Prove $sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta $












4












$begingroup$


Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$



I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.



After some help I found that you move the identity around, so:



$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$



Then,



$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$



the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?






trigonometry







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$endgroup$












  • $begingroup$
    Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
    $endgroup$
    – John Joy
    Aug 30 '15 at 0:23


















4












$begingroup$


Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$



I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.



After some help I found that you move the identity around, so:



$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$



Then,



$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$



the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
    $endgroup$
    – John Joy
    Aug 30 '15 at 0:23
















4












4








4


2



$begingroup$


Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$



I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.



After some help I found that you move the identity around, so:



$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$



Then,



$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$



the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?






trigonometry







share|cite|improve this question











$endgroup$




Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$



I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.



After some help I found that you move the identity around, so:



$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$



Then,



$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$



the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?






trigonometry




trigonometry






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edited Nov 11 '17 at 5:21









Paramanand Singh

51.1k557168




51.1k557168










asked Aug 29 '15 at 23:30









Sunny MannSunny Mann

36118




36118












  • $begingroup$
    Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
    $endgroup$
    – John Joy
    Aug 30 '15 at 0:23




















  • $begingroup$
    Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
    $endgroup$
    – John Joy
    Aug 30 '15 at 0:23


















$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23






$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23












7 Answers
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$begingroup$

I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $



We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $



$sin^2theta + cos^4theta = cos^2theta + sin^4theta $



$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $



$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $



$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $



$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $



$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $



$sin^4theta+cos^2theta=cos^2theta + sin^4theta $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (1−2sin2θ+sin4θ) where did you get this expression from?
    $endgroup$
    – Sunny Mann
    Aug 30 '15 at 0:30






  • 1




    $begingroup$
    Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
    $endgroup$
    – Leo
    Aug 30 '15 at 0:30












  • $begingroup$
    you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
    $endgroup$
    – Sunny Mann
    Aug 30 '15 at 0:34












  • $begingroup$
    We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
    $endgroup$
    – Leo
    Aug 30 '15 at 0:37












  • $begingroup$
    it's supposed to be $1- sin^2(theta), no?
    $endgroup$
    – Sunny Mann
    Aug 30 '15 at 0:40



















13












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Rewrite this as



$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How does this prove the identity?
    $endgroup$
    – Sunny Mann
    Aug 29 '15 at 23:34






  • 2




    $begingroup$
    Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
    $endgroup$
    – Micah
    Aug 29 '15 at 23:34










  • $begingroup$
    could you factor any side actually both of them look like difference of squares?
    $endgroup$
    – Sunny Mann
    Aug 29 '15 at 23:39










  • $begingroup$
    You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
    $endgroup$
    – Micah
    Aug 29 '15 at 23:40










  • $begingroup$
    Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
    $endgroup$
    – Sunny Mann
    Aug 29 '15 at 23:44



















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As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,



$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Here would be the other points to remember:



    $sin^2theta+cos^2theta=1$



    $x^4-y^4=(x^2-y^2)(x^2+y^2)$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:



      Consider the function



      $$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$



      Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.



      Hope you find this useful/interesting.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        A "forwards" proof:



        Render



        $sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$



        Regroup the terms on the right as



        $(sin^2theta+1-2sin^2theta)+sin^4theta$



        and put $1-sin^2theta=cos^2theta$.






        share|cite|improve this answer











        $endgroup$





















          0












          $begingroup$

          $$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            could you elaborate please?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:14










          • $begingroup$
            Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
            $endgroup$
            – John Joy
            Aug 30 '15 at 0:20










          • $begingroup$
            $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:26












          • $begingroup$
            What do you get when you expand $(1-sin^2theta)^2$?
            $endgroup$
            – John Joy
            Aug 30 '15 at 0:28












          • $begingroup$
            $1- sin^4(theta) or cos^4(theta)$
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:36











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          7 Answers
          7






          active

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          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $



          We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $



          $sin^2theta + cos^4theta = cos^2theta + sin^4theta $



          $sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $



          $sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $



          $sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $



          $sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $



          $sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $



          $sin^4theta+cos^2theta=cos^2theta + sin^4theta $






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (1−2sin2θ+sin4θ) where did you get this expression from?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:30






          • 1




            $begingroup$
            Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
            $endgroup$
            – Leo
            Aug 30 '15 at 0:30












          • $begingroup$
            you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:34












