










                      New contributor




                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited yesterday









                      KM101

                      5,4911423




                      5,4911423






                      New contributor




                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      answered yesterday









                      M. C.

                      213




                      213




                      New contributor




                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      New contributor





                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.























                          1














                          $x^2-y^2$ 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


























                            1














                            $x^2-y^2$ 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
























                              1












                              1








                              1






                              $x^2-y^2$ 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












                              $x^2-y^2$ 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.sacamento

                              2,4141819




                              2,4141819























                                  0














                                  $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


























                                    0














                                    $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
























                                      0












                                      0








                                      0






                                      $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












                                      $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 yesterday









                                      marty cohen

                                      72.7k549128




                                      72.7k549128






























                                          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.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • 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.


                                          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

                                          Mario Kart Wii

                                          Image
                                          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                                          Read more

                                          The Binding of Isaac: Rebirth/Afterbirth

                                          Image
                                          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                                          Read more

                                          What does “Dominus providebit” mean?

                                          Image
                                          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                                          Read more