Are functions $f(x)=sin^2x-sin^4x$, $g(x)=cos^2x-cos^4x$ equal or not? Why?












1












$begingroup$


We have two functions named $f$ and $g$.
$$f(x)=sin^2x-sin^4x$$
$$g(x)=cos^2x-cos^4x$$
find out that they are equal functions or not.
My problem is with:




  1. finding their domain

  2. if we choose any $x$ from $D_f=D_g$ , then we have $$f(x)=g(x)$$


Please help me with the whole entry!
I dont know how to calculate their doamins and having trouble in step 2.






algebra-precalculus functions







share|cite|improve this question











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    1












    $begingroup$


    We have two functions named $f$ and $g$.
    $$f(x)=sin^2x-sin^4x$$
    $$g(x)=cos^2x-cos^4x$$
    find out that they are equal functions or not.
    My problem is with:




    1. finding their domain

    2. if we choose any $x$ from $D_f=D_g$ , then we have $$f(x)=g(x)$$


    Please help me with the whole entry!
    I dont know how to calculate their doamins and having trouble in step 2.






    algebra-precalculus functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      We have two functions named $f$ and $g$.
      $$f(x)=sin^2x-sin^4x$$
      $$g(x)=cos^2x-cos^4x$$
      find out that they are equal functions or not.
      My problem is with:




      1. finding their domain

      2. if we choose any $x$ from $D_f=D_g$ , then we have $$f(x)=g(x)$$


      Please help me with the whole entry!
      I dont know how to calculate their doamins and having trouble in step 2.






      algebra-precalculus functions







      share|cite|improve this question











      $endgroup$




      We have two functions named $f$ and $g$.
      $$f(x)=sin^2x-sin^4x$$
      $$g(x)=cos^2x-cos^4x$$
      find out that they are equal functions or not.
      My problem is with:




      1. finding their domain

      2. if we choose any $x$ from $D_f=D_g$ , then we have $$f(x)=g(x)$$


      Please help me with the whole entry!
      I dont know how to calculate their doamins and having trouble in step 2.






      algebra-precalculus functions




      algebra-precalculus functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 21:53









      Dando18

      4,68241235




      4,68241235










      asked Dec 8 '18 at 21:44









      user602338user602338

      1637




      1637






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          We have, by simple algebra,



          $sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin^2 x - cos^2 x)(1) = sin^2 x - cos^2 x, tag 1$



          and thus, again by simple algebra



          $f(x) = sin^2 x - sin^4 x = cos^2 x - cos^4 x = g(x), tag 2$



          wherever both $cos x$ and $sin x$ are defined; since the domain of each is $Bbb R$, we have



          $D_f = Bbb R = D_g tag 3$



          as well, and we have seen above that



          $forall x in Bbb R, ; f(x) = g(x). tag 4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            I want to chat with someone and gain information about euroup.
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
            $endgroup$
            – Robert Lewis
            Dec 8 '18 at 22:13












          • $begingroup$
            If you are in america not important i want gain some information about there too
            $endgroup$
            – user602338
            Dec 8 '18 at 22:13






          • 1




            $begingroup$
            I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
            $endgroup$
            – user602338
            Dec 8 '18 at 22:14



















          2












          $begingroup$

          Hint
          Use the identity
          $$
          cos^2x+sin^2x=1
          $$
          to show that $$f(x)=sin^2 xcos^2 x=g(x).$$ Both $f$ and $g$ have as their domain $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
            $endgroup$
            – user602338
            Dec 8 '18 at 21:50











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          We have, by simple algebra,



          $sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin^2 x - cos^2 x)(1) = sin^2 x - cos^2 x, tag 1$



          and thus, again by simple algebra



          $f(x) = sin^2 x - sin^4 x = cos^2 x - cos^4 x = g(x), tag 2$



          wherever both $cos x$ and $sin x$ are defined; since the domain of each is $Bbb R$, we have



          $D_f = Bbb R = D_g tag 3$



          as well, and we have seen above that



          $forall x in Bbb R, ; f(x) = g(x). tag 4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            I want to chat with someone and gain information about euroup.
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
            $endgroup$
            – Robert Lewis
            Dec 8 '18 at 22:13












          • $begingroup$
            If you are in america not important i want gain some information about there too
            $endgroup$
            – user602338
            Dec 8 '18 at 22:13






          • 1




            $begingroup$
            I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
            $endgroup$
            – user602338
            Dec 8 '18 at 22:14
















          2












          $begingroup$

          We have, by simple algebra,



          $sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin^2 x - cos^2 x)(1) = sin^2 x - cos^2 x, tag 1$



          and thus, again by simple algebra



          $f(x) = sin^2 x - sin^4 x = cos^2 x - cos^4 x = g(x), tag 2$



          wherever both $cos x$ and $sin x$ are defined; since the domain of each is $Bbb R$, we have



          $D_f = Bbb R = D_g tag 3$



          as well, and we have seen above that



          $forall x in Bbb R, ; f(x) = g(x). tag 4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            I want to chat with someone and gain information about euroup.
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
            $endgroup$
            – Robert Lewis
            Dec 8 '18 at 22:13












          • $begingroup$
            If you are in america not important i want gain some information about there too
            $endgroup$
            – user602338
            Dec 8 '18 at 22:13






          • 1




            $begingroup$
            I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
            $endgroup$
            – user602338
            Dec 8 '18 at 22:14














