Simplification of $cos^4(x) + sin^4(x)$












0












$begingroup$


$$begin{align}
(sin x)^4+(cos x)^4
&= (1-cos2x)^2/4 + (1+cos2x)^2/4\
&= (1-2cos2x+(cos2x)^2+1+2cos2x+(cos2x)^2)/4\
&= (1+(cos2x)^2)/2\
end{align}$$



Is this correct?






trigonometry







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $$begin{align}
    (sin x)^4+(cos x)^4
    &= (1-cos2x)^2/4 + (1+cos2x)^2/4\
    &= (1-2cos2x+(cos2x)^2+1+2cos2x+(cos2x)^2)/4\
    &= (1+(cos2x)^2)/2\
    end{align}$$



    Is this correct?






    trigonometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      $$begin{align}
      (sin x)^4+(cos x)^4
      &= (1-cos2x)^2/4 + (1+cos2x)^2/4\
      &= (1-2cos2x+(cos2x)^2+1+2cos2x+(cos2x)^2)/4\
      &= (1+(cos2x)^2)/2\
      end{align}$$



      Is this correct?






      trigonometry







      share|cite|improve this question











      $endgroup$




      $$begin{align}
      (sin x)^4+(cos x)^4
      &= (1-cos2x)^2/4 + (1+cos2x)^2/4\
      &= (1-2cos2x+(cos2x)^2+1+2cos2x+(cos2x)^2)/4\
      &= (1+(cos2x)^2)/2\
      end{align}$$



      Is this correct?






      trigonometry




      trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 10:48







      Andes Lam

















      asked Dec 26 '18 at 10:30









      Andes LamAndes Lam

      213




      213






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          A faster way:
          begin{align}
          sin^4x+cos^4x&=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\
          &=1-frac12(2sin xcos x)^2=1-frac12sin^22x \[1ex]
          &=1-frac12frac{1-cos 4x}2=frac{3+cos 4x}4.
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:45






          • 1




            $begingroup$
            That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
            $endgroup$
            – Bernard
            Dec 26 '18 at 10:52



















          0












          $begingroup$

          $$cos2y=cos^2y-sin^2y=1-2sin^2y=2cos^2y-1$$
          $$impliessin^2y= dfrac{1-cos2y}2 ,cos^2y=dfrac{1+cos2y}2$$



          First $y=x$



          Apply the same $(y=2x)$ for $cos^22x=dfrac{1+cos4x}2$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:43












          • $begingroup$
            @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
            $endgroup$
            – lab bhattacharjee
            Dec 26 '18 at 10:47










          • $begingroup$
            Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:51










          • $begingroup$
            thanks lab, I get it and I make it thru now.
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:53



















          0












          $begingroup$

          $$
          left(frac{e^{ix}-e^{-ix}}{2i}right)^4+left(frac{e^{ix}+e^{-ix}}{2}right)^4 = frac{1}{2^4}left(left(e^{ix}-e^{-ix}right)^4+left(e^{ix}+e^{-ix}right)^4right)
          $$



          but



          $$
          (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4)
          $$



          hence



          $$
          sin^4 x+cos^4 x = frac{1}{2^3}left(2cos(4x)+6right) = frac{1}{4}left(cos(4x)+3right)
          $$






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            A faster way:
            begin{align}
            sin^4x+cos^4x&=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\
            &=1-frac12(2sin xcos x)^2=1-frac12sin^22x \[1ex]
            &=1-frac12frac{1-cos 4x}2=frac{3+cos 4x}4.
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:45






            • 1




              $begingroup$
              That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
              $endgroup$
              – Bernard
              Dec 26 '18 at 10:52
















            3












            $begingroup$

            A faster way:
            begin{align}
            sin^4x+cos^4x&=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\
            &=1-frac12(2sin xcos x)^2=1-frac12sin^22x \[1ex]
            &=1-frac12frac{1-cos 4x}2=frac{3+cos 4x}4.
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:45






            • 1




              $begingroup$
              That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
              $endgroup$
              – Bernard
              Dec 26 '18 at 10:52














