# How do I verify sin^2-sin^4=cos^2-cos^4?

Mar 9, 2018

We're trying to prove ${\sin}^{2} x - {\sin}^{4} x = {\cos}^{2} x - {\cos}^{4} x$.

Start with the left side and manipulate it until it looks like the right side. First, factor out the ${\sin}^{2} x$ term, then use the Pythagorean identity:

${\sin}^{2} x + {\cos}^{2} x = 1$

Here are some alternative (but equivalent) identities:

${\sin}^{2} x = 1 - {\cos}^{2} x$

${\cos}^{2} x = 1 - {\sin}^{2} x$

Now, here's the actual proof:

$L H S = \textcolor{p u r p \le}{{\sin}^{2} x} - \textcolor{p u r p \le}{{\sin}^{4} x}$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{{\sin}^{2} x} \cdot \textcolor{red}{1} - \textcolor{b l u e}{{\sin}^{2} x} \cdot \textcolor{red}{{\sin}^{2} x}$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{{\sin}^{2} x} \left(\textcolor{red}{1 - {\sin}^{2} x}\right)$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{\left(1 - {\cos}^{2} x\right)} \left(\textcolor{red}{1 - {\sin}^{2} x}\right)$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{\left(1 - {\cos}^{2} x\right)} \left(\textcolor{red}{{\cos}^{2} x}\right)$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{\left(1 - {\cos}^{2} x\right)} \left(\textcolor{red}{{\cos}^{2} x}\right)$

$\textcolor{w h i t e}{L H S} = \textcolor{b l u e}{1} \cdot \textcolor{red}{{\cos}^{2} x} - \textcolor{b l u e}{{\cos}^{2} x} \cdot \textcolor{red}{{\cos}^{2} x}$

$\textcolor{w h i t e}{L H S} = \textcolor{p u r p \le}{{\cos}^{2} x} - \textcolor{p u r p \le}{{\cos}^{4} x}$

$\textcolor{w h i t e}{L H S} = R H S$