Vtyjkyuk

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?

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 $

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.

As $cos^2 theta +sin^2 theta= 1$ we have $$sin^4 theta -cos^4 theta =(cos^2 theta -sin^2 theta)colorred(cos^2 theta +sin^2 theta)= cos^2 theta -sin^2 theta$$
That is,

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

Here would be the other points to remember:

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

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

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

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

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.

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

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