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

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?

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

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

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

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

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

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

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

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

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

• (1−2sin2θ+sin4θ) where did you get this expression from? – Sunny Mann Aug 30 '15 at 0:30
• 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 – Leo Aug 30 '15 at 0:30
• you jumped from +$\cos^4(\theta)$ to ($1-2\sin^2(\theta)+\sin^4(\theta))$ How? – Sunny Mann Aug 30 '15 at 0:34
• We can write $\cos^4\theta$ as $\cos^2\theta\cos^2\theta$. Imagine it as $(\cos\theta)^4 = (\cos\theta)^2(\cos\theta)^2$. – Leo Aug 30 '15 at 0:37
• it's supposed to be $1- \sin^2(\theta), no? – Sunny Mann Aug 30 '15 at 0:40 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. • How does this prove the identity? – Sunny Mann Aug 29 '15 at 23:34 • Try it and see! What do you get when you factor the right-hand side as a difference of two squares? – Micah Aug 29 '15 at 23:34 • could you factor any side actually both of them look like difference of squares? – Sunny Mann Aug 29 '15 at 23:39 • You could factor either side, but factoring the right side will be helpful and factoring the left side will not. – Micah Aug 29 '15 at 23:40 • Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)). – Sunny Mann Aug 29 '15 at 23:44 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, $$\sin^2 \theta +\cos^4 \theta =\cos^2 \theta +\sin^4 \theta$$ Here would be the other points to remember:$sin^2\theta+cos^2\theta=1x^4-y^4=(x^2-y^2)(x^2+y^2)$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. 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$1-\sin^2\theta=\cos^2\theta$. $$\sin^2\theta+\cos^4\theta=\sin^2\theta+\bigg(\cos^2\theta\bigg)^2=\sin^2\theta+\bigg(1-\sin^2\theta\bigg)^2=\dots$$ • could you elaborate please? – Sunny Mann Aug 30 '15 at 0:14 • 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. – John Joy Aug 30 '15 at 0:20 •$\sin^2(\theta) becomes \sin^4(\theta)$after putting the expression to the power of two, but happens to the 1-? – Sunny Mann Aug 30 '15 at 0:26 • What do you get when you expand$(1-\sin^2\theta)^2$? – John Joy Aug 30 '15 at 0:28 •$1- \sin^4(\theta) or \cos^4(\theta)\$ – Sunny Mann Aug 30 '15 at 0:36

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.

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

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

$$\Longleftrightarrow$$

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

$$\Longleftrightarrow$$

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

$$\Longleftrightarrow$$

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

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