Proof that $cos(2theta) = cos^2theta-sin^2theta$ by showing equivalence of derivatives












5












$begingroup$


It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.



But if we let $f(x) = sin(2x)$, we can differentiate two ways:



1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$



2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$



Is this a viable proof?



Thanks in Advance!






derivatives trigonometry proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
    $endgroup$
    – Janitha357
    Dec 6 '18 at 4:56






  • 1




    $begingroup$
    It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
    $endgroup$
    – timtfj
    Jan 9 at 4:38
















5












$begingroup$


It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.



But if we let $f(x) = sin(2x)$, we can differentiate two ways:



1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$



2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$



Is this a viable proof?



Thanks in Advance!






derivatives trigonometry proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
    $endgroup$
    – Janitha357
    Dec 6 '18 at 4:56






  • 1




    $begingroup$
    It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
    $endgroup$
    – timtfj
    Jan 9 at 4:38














5












5








5


1



$begingroup$


It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.



But if we let $f(x) = sin(2x)$, we can differentiate two ways:



1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$



2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$



Is this a viable proof?



Thanks in Advance!






derivatives trigonometry proof-verification proof-writing







share|cite|improve this question











$endgroup$




It can be shown using trig identities that $cos(2theta) = cos^2theta-sin^2theta$.



But if we let $f(x) = sin(2x)$, we can differentiate two ways:



1) $$f(x) = sin(2x) rightarrow f(x) = 2sin(x)cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[cos(x)cos(x)-sin(x)sin(x)] = 2[cos^2(x)-sin^2(x)]$$



2) If we differentiate $f(x)$ as is, then :
$$frac{d}{dx}f(x) = 2cos(2x) $$
Therefore: $2cos(2x) = 2[cos^2(x) - sin^2(x)] rightarrow cos(2x) = cos^2(x) -sin^2(x)$



Is this a viable proof?



Thanks in Advance!






derivatives trigonometry proof-verification proof-writing




derivatives trigonometry proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 4:30









Blue

47.8k870152




47.8k870152










asked Dec 6 '18 at 4:48









Aniruddh VenkatesanAniruddh Venkatesan

154112




154112








  • 1




    $begingroup$
    I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
    $endgroup$
    – Janitha357
    Dec 6 '18 at 4:56






  • 1




    $begingroup$
    It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
    $endgroup$
    – timtfj
    Jan 9 at 4:38














  • 1




    $begingroup$
    I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
    $endgroup$
    – Janitha357
    Dec 6 '18 at 4:56






  • 1




    $begingroup$
    It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
    $endgroup$
    – timtfj
    Jan 9 at 4:38








1




1




$begingroup$
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
$endgroup$
– Janitha357
Dec 6 '18 at 4:56




$begingroup$
I think the only thing that you can infer by derivatives of two functions being equal is that the two functions differ by a constant.
$endgroup$
– Janitha357
Dec 6 '18 at 4:56




1




1




$begingroup$
It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
$endgroup$
– timtfj
Jan 9 at 4:38




$begingroup$
It doesn't involve the derivatives of two functions—it calculates the derivative of one function two ways. So the resulting expressions represent the same thing and therefore have to be equal.
$endgroup$
– timtfj
Jan 9 at 4:38










3 Answers
3






active

oldest

votes


















3












$begingroup$

Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.



A bit overkill, obviously, but 100% valid, as with your method.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The ideas and work shown are valid.



    In order to format the work here into the format of a proof, we need to identify our hypotheses.



    The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.



    Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The nethod is valid and its logic is:





      • $f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

      • One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

      • Since $f'(x)$ is unique, $g(x)=h(x)$.


      It would only be invalid if, say, one of the differentiation methods was wrong.






      share|cite|improve this answer









      $endgroup$













        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%2f3028046%2fproof-that-cos2-theta-cos2-theta-sin2-theta-by-showing-equivalence-of%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












        $begingroup$

        Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.



        A bit overkill, obviously, but 100% valid, as with your method.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.



          A bit overkill, obviously, but 100% valid, as with your method.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.



            A bit overkill, obviously, but 100% valid, as with your method.






            share|cite|improve this answer









            $endgroup$



            Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.



            A bit overkill, obviously, but 100% valid, as with your method.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 4:50









            Eevee TrainerEevee Trainer

            5,4691936




            5,4691936























                3












                $begingroup$

                The ideas and work shown are valid.



                In order to format the work here into the format of a proof, we need to identify our hypotheses.



                The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.



                Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  The ideas and work shown are valid.



                  In order to format the work here into the format of a proof, we need to identify our hypotheses.



                  The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.



                  Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    The ideas and work shown are valid.



                    In order to format the work here into the format of a proof, we need to identify our hypotheses.



                    The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.



                    Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.






                    share|cite|improve this answer









                    $endgroup$



                    The ideas and work shown are valid.



                    In order to format the work here into the format of a proof, we need to identify our hypotheses.



                    The things that are assumed here are that $sin{2x}=2sin{x}cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.



                    Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 6 '18 at 5:18









                    John BJohn B

                    1766




                    1766























                        0












                        $begingroup$

                        The nethod is valid and its logic is:





                        • $f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

                        • One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

                        • Since $f'(x)$ is unique, $g(x)=h(x)$.


                        It would only be invalid if, say, one of the differentiation methods was wrong.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The nethod is valid and its logic is:





                          • $f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

                          • One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

                          • Since $f'(x)$ is unique, $g(x)=h(x)$.


                          It would only be invalid if, say, one of the differentiation methods was wrong.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The nethod is valid and its logic is:





                            • $f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

                            • One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

                            • Since $f'(x)$ is unique, $g(x)=h(x)$.


                            It would only be invalid if, say, one of the differentiation methods was wrong.






                            share|cite|improve this answer









                            $endgroup$



                            The nethod is valid and its logic is:





                            • $f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

                            • One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

                            • Since $f'(x)$ is unique, $g(x)=h(x)$.


                            It would only be invalid if, say, one of the differentiation methods was wrong.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 9 at 4:22









                            timtfjtimtfj

                            1,318318




                            1,318318






























                                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.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028046%2fproof-that-cos2-theta-cos2-theta-sin2-theta-by-showing-equivalence-of%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

                                Mario Kart Wii

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                                Read more

                                What does “Dominus providebit” mean?

                                Image
                                5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                                Read more

                                File:Tiny Toon Adventures Wacky Sports JP Title.png

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
                                Read more