# How do you verify cos^4x - sin^4x = 1 - 2sin^2x?

Mar 19, 2018

As below:

#### Explanation:

Note that ${\cos}^{4} x = {\left({\cos}^{2} x\right)}^{2}$

$L . H . S .$
$= {\cos}^{4} x - {\sin}^{4} x$

$= {\left({\cos}^{2} x\right)}^{2} - {\left({\sin}^{2} x\right)}^{2}$

$= \left({\cos}^{2} x + {\sin}^{2} x\right) \left({\cos}^{2} x - {\sin}^{2} x\right)$

$= \left({\cos}^{2} x + {\sin}^{2} x\right) \left({\cos}^{2} x + {\sin}^{2} x - 2 {\sin}^{2} x\right)$

$= \left(1\right) \left(1 - 2 {\sin}^{2} x\right)$

$= 1 - 2 {\sin}^{2} x$
$= R . H . S .$

Mar 19, 2018

To prove that ${\cos}^{4} x - {\sin}^{4} x = 1 - 2 {\sin}^{2} x$, we'll need the Pythagorean identity and a variation on the Pythagorean identity:

$\textcolor{w h i t e}{\implies} {\cos}^{2} x + {\sin}^{2} x = 1$

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

I'll start with the left-hand side and manipulate it until it looks like the right-hand side using these two identities:

$L H S = {\cos}^{4} x - {\sin}^{4} x$

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

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

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

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

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

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

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

That's the proof. Hope this helped!

Mar 19, 2018

See below

#### Explanation:

To verify: ${\cos}^{4} x - {\sin}^{4} x = 1 - 2 {\sin}^{2} x$

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

$\therefore L H S = {\left(1 - \theta\right)}^{2} - {\theta}^{2}$

$= 1 - 2 \theta + {\theta}^{2} - {\theta}^{2}$

$= 1 - 2 \theta$

Undo substitution:

$L H S = 1 - 2 {\sin}^{2} x = R H S$