          • $begingroup$
            We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
            $endgroup$
            – Leo
            Aug 30 '15 at 0:37












          • $begingroup$
            it's supposed to be $1- sin^2(theta), no?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:40
















          4












          $begingroup$

          I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $



          We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $



          $sin^2theta + cos^4theta = cos^2theta + sin^4theta $



          $sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $



          $sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $



          $sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $



          $sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $



          $sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $



          $sin^4theta+cos^2theta=cos^2theta + sin^4theta $






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (1−2sin2θ+sin4θ) where did you get this expression from?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:30






          • 1




            $begingroup$
            Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
            $endgroup$
            – Leo
            Aug 30 '15 at 0:30












          • $begingroup$
            you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:34












          • $begingroup$
            We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
            $endgroup$
            – Leo
            Aug 30 '15 at 0:37












          • $begingroup$
            it's supposed to be $1- sin^2(theta), no?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:40














          4












          4








          4





          $begingroup$

          I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $



          We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $



          $sin^2theta + cos^4theta = cos^2theta + sin^4theta $



          $sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $



          $sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $



          $sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $



          $sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $



          $sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $



          $sin^4theta+cos^2theta=cos^2theta + sin^4theta $






          share|cite|improve this answer









          $endgroup$



          I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $



          We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $



          $sin^2theta + cos^4theta = cos^2theta + sin^4theta $



          $sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $



          $sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $



          $sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $



          $sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $



          $sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $



          $sin^4theta+cos^2theta=cos^2theta + sin^4theta $







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 30 '15 at 0:22









          LeoLeo

          908




          908












          • $begingroup$
            (1−2sin2θ+sin4θ) where did you get this expression from?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:30






          • 1




            $begingroup$
            Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
            $endgroup$
            – Leo
            Aug 30 '15 at 0:30












          • $begingroup$
            you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:34












          • $begingroup$
            We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
            $endgroup$
            – Leo
            Aug 30 '15 at 0:37












          • $begingroup$
            it's supposed to be $1- sin^2(theta), no?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:40


















          • $begingroup$
            (1−2sin2θ+sin4θ) where did you get this expression from?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:30






          • 1




            $begingroup$
            Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
            $endgroup$
            – Leo
            Aug 30 '15 at 0:30












          • $begingroup$
            you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:34












          • $begingroup$
            We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
            $endgroup$
            – Leo
            Aug 30 '15 at 0:37












          • $begingroup$
            it's supposed to be $1- sin^2(theta), no?
            $endgroup$
            – Sunny Mann
            Aug 30 '15 at 0:40
















          $begingroup$
          (1−2sin2θ+sin4θ) where did you get this expression from?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:30




          $begingroup$
          (1−2sin2θ+sin4θ) where did you get this expression from?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:30




          1




          1




          $begingroup$
          Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
          $endgroup$
          – Leo
          Aug 30 '15 at 0:30






          $begingroup$
          Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
          $endgroup$
          – Leo
          Aug 30 '15 at 0:30














          $begingroup$
          you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:34






          $begingroup$
          you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:34














          $begingroup$
          We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
          $endgroup$
          – Leo
          Aug 30 '15 at 0:37






          $begingroup$
          We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
          $endgroup$
          – Leo
          Aug 30 '15 at 0:37














          $begingroup$
          it's supposed to be $1- sin^2(theta), no?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:40




          $begingroup$
          it's supposed to be $1- sin^2(theta), no?
          $endgroup$
          – Sunny Mann
          Aug 30 '15 at 0:40











          13












          $begingroup$

          Rewrite this as



          $$
          sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
          $$
          and then factor the right-hand side as a difference of two squares.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does this prove the identity?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:34






          • 2




            $begingroup$
            Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
            $endgroup$
            – Micah
            Aug 29 '15 at 23:34










          • $begingroup$
            could you factor any side actually both of them look like difference of squares?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:39










          • $begingroup$
            You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
            $endgroup$
            – Micah
            Aug 29 '15 at 23:40










          • $begingroup$
            Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:44
















          13












          $begingroup$

          Rewrite this as



          $$
          sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
          $$
          and then factor the right-hand side as a difference of two squares.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How does this prove the identity?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:34






          • 2




            $begingroup$
            Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
            $endgroup$
            – Micah
            Aug 29 '15 at 23:34










          • $begingroup$
            could you factor any side actually both of them look like difference of squares?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:39










