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

Question

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 00:23

score 17 · Answer 1 · answered Aug 29 '15 at 23:32

17

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.

answered Aug 29 '15 at 23:32

Micah

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How does this prove the identity? – Sunny Mann Aug 29 '15 at 23:34
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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
Yes. One of those factors is identical to the left-hand side; the other one should look familiar for other reasons... – Micah Aug 29 '15 at 23:47
ok so the sum of squares simplifies to 1 and the other becomes the same as the left side, so..that brings me back to the original question, is it proven? I'm confused, I'm very sorry – Sunny Mann Aug 29 '15 at 23:50

score 7 · Answer 2 · edited Jan 14 '19 at 23:10

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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,

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

edited Jan 14 '19 at 23:10

J. W. Tanner

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answered Nov 10 '17 at 15:59

Guy Fsone

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score 5 · Accepted Answer · answered Aug 30 '15 at 00:22

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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 $

answered Aug 30 '15 at 00:22

Leo

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(1−2sin2θ+sin4θ) where did you get this expression from? – Sunny Mann Aug 30 '15 at 00:30
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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 00:30
you jumped from +$\cos^4(\theta)$ to ($1-2\sin^2(\theta)+\sin^4(\theta))$ How? – Sunny Mann Aug 30 '15 at 00:34
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 00:37
it's supposed to be $1- \sin^2(\theta), no? – Sunny Mann Aug 30 '15 at 00:40
Then because $ (\cos\theta)^2 = \cos^2\theta = 1 - \sin^2\theta $ which is rearranged from the trigonometric identity: $\sin^2\theta + \cos^2\theta = 1$. You can substitude the $\cos^2\theta $ in our $\cos^2\theta\cos^2\theta $ to give $ (1 - \sin^2\theta)(1-\sin^2\theta) $ and then expand from there :) Look to see the second paragraph in the answer. – Leo Aug 30 '15 at 00:43
Yes it is! The $\cos^2\theta $ can be substituted for $ 1 - \sin^2\theta $ but we have two lots of them from the $ \cos^4\theta = (\cos^2\theta)(\cos^2\theta) $! – Leo Aug 30 '15 at 00:48

score 2 · Answer 4 · answered Aug 29 '15 at 23:40

2

Here would be the other points to remember:

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

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

answered Aug 29 '15 at 23:40

JB King

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score 1 · Answer 5 · answered Oct 11 '17 at 19:50

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.

Oscar Lanzi · Answer 6 · 2017-11-10T16:40:27.703

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

edited Nov 10 '17 at 16:40

answered Nov 10 '17 at 16:33

Oscar Lanzi

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score 0 · Answer 7 · answered Aug 30 '15 at 00:13

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

answered Aug 30 '15 at 00:13

John Joy

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could you elaborate please? – Sunny Mann Aug 30 '15 at 00:14
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 00:20
$\sin^2(\theta) becomes \sin^4(\theta)$ after putting the expression to the power of two, but happens to the 1-? – Sunny Mann Aug 30 '15 at 00:26
What do you get when you expand $(1-\sin^2\theta)^2$? – John Joy Aug 30 '15 at 00:28
$1- \sin^4(\theta) or \cos^4(\theta)$ – Sunny Mann Aug 30 '15 at 00:36
$\cos^4(/theta)$ is not required on the right hand side of the identity. – Sunny Mann Aug 30 '15 at 00:51
How did you come up with $1-\sin^4\theta$? Show your reasoning. – John Joy Aug 30 '15 at 01:21
I thought it was the power of a power exponent rule with 2^2, is it a difference of squares? – Sunny Mann Aug 30 '15 at 01:27
You're implying that $$(1-\sin^2\theta)^2 = 1-\sin^4\theta$$ How did you deduce that? – John Joy Aug 30 '15 at 01:29
jay I multiplied the exponents, I think after a second thought I think I'm wrong, but its not $1-\sin^4(\theta), then it must be ($1-\sin^4(\theta))($1-\sin^4(\theta)) – Sunny Mann Aug 30 '15 at 01:33
Try reviewing the chapter on factoring and expanding quadratics. $$(1+t)^2=(1+t)(1+t)=1+2t+t^2 \implies (1-\sin^2\theta)^2=1-2\sin^2\theta+\sin^4\theta$$ See Leo's answer. – John Joy Aug 30 '15 at 01:38
Ah perfect square trinominal, I got confused seeing double exponent one inside the bracket, one outside, sorry John, and it's the same as Leo described above? I really really honestly appreciate the guidance. – Sunny Mann Aug 30 '15 at 01:43
exactly, think of $-\sin^2\theta$ as a number, then what we have is $(1+\text{a number})^2$ – John Joy Aug 30 '15 at 01:47
That was exactly the issue, and that proves the identity, much appreciated. – Sunny Mann Aug 30 '15 at 01:50

Greg Nisbet · Answer 8 · 2019-07-30T04:03:25.730

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.

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

score 0 · Answer 9 · answered Jul 30 '19 at 04:30

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

$$ \Longleftrightarrow $$

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

$$ \Longleftrightarrow $$

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

$$ \Longleftrightarrow $$

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

score 0 · Answer 10 · answered Jul 30 '19 at 05:07

0

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

answered Jul 30 '19 at 05:07

Narasimham

40,495

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

10 Answers10