Proving $1-cos2a+2sin a sin3a = 2 sin^22a$ and $sin^22a - sin^2a = sin3asin a$











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Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.




1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$




I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
$$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
but I am lost after this point.




2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$




I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.



That brings me to:
$$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
or
$$sin^2a(4 cos^2a -1) = sin3asin a$$
but I cannot find the solution.




So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?




Thanks for your attention. I’m looking forward to your reply.






trigonometry







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    up vote
    -1
    down vote

    favorite












    Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.




    1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$




    I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
    $$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
    but I am lost after this point.




    2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$




    I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.



    That brings me to:
    $$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
    or
    $$sin^2a(4 cos^2a -1) = sin3asin a$$
    but I cannot find the solution.




    So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?




    Thanks for your attention. I’m looking forward to your reply.






    trigonometry







    share|cite|improve this question


























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.




      1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$




      I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
      $$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
      but I am lost after this point.




      2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$




      I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.



      That brings me to:
      $$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
      or
      $$sin^2a(4 cos^2a -1) = sin3asin a$$
      but I cannot find the solution.




      So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?




      Thanks for your attention. I’m looking forward to your reply.






      trigonometry







      share|cite|improve this question















      Thanks for viewing this post. I got an assignment but have a hard time solving a few questions.




      1) Prove: $$1 - cos2a + 2 sin a, sin3a = 2 sin^22a$$




      I started off with rewriting $1 - cos2a$ to $2 sin^2a$. That gets me to:
      $$2 sin^2a + 2sin a, sin3a = 2 sin^22a$$
      but I am lost after this point.




      2) Prove: $$sin^22a - sin^2a = sin3a ,sin a$$




      I have begun with replacing $sin^22a$ with $4 sin^2a ,cos^2a$.



      That brings me to:
      $$4 sin^2a,cos^2a - sin^2a = sin3a, sin a$$
      or
      $$sin^2a(4 cos^2a -1) = sin3asin a$$
      but I cannot find the solution.




      So, can anyone give me hints on how to prove these 2 tasks, and also correct me if I started the exercise the wrong way?




      Thanks for your attention. I’m looking forward to your reply.






      trigonometry




      trigonometry






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      edited Nov 28 at 22:38









      Blue

      47.3k870149




      47.3k870149










      asked Nov 28 at 21:33









      Toa Narumi

      91




      91






















          4 Answers
          4






          active

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          up vote
          0
          down vote













          Hint:



          Linearise the r.h.s.:
          $$2sin^22a=1-cos 4a,$$



          and, in the l.h.s., $;2sin asin 3a$.






          share|cite|improve this answer




























            up vote
            0
            down vote














            1. Hint:


            Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
            $$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
            $$sin(3alpha)=3sinalpha-4sin^3alpha$$



            SOLUTION




            $$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$





            1. SOLUTION


              $$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$









            share|cite|improve this answer




























              up vote
              0
              down vote













              Using http://mathworld.wolfram.com/WernerFormulas.html,



              $$2sin asin3a=cos(3-1)a-cos(3+1)a$$
              Then use $cos2y=1-2sin^2y$



              For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



              See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$






              share|cite|improve this answer




























                up vote
                -1
                down vote













                HINT



                For the first one use that



                $$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$



                then



                $$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$



                $$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
                $$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$



                and recall that $cos (2x)=1-2sin^2 x$.



                Can you proceed also for the second one, using the same identities?






                share|cite|improve this answer























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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes








                  up vote
                  0
                  down vote













                  Hint:



                  Linearise the r.h.s.:
                  $$2sin^22a=1-cos 4a,$$



                  and, in the l.h.s., $;2sin asin 3a$.






                  share|cite|improve this answer

























                    up vote
                    0
                    down vote













                    Hint:



                    Linearise the r.h.s.:
                    $$2sin^22a=1-cos 4a,$$



                    and, in the l.h.s., $;2sin asin 3a$.






                    share|cite|improve this answer























                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      Hint:



                      Linearise the r.h.s.:
                      $$2sin^22a=1-cos 4a,$$



                      and, in the l.h.s., $;2sin asin 3a$.






                      share|cite|improve this answer












                      Hint:



                      Linearise the r.h.s.:
                      $$2sin^22a=1-cos 4a,$$



                      and, in the l.h.s., $;2sin asin 3a$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 28 at 21:47









                      Bernard

                      117k637110




                      117k637110






















                          up vote
                          0
                          down vote














                          1. Hint:


                          Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
                          $$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
                          $$sin(3alpha)=3sinalpha-4sin^3alpha$$



                          SOLUTION




                          $$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$





                          1. SOLUTION


                            $$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$









                          share|cite|improve this answer

























                            up vote
                            0
                            down vote














                            1. Hint:


                            Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
                            $$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
                            $$sin(3alpha)=3sinalpha-4sin^3alpha$$



                            SOLUTION




                            $$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$





                            1. SOLUTION


                              $$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$









                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote










                              1. Hint:


                              Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
                              $$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
                              $$sin(3alpha)=3sinalpha-4sin^3alpha$$