            3












            3








            3





            $begingroup$

            A faster way:
            begin{align}
            sin^4x+cos^4x&=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\
            &=1-frac12(2sin xcos x)^2=1-frac12sin^22x \[1ex]
            &=1-frac12frac{1-cos 4x}2=frac{3+cos 4x}4.
            end{align}






            share|cite|improve this answer









            $endgroup$



            A faster way:
            begin{align}
            sin^4x+cos^4x&=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x\
            &=1-frac12(2sin xcos x)^2=1-frac12sin^22x \[1ex]
            &=1-frac12frac{1-cos 4x}2=frac{3+cos 4x}4.
            end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 26 '18 at 10:40









            BernardBernard

            123k741116




            123k741116












            • $begingroup$
              That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:45






            • 1




              $begingroup$
              That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
              $endgroup$
              – Bernard
              Dec 26 '18 at 10:52


















            • $begingroup$
              That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:45






            • 1




              $begingroup$
              That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
              $endgroup$
              – Bernard
              Dec 26 '18 at 10:52
















            $begingroup$
            That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:45




            $begingroup$
            That's exactly the answer I found in other posts. I just wonder why the double angle formula couldn't arrive at the same conclusion at yours...
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:45




            1




            1




            $begingroup$
            That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
            $endgroup$
            – Bernard
            Dec 26 '18 at 10:52




            $begingroup$
            That's because you didn't go till the end of the linearisation process($cos^22x=frac{1+cos4x}2$).
            $endgroup$
            – Bernard
            Dec 26 '18 at 10:52











            0












            $begingroup$

            $$cos2y=cos^2y-sin^2y=1-2sin^2y=2cos^2y-1$$
            $$impliessin^2y= dfrac{1-cos2y}2 ,cos^2y=dfrac{1+cos2y}2$$



            First $y=x$



            Apply the same $(y=2x)$ for $cos^22x=dfrac{1+cos4x}2$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:43












            • $begingroup$
              @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
              $endgroup$
              – lab bhattacharjee
              Dec 26 '18 at 10:47










            • $begingroup$
              Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:51










            • $begingroup$
              thanks lab, I get it and I make it thru now.
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:53
















            0












            $begingroup$

            $$cos2y=cos^2y-sin^2y=1-2sin^2y=2cos^2y-1$$
            $$impliessin^2y= dfrac{1-cos2y}2 ,cos^2y=dfrac{1+cos2y}2$$



            First $y=x$



            Apply the same $(y=2x)$ for $cos^22x=dfrac{1+cos4x}2$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:43












            • $begingroup$
              @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
              $endgroup$
              – lab bhattacharjee
              Dec 26 '18 at 10:47










            • $begingroup$
              Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:51










            • $begingroup$
              thanks lab, I get it and I make it thru now.
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:53














            0












            0








            0





            $begingroup$

            $$cos2y=cos^2y-sin^2y=1-2sin^2y=2cos^2y-1$$
            $$impliessin^2y= dfrac{1-cos2y}2 ,cos^2y=dfrac{1+cos2y}2$$



            First $y=x$



            Apply the same $(y=2x)$ for $cos^22x=dfrac{1+cos4x}2$






            share|cite|improve this answer











            $endgroup$



            $$cos2y=cos^2y-sin^2y=1-2sin^2y=2cos^2y-1$$
            $$impliessin^2y= dfrac{1-cos2y}2 ,cos^2y=dfrac{1+cos2y}2$$



            First $y=x$



            Apply the same $(y=2x)$ for $cos^22x=dfrac{1+cos4x}2$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 26 '18 at 10:46

























            answered Dec 26 '18 at 10:32









            lab bhattacharjeelab bhattacharjee

            227k15158275




            227k15158275












            • $begingroup$
              An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:43












            • $begingroup$
              @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
              $endgroup$
              – lab bhattacharjee
              Dec 26 '18 at 10:47










            • $begingroup$
              Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:51










            • $begingroup$
              thanks lab, I get it and I make it thru now.
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:53


















            • $begingroup$
              An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:43












