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







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$endgroup$








  • 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




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edited Jan 9 at 4:30









Blue

47.8k870152




47.8k870152










asked Dec 6 '18 at 4:48









Aniruddh VenkatesanAniruddh Venkatesan

127110




127110








  • 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









      $endgroup$













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        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

            5,4491936




            5,4491936























                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






























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