          2












          2








          2





          $begingroup$

          We have, by simple algebra,



          $sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin^2 x - cos^2 x)(1) = sin^2 x - cos^2 x, tag 1$



          and thus, again by simple algebra



          $f(x) = sin^2 x - sin^4 x = cos^2 x - cos^4 x = g(x), tag 2$



          wherever both $cos x$ and $sin x$ are defined; since the domain of each is $Bbb R$, we have



          $D_f = Bbb R = D_g tag 3$



          as well, and we have seen above that



          $forall x in Bbb R, ; f(x) = g(x). tag 4$






          share|cite|improve this answer









          $endgroup$



          We have, by simple algebra,



          $sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x) = (sin^2 x - cos^2 x)(1) = sin^2 x - cos^2 x, tag 1$



          and thus, again by simple algebra



          $f(x) = sin^2 x - sin^4 x = cos^2 x - cos^4 x = g(x), tag 2$



          wherever both $cos x$ and $sin x$ are defined; since the domain of each is $Bbb R$, we have



          $D_f = Bbb R = D_g tag 3$



          as well, and we have seen above that



          $forall x in Bbb R, ; f(x) = g(x). tag 4$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 22:07









          Robert LewisRobert Lewis

          46.6k23067




          46.6k23067












          • $begingroup$
            Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            I want to chat with someone and gain information about euroup.
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
            $endgroup$
            – Robert Lewis
            Dec 8 '18 at 22:13












          • $begingroup$
            If you are in america not important i want gain some information about there too
            $endgroup$
            – user602338
            Dec 8 '18 at 22:13






          • 1




            $begingroup$
            I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
            $endgroup$
            – user602338
            Dec 8 '18 at 22:14


















          • $begingroup$
            Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            I want to chat with someone and gain information about euroup.
            $endgroup$
            – user602338
            Dec 8 '18 at 22:10










          • $begingroup$
            @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
            $endgroup$
            – Robert Lewis
            Dec 8 '18 at 22:13












          • $begingroup$
            If you are in america not important i want gain some information about there too
            $endgroup$
            – user602338
            Dec 8 '18 at 22:13






          • 1




            $begingroup$
            I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
            $endgroup$
            – user602338
            Dec 8 '18 at 22:14
















          $begingroup$
          Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
          $endgroup$
          – user602338
          Dec 8 '18 at 22:10




          $begingroup$
          Thanks a lot Robert for your explanations. Robert it might be a personal question but i am eager to know that are you in euroup?
          $endgroup$
          – user602338
          Dec 8 '18 at 22:10












          $begingroup$
          I want to chat with someone and gain information about euroup.
          $endgroup$
          – user602338
          Dec 8 '18 at 22:10




          $begingroup$
          I want to chat with someone and gain information about euroup.
          $endgroup$
          – user602338
          Dec 8 '18 at 22:10












          $begingroup$
          @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
          $endgroup$
          – Robert Lewis
          Dec 8 '18 at 22:13






          $begingroup$
          @user602338: No, sorry. I'm in Santa Barbara, California, on the west coast of the USA.
          $endgroup$
          – Robert Lewis
          Dec 8 '18 at 22:13














          $begingroup$
          If you are in america not important i want gain some information about there too
          $endgroup$
          – user602338
          Dec 8 '18 at 22:13




          $begingroup$
          If you are in america not important i want gain some information about there too
          $endgroup$
          – user602338
          Dec 8 '18 at 22:13




          1




          1




          $begingroup$
          I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
          $endgroup$
          – user602338
          Dec 8 '18 at 22:14




          $begingroup$
          I know Mr.Lewis you are now 68 and a very experienced person! You are rare here i think!
          $endgroup$
          – user602338
          Dec 8 '18 at 22:14











          2












          $begingroup$

          Hint
          Use the identity
          $$
          cos^2x+sin^2x=1
          $$
          to show that $$f(x)=sin^2 xcos^2 x=g(x).$$ Both $f$ and $g$ have as their domain $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
            $endgroup$
            – user602338
            Dec 8 '18 at 21:50
















          2












          $begingroup$

          Hint
          Use the identity
          $$
          cos^2x+sin^2x=1
          $$
          to show that $$f(x)=sin^2 xcos^2 x=g(x).$$ Both $f$ and $g$ have as their domain $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
            $endgroup$
            – user602338
            Dec 8 '18 at 21:50














          2












          2








          2





          $begingroup$

          Hint
          Use the identity
          $$
          cos^2x+sin^2x=1
          $$
          to show that $$f(x)=sin^2 xcos^2 x=g(x).$$ Both $f$ and $g$ have as their domain $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$



          Hint
          Use the identity
          $$
          cos^2x+sin^2x=1
          $$
          to show that $$f(x)=sin^2 xcos^2 x=g(x).$$ Both $f$ and $g$ have as their domain $mathbb{R}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 21:47









          Foobaz JohnFoobaz John

          22k41352




          22k41352








          • 1




            $begingroup$
            Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
            $endgroup$
            – user602338
            Dec 8 '18 at 21:50














          • 1




            $begingroup$
            Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
            $endgroup$
            – user602338
            Dec 8 '18 at 21:50








          1




          1




          $begingroup$
          Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
          $endgroup$
          – user602338
          Dec 8 '18 at 21:50




          $begingroup$
          Really sorry but i didn't understand. Why f(x)=sin^2xcos^2x ???
          $endgroup$
          – user602338
          Dec 8 '18 at 21:50


















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