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





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052817%2fsimplification-of-cos4x-sin4x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





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


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052817%2fsimplification-of-cos4x-sin4x%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Human spaceflight

                    Image
                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                    Read more

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    Image
                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                    Read more

                    File:DeusFollowingSea.jpg

                    Image
                    File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                    Read more