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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?

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  • $\begingroup$ Take the last step, and argue in reverse. e.g. $$\sin^2\theta-\cos^2\theta = \sin^2\theta-\cos^2\theta\implies\sin^2\theta-\cos^2\theta=(\sin^2\theta+\cos^2 \theta)(\sin^2\theta-\cos^2\theta)=\dots$$ $\endgroup$
    – John Joy
    Aug 30, 2015 at 0:23

10 Answers 10

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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.

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  • $\begingroup$ How does this prove the identity? $\endgroup$
    – Sunny Mann
    Aug 29, 2015 at 23:34
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    $\begingroup$ 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, 2015 at 23:34
  • $\begingroup$ could you factor any side actually both of them look like difference of squares? $\endgroup$
    – Sunny Mann
    Aug 29, 2015 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, 2015 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, 2015 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,

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

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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^2\theta + \cos^2\theta = 1 $ rearranged into $\sin^2\theta = 1 - \cos^2\theta $ and $\cos^2\theta = 1 - \sin^2\theta $

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

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

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

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

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

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

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

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

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  • $\begingroup$ (1−2sin2θ+sin4θ) where did you get this expression from? $\endgroup$
    – Sunny Mann
    Aug 30, 2015 at 0:30
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    $\begingroup$ 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, 2015 at 0:30
  • $\begingroup$ you jumped from +$\cos^4(\theta)$ to ($1-2\sin^2(\theta)+\sin^4(\theta))$ How? $\endgroup$
    – Sunny Mann
    Aug 30, 2015 at 0:34
  • $\begingroup$ We can write $\cos^4\theta $ as $\cos^2\theta\cos^2\theta $. Imagine it as $(\cos\theta)^4 = (\cos\theta)^2(\cos\theta)^2 $. $\endgroup$
    – Leo
    Aug 30, 2015 at 0:37
  • $\begingroup$ it's supposed to be $1- \sin^2(\theta), no? $\endgroup$
    – Sunny Mann
    Aug 30, 2015 at 0:40
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Here would be the other points to remember:

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

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

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

Hope you find this useful/interesting.

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A "forwards" proof:

Render

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

Regroup the terms on the right as

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

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

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

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  • $\begingroup$ could you elaborate please? $\endgroup$
    – Sunny Mann
    Aug 30, 2015 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, 2015 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, 2015 at 0:26
  • $\begingroup$ What do you get when you expand $(1-\sin^2\theta)^2$? $\endgroup$
    – John Joy
    Aug 30, 2015 at 0:28
  • $\begingroup$ $1- \sin^4(\theta) or \cos^4(\theta)$ $\endgroup$
    – Sunny Mann
    Aug 30, 2015 at 0:36
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You can prove this fact by showing that the LHS and RHS agree at the point 0 and that their derivatives are equal everywhere.

Let's use the following as abbreviations for sine and cosine.

$$ y(t) = \sin(t) $$ $$ x(t) = \cos(t) $$

We want to show (Orig).

$$ y^2 + x^4 = x^2 + y^4 \tag{Orig} $$

Negate the goal.

$$ y^2 + x^4 \neq x^2 + y^4 \tag{NG} $$

If we can show that $y(0)^2 + x(0)^4 \neq x(0)^2 + y(0)^4 $ (101) and that the derivatives of both sides are unequal (105), that's equivalent to (NG).

$$ y(0)^2 + x(0)^4 \neq x(0)^2 + y(0)^4 \tag{101} $$ $$ 0 + 1 \neq 1 + 0 \tag{102} $$ $$ 1 \neq 1 \tag{103} $$ $$ \bot \tag{104} $$

And the derivative

$$ (y^2 + x^4)' \neq (x^2 + y^4)' \tag{105} $$ $$ 2yy' + 4x^3x' \neq 2xx' + 4y^3y' \tag{106} $$

Normalize $y' \mapsto x$ and $x' \mapsto -y$ .

$$ 2yx - 4x^3y \neq -2yx + 4y^3x \tag{107} $$

Combine like terms.

$$ 4yx \neq 4y^3x + 4x^3y \tag{108} $$ $$ 4yx \neq 4yx\cdot(y^2 + x^2) \tag{109} $$ $$ 4yx \neq 4yx\cdot 1 \tag{110} $$ $$ 4yx \neq 4yx \tag{111} $$ $$ \bot \tag{112} $$

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

$$ \Longleftrightarrow $$

$$ \sin^2\theta - \cos^2\theta = \sin^4\theta - \cos^4\theta$$

$$ \Longleftrightarrow $$

$$ \sin^2\theta - \cos^2\theta = (\underbrace{\sin^2\theta + \cos^2\theta}_{1})(\sin^2\theta - \cos^2\theta) $$

$$ \Longleftrightarrow $$

$$ \sin^2\theta - \cos^2\theta = \sin^2\theta - \cos^2\theta $$

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Bringing everything in terms of $\cos \theta=c: $ $$LHS=1-c^2+c^4\,;$$ $$RHS= c^2+(1-c^2)^2= c^2 +1-2 c^2 +c^4=1-c^2+c^4\,=LHS\,$$

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