          • $begingroup$
            You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
            $endgroup$
            – Micah
            Aug 29 '15 at 23:40










          • $begingroup$
            Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:44














          13












          13








          13





          $begingroup$

          Rewrite this as



          $$
          sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
          $$
          and then factor the right-hand side as a difference of two squares.






          share|cite|improve this answer









          $endgroup$



          Rewrite this as



          $$
          sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
          $$
          and then factor the right-hand side as a difference of two squares.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 29 '15 at 23:32









          MicahMicah

          30.2k1364106




          30.2k1364106












          • $begingroup$
            How does this prove the identity?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:34






          • 2




            $begingroup$
            Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
            $endgroup$
            – Micah
            Aug 29 '15 at 23:34










          • $begingroup$
            could you factor any side actually both of them look like difference of squares?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:39










          • $begingroup$
            You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
            $endgroup$
            – Micah
            Aug 29 '15 at 23:40










          • $begingroup$
            Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:44


















          • $begingroup$
            How does this prove the identity?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:34






          • 2




            $begingroup$
            Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
            $endgroup$
            – Micah
            Aug 29 '15 at 23:34










          • $begingroup$
            could you factor any side actually both of them look like difference of squares?
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:39










          • $begingroup$
            You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
            $endgroup$
            – Micah
            Aug 29 '15 at 23:40










          • $begingroup$
            Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
            $endgroup$
            – Sunny Mann
            Aug 29 '15 at 23:44
















          $begingroup$
          How does this prove the identity?
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:34




          $begingroup$
          How does this prove the identity?
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:34




          2




          2




          $begingroup$
          Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
          $endgroup$
          – Micah
          Aug 29 '15 at 23:34




          $begingroup$
          Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
          $endgroup$
          – Micah
          Aug 29 '15 at 23:34












          $begingroup$
          could you factor any side actually both of them look like difference of squares?
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:39




          $begingroup$
          could you factor any side actually both of them look like difference of squares?
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:39












          $begingroup$
          You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
          $endgroup$
          – Micah
          Aug 29 '15 at 23:40




          $begingroup$
          You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
          $endgroup$
          – Micah
          Aug 29 '15 at 23:40












          $begingroup$
          Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:44




          $begingroup$
          Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
          $endgroup$
          – Sunny Mann
          Aug 29 '15 at 23:44











          3












          $begingroup$

          As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
          That is,



          $$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$






          share|cite|improve this answer











          $endgroup$


















            3












            $begingroup$

            As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
            That is,



            $$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$






            share|cite|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
              That is,



              $$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$






              share|cite|improve this answer











              $endgroup$



              As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
              That is,



              $$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 14 at 23:10









              J. W. Tanner

              3,9071320




              3,9071320










              answered Nov 10 '17 at 15:59









              Guy FsoneGuy Fsone

              17.3k43074




              17.3k43074























                  2












                  $begingroup$

                  Here would be the other points to remember:



                  $sin^2theta+cos^2theta=1$



                  $x^4-y^4=(x^2-y^2)(x^2+y^2)$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Here would be the other points to remember:



                    $sin^2theta+cos^2theta=1$



                    $x^4-y^4=(x^2-y^2)(x^2+y^2)$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Here would be the other points to remember:



                      $sin^2theta+cos^2theta=1$



                      $x^4-y^4=(x^2-y^2)(x^2+y^2)$






                      share|cite|improve this answer









                      $endgroup$



                      Here would be the other points to remember:



                      $sin^2theta+cos^2theta=1$



                      $x^4-y^4=(x^2-y^2)(x^2+y^2)$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 29 '15 at 23:40









                      JB KingJB King

                      3,50911014




                      3,50911014























                          1












                          $begingroup$

                          Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:



                          Consider the function



                          $$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$



                          Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.



                          Hope you find this useful/interesting.






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:



                            Consider the function



                            $$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$



                            Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.



                            Hope you find this useful/interesting.






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:



                              Consider the function



                              $$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$



                              Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.



                              Hope you find this useful/interesting.






                              share|cite|improve this answer









                              $endgroup$



                              Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:



                              Consider the function



                              $$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$



                              Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.



