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







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$endgroup$












  • $begingroup$
    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

48.3k870153




48.3k870153










asked May 8 '14 at 22:27









Ila IsabelleIla Isabelle

50310




50310












  • $begingroup$
    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









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          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

              63.7k24191




              63.7k24191























                  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

                      2,4261819




                      2,4261819























                          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

                              5,9251524




                              5,9251524










                              answered Jan 4 at 22:47









                              M. C.M. C.

                              489




                              489























                                  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

                                      73.5k549128




                                      73.5k549128






























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                                          張江高科駅

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                                          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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