            • $begingroup$
              @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
              $endgroup$
              – lab bhattacharjee
              Dec 26 '18 at 10:47










            • $begingroup$
              Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
              $endgroup$
              – Eevee Trainer
              Dec 26 '18 at 10:51










            • $begingroup$
              thanks lab, I get it and I make it thru now.
              $endgroup$
              – Andes Lam
              Dec 26 '18 at 10:53
















            $begingroup$
            An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:43






            $begingroup$
            An error: $$cos^2(y) = frac{1 + cos(2y)}{2}$$ What you have suggests $sin^2y = cos^2y$, which is patently false. Neglecting, of course, that you're not at all addressing OP's question. Where does $x$ even come from in your answer? It's impossible to follow your logic.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:43














            $begingroup$
            @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
            $endgroup$
            – lab bhattacharjee
            Dec 26 '18 at 10:47




            $begingroup$
            @EeveeTrainer, That was a typo. I introduced to $y$ to replace it with $x,2x$ one by one
            $endgroup$
            – lab bhattacharjee
            Dec 26 '18 at 10:47












            $begingroup$
            Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:51




            $begingroup$
            Okay, now I see where you're coming from. Though personally I would elaborate a little more in your answer to demonstrate the utility of this myself. A little elaboration and explanation can go a long way in terms of an answer's usefulness, because if it's impenetrable for other MSE users (IIRC I and someone else had like issues with another of your answers tonight) it's probably not very helpful to a question's OP in case of questions like these. Just my two cents.
            $endgroup$
            – Eevee Trainer
            Dec 26 '18 at 10:51












            $begingroup$
            thanks lab, I get it and I make it thru now.
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:53




            $begingroup$
            thanks lab, I get it and I make it thru now.
            $endgroup$
            – Andes Lam
            Dec 26 '18 at 10:53











            0












            $begingroup$

            $$
            left(frac{e^{ix}-e^{-ix}}{2i}right)^4+left(frac{e^{ix}+e^{-ix}}{2}right)^4 = frac{1}{2^4}left(left(e^{ix}-e^{-ix}right)^4+left(e^{ix}+e^{-ix}right)^4right)
            $$



            but



            $$
            (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4)
            $$



            hence



            $$
            sin^4 x+cos^4 x = frac{1}{2^3}left(2cos(4x)+6right) = frac{1}{4}left(cos(4x)+3right)
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $$
              left(frac{e^{ix}-e^{-ix}}{2i}right)^4+left(frac{e^{ix}+e^{-ix}}{2}right)^4 = frac{1}{2^4}left(left(e^{ix}-e^{-ix}right)^4+left(e^{ix}+e^{-ix}right)^4right)
              $$



              but



              $$
              (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4)
              $$



              hence



              $$
              sin^4 x+cos^4 x = frac{1}{2^3}left(2cos(4x)+6right) = frac{1}{4}left(cos(4x)+3right)
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $$
                left(frac{e^{ix}-e^{-ix}}{2i}right)^4+left(frac{e^{ix}+e^{-ix}}{2}right)^4 = frac{1}{2^4}left(left(e^{ix}-e^{-ix}right)^4+left(e^{ix}+e^{-ix}right)^4right)
                $$



                but



                $$
                (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4)
                $$



                hence



                $$
                sin^4 x+cos^4 x = frac{1}{2^3}left(2cos(4x)+6right) = frac{1}{4}left(cos(4x)+3right)
                $$






                share|cite|improve this answer









                $endgroup$



                $$
                left(frac{e^{ix}-e^{-ix}}{2i}right)^4+left(frac{e^{ix}+e^{-ix}}{2}right)^4 = frac{1}{2^4}left(left(e^{ix}-e^{-ix}right)^4+left(e^{ix}+e^{-ix}right)^4right)
                $$



                but



                $$
                (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4)
                $$



                hence



                $$
                sin^4 x+cos^4 x = frac{1}{2^3}left(2cos(4x)+6right) = frac{1}{4}left(cos(4x)+3right)
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 26 '18 at 11:56









                CesareoCesareo

                9,3613517




                9,3613517






























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