                              Hope you find this useful/interesting.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Oct 11 '17 at 19:50









                              DmorenoDmoreno

                              6,53231240




                              6,53231240























                                  1












                                  $begingroup$

                                  A "forwards" proof:



                                  Render



                                  $sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$



                                  Regroup the terms on the right as



                                  $(sin^2theta+1-2sin^2theta)+sin^4theta$



                                  and put $1-sin^2theta=cos^2theta$.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    A "forwards" proof:



                                    Render



                                    $sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$



                                    Regroup the terms on the right as



                                    $(sin^2theta+1-2sin^2theta)+sin^4theta$



                                    and put $1-sin^2theta=cos^2theta$.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      A "forwards" proof:



                                      Render



                                      $sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$



                                      Regroup the terms on the right as



                                      $(sin^2theta+1-2sin^2theta)+sin^4theta$



                                      and put $1-sin^2theta=cos^2theta$.






                                      share|cite|improve this answer











                                      $endgroup$



                                      A "forwards" proof:



                                      Render



                                      $sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$



                                      Regroup the terms on the right as



                                      $(sin^2theta+1-2sin^2theta)+sin^4theta$



                                      and put $1-sin^2theta=cos^2theta$.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Nov 10 '17 at 16:40

























                                      answered Nov 10 '17 at 16:33









                                      Oscar LanziOscar Lanzi

                                      13.3k12136




                                      13.3k12136























                                          0












                                          $begingroup$

                                          $$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            could you elaborate please?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:14










                                          • $begingroup$
                                            Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:20










                                          • $begingroup$
                                            $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:26












                                          • $begingroup$
                                            What do you get when you expand $(1-sin^2theta)^2$?
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:28












                                          • $begingroup$
                                            $1- sin^4(theta) or cos^4(theta)$
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:36
















                                          0












                                          $begingroup$

                                          $$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$






                                          share|cite|improve this answer









                                          $endgroup$













                                          • $begingroup$
                                            could you elaborate please?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:14










                                          • $begingroup$
                                            Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:20










                                          • $begingroup$
                                            $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:26












                                          • $begingroup$
                                            What do you get when you expand $(1-sin^2theta)^2$?
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:28












                                          • $begingroup$
                                            $1- sin^4(theta) or cos^4(theta)$
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:36














                                          0












                                          0








                                          0





                                          $begingroup$

                                          $$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$






                                          share|cite|improve this answer









                                          $endgroup$



                                          $$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered Aug 30 '15 at 0:13









                                          John JoyJohn Joy

                                          6,29911727




                                          6,29911727












                                          • $begingroup$
                                            could you elaborate please?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:14










                                          • $begingroup$
                                            Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:20










                                          • $begingroup$
                                            $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:26












                                          • $begingroup$
                                            What do you get when you expand $(1-sin^2theta)^2$?
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:28












                                          • $begingroup$
                                            $1- sin^4(theta) or cos^4(theta)$
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:36


















                                          • $begingroup$
                                            could you elaborate please?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:14










                                          • $begingroup$
                                            Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:20










                                          • $begingroup$
                                            $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:26












                                          • $begingroup$
                                            What do you get when you expand $(1-sin^2theta)^2$?
                                            $endgroup$
                                            – John Joy
                                            Aug 30 '15 at 0:28












                                          • $begingroup$
                                            $1- sin^4(theta) or cos^4(theta)$
                                            $endgroup$
                                            – Sunny Mann
                                            Aug 30 '15 at 0:36
















                                          $begingroup$
                                          could you elaborate please?
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:14




                                          $begingroup$
                                          could you elaborate please?
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:14












                                          $begingroup$
                                          Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                          $endgroup$
                                          – John Joy
                                          Aug 30 '15 at 0:20




                                          $begingroup$
                                          Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
                                          $endgroup$
                                          – John Joy
                                          Aug 30 '15 at 0:20












                                          $begingroup$
                                          $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:26






                                          $begingroup$
                                          $sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:26














                                          $begingroup$
                                          What do you get when you expand $(1-sin^2theta)^2$?
                                          $endgroup$
                                          – John Joy
                                          Aug 30 '15 at 0:28






                                          $begingroup$
                                          What do you get when you expand $(1-sin^2theta)^2$?
                                          $endgroup$
                                          – John Joy
                                          Aug 30 '15 at 0:28














                                          $begingroup$
                                          $1- sin^4(theta) or cos^4(theta)$
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:36




                                          $begingroup$
                                          $1- sin^4(theta) or cos^4(theta)$
                                          $endgroup$
                                          – Sunny Mann
                                          Aug 30 '15 at 0:36


















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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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