# How do you prove cos4x = 8cos^4x - 8cos^2x + 1?

Mar 1, 2018

$\text{Please see proof below.}$

#### Explanation:

$\text{This can be done by twice using the double angle formula for cos.}$
$\text{We have:}$

 \qquad \qquad cos2x \ = \ cos^2x - sin^2 x; \qquad \quad \ color{blue}{"double angle formula for cos"}

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus = \setminus {\cos}^{2} x - \left(1 - {\cos}^{2} x\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus = \setminus {\cos}^{2} x - 1 + {\cos}^{2} x$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus = \setminus 2 {\cos}^{2} x - 1.$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \cos 2 x \setminus = \setminus 2 {\cos}^{2} x - 1. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \left(I\right)$

$\text{So now:}$

 \qquad \qquad cos4x \ = \ cos( 2 cdot (2 x ) ); \qquad \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{"now let:" \qquad A = 2x"}

 \qquad \qquad \qquad= \ cos( 2 A );

 \qquad \qquad \qquad= \ 2 cos^2A - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ color{blue}{"by (I)"}

 \qquad \qquad \qquad= \ 2 ( cosA )^2 - 1;

 \qquad \qquad \qquad= \ 2 ( cos 2 x )^2 - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad color{blue}{"since:" \qquad A = 2x"}

 \qquad \qquad \qquad= \ 2 ( 2 cos^2x - 1 )^2 - 1; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ color{blue}{"by (I)"}

 \qquad \qquad \qquad= \ 2 ( [ 2 cos^2x ]^2 - 2 [ 2 cos^2x ] + 1 ) - 1;
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \textcolor{b l u e}{\text{as:} q \quad {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}}$

 \qquad \qquad \qquad= \ 2 ( 4 cos^4x - 4 cos^2x + 1 ) - 1;

 \qquad \qquad \qquad= \ 8 cos^4x - 8 cos^2x + 2 - 1;

$\setminus q \quad \setminus q \quad \setminus q \quad = \setminus 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1.$

$\text{This is what we wanted to show !!}$

$\text{Summarizing, we have shown:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \cos 4 x \setminus = \setminus 8 {\cos}^{4} x - 8 {\cos}^{2} x + 1.$