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












0














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



Is this correct?






trigonometry







share|cite|improve this question





























    0














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



    Is this correct?






    trigonometry







    share|cite|improve this question



























      0












      0








      0


      1





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



      Is this correct?






      trigonometry







      share|cite|improve this question















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



      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 at 10:48

























      asked Dec 26 at 10:30









      Andes Lam

      32




      32






















          3 Answers
          3






          active

          oldest

          votes


















          3














          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





















          • 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...
            – Andes Lam
            Dec 26 at 10:45






          • 1




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



















          0














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



          First $y=x$



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






          share|cite|improve this answer























          • 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.
            – Eevee Trainer
            Dec 26 at 10:43












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










          • 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.
            – Eevee Trainer
            Dec 26 at 10:51










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



















          0














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



          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





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            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





















            • 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...
              – Andes Lam
              Dec 26 at 10:45






            • 1




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
















            3














            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





















            • 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...
              – Andes Lam
              Dec 26 at 10:45






            • 1




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














            3












            3








            3






            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












            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 at 10:40









            Bernard

            118k639112




            118k639112












            • 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...
              – Andes Lam
              Dec 26 at 10:45






            • 1




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


















            • 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...
              – Andes Lam
              Dec 26 at 10:45






            • 1




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
















            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...
            – Andes Lam
            Dec 26 at 10:45




            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...
            – Andes Lam
            Dec 26 at 10:45




            1




            1




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




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











            0














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



            First $y=x$



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






            share|cite|improve this answer























            • 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.
              – Eevee Trainer
              Dec 26 at 10:43












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










            • 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.
              – Eevee Trainer
              Dec 26 at 10:51










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
















            0














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



            First $y=x$



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






            share|cite|improve this answer























            • 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.
              – Eevee Trainer
              Dec 26 at 10:43












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










            • 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.
              – Eevee Trainer
              Dec 26 at 10:51










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














            0












            0








            0






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



            First $y=x$



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






            share|cite|improve this answer














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



            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 at 10:46

























            answered Dec 26 at 10:32









            lab bhattacharjee

            223k15156274




            223k15156274












            • 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.
              – Eevee Trainer
              Dec 26 at 10:43












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










            • 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.
              – Eevee Trainer
              Dec 26 at 10:51










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


















            • 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.
              – Eevee Trainer
              Dec 26 at 10:43












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










            • 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.
              – Eevee Trainer
              Dec 26 at 10:51










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
















            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.
            – Eevee Trainer
            Dec 26 at 10:43






            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.
            – Eevee Trainer
            Dec 26 at 10:43














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




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












            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.
            – Eevee Trainer
            Dec 26 at 10:51




            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.
            – Eevee Trainer
            Dec 26 at 10:51












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




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











            0














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



            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


























              0














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



              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
























                0












                0








                0






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



                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












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



                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 at 11:56









                Cesareo

                8,1983516




                8,1983516






























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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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