                              SOLUTION




                              $$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$





                              1. SOLUTION


                                $$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$









                              share|cite|improve this answer













                              1. Hint:


                              Consider the trigonometric identities $$sin^2alpha+cos^2alpha=1$$
                              $$sin(2alpha)=2sinalpha cosalphaquad cos(2alpha)=cos^2alpha-sin^2alpha$$
                              $$sin(3alpha)=3sinalpha-4sin^3alpha$$



                              SOLUTION




                              $$sin^2alpha(1-sin^2alpha)=sin^2alphacos^2alpha$$ $$= sinalpha(sinalpha-sin^3alpha)quad mid·4$$ $$Rightarrow sinalpha(sinalpha+3sinalpha-4sin^3alpha)=4sin^2alpha ·cos^2alpha$$ $$sinalpha(sinalpha+sin(3alpha))=sin^2alpha+sinalpha sin(3alpha)=4sin^2alpha ·cos^2alpha=sin^2(2alpha)$$ $$Rightarrow 2sin^2alpha+2sinalphasin(3alpha)=bigl(1-cos^2alpha+sin^2alphabigr)+2sinalphasin(3alpha)=2sin^2alpha$$ Hence $$1-cos(2alpha)+2sinalphasin(3alpha)=2sin^2alpha$$





                              1. SOLUTION


                                $$sinalpha(3-4sin^2alpha)=sin(3alpha)$$ Thus $$sinalpha(4-4sin^2alpha-1)=sinalpha(4cos^2alpha-1)=4sinalphacos^2alpha-sinalpha=sin(3alpha)$$ $$Rightarrow4sin^2alphacos^2alpha-sin^2alpha=sin^2(2alpha)-sin^2alpha=sin(3alpha)sin(alpha)$$










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                              answered Nov 28 at 22:06









                              Dr. Mathva

                              908315




                              908315






















                                  up vote
                                  0
                                  down vote













                                  Using http://mathworld.wolfram.com/WernerFormulas.html,



                                  $$2sin asin3a=cos(3-1)a-cos(3+1)a$$
                                  Then use $cos2y=1-2sin^2y$



                                  For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



                                  See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    Using http://mathworld.wolfram.com/WernerFormulas.html,



                                    $$2sin asin3a=cos(3-1)a-cos(3+1)a$$
                                    Then use $cos2y=1-2sin^2y$



                                    For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



                                    See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Using http://mathworld.wolfram.com/WernerFormulas.html,



                                      $$2sin asin3a=cos(3-1)a-cos(3+1)a$$
                                      Then use $cos2y=1-2sin^2y$



                                      For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



                                      See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$






                                      share|cite|improve this answer












                                      Using http://mathworld.wolfram.com/WernerFormulas.html,



                                      $$2sin asin3a=cos(3-1)a-cos(3+1)a$$
                                      Then use $cos2y=1-2sin^2y$



                                      For the second, use Prove $ sin(A+B)sin(A-B)=sin^2A-sin^2B $



                                      See also: Prove that $cos (A + B)cos (A - B) = {cos ^2}A - {sin ^2}B$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 29 at 4:58









                                      lab bhattacharjee

                                      222k15155273




                                      222k15155273






















                                          up vote
                                          -1
                                          down vote













                                          HINT



                                          For the first one use that



                                          $$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$



                                          then



                                          $$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$



                                          $$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
                                          $$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$



                                          and recall that $cos (2x)=1-2sin^2 x$.



                                          Can you proceed also for the second one, using the same identities?






                                          share|cite|improve this answer



























                                            up vote
                                            -1
                                            down vote













                                            HINT



                                            For the first one use that



                                            $$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$



                                            then



                                            $$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$



                                            $$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
                                            $$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$



                                            and recall that $cos (2x)=1-2sin^2 x$.



                                            Can you proceed also for the second one, using the same identities?






                                            share|cite|improve this answer

























                                              up vote
                                              -1
                                              down vote










                                              up vote
                                              -1
                                              down vote









                                              HINT



                                              For the first one use that



                                              $$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$



                                              then



                                              $$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$



                                              $$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
                                              $$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$



                                              and recall that $cos (2x)=1-2sin^2 x$.



                                              Can you proceed also for the second one, using the same identities?






                                              share|cite|improve this answer














                                              HINT



                                              For the first one use that



                                              $$2sin theta sin varphi = {{cos(theta - varphi) - cos(theta + varphi)} }$$



                                              then



                                              $$1 - cos (2a) + 2 sin a, sin (3a) = 2 sin^2 (2a=$$



                                              $$1 - cos (2a) + cos(-2a)-cos (4a) = 2 sin^2 (2a)$$
                                              $$1 - color{red}{cos (2a) + cos(2a)}-cos (4a) = 2 sin^2 (2a)$$



                                              and recall that $cos (2x)=1-2sin^2 x$.



                                              Can you proceed also for the second one, using the same identities?







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited Nov 28 at 21:59

























                                              answered Nov 28 at 21:41









                                              gimusi

                                              1




                                              1






























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                                                  Sphinx de Gizeh

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                                                  Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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                                                  Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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                                                  3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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