# How do I proof this equation is an identity? 1-sec(x)cos³(x)=sin²(x)

Mar 20, 2018

Recall that $\sec x = \frac{1}{\cos} x$. Therefore

$1 - \left(\frac{1}{\cos} x\right) {\cos}^{3} x = {\sin}^{2} x$

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

And since ${\sin}^{2} x + {\cos}^{2} x = 1$, then ${\sin}^{2} x = 1 - {\cos}^{2} x$.

${\sin}^{2} x = {\sin}^{2} x$

$L H S = R H S$

Hopefully this helps!

Mar 20, 2018

Pythagorean Identity:
$1 - {\cos}^{2} x = {\sin}^{2} x$
And reciprocal identity:
$\frac{1}{\sec} x = \cos x$

Applied to:
1-sec(x)cos³(x)=sin²(x)
1-1/cancel(cosx)*coscancel(³)(x)=sin²(x)
$1 - {\cos}^{2} x = {\sin}^{2} x$
${\sin}^{2} x = {\sin}^{2} x$

Mar 20, 2018

Use these three identities (the third one is pretty much the same as the second, but switched around a little):

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

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

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

To complete the proof, I'll start with the left-hand side of the equation and manipulate it until it equals the right-hand side of the equation:

$L H S = 1 - \sec \left(x\right) {\cos}^{3} \left(x\right)$

$\textcolor{w h i t e}{L H S} = 1 - \sec \left(x\right) \cdot {\cos}^{3} \left(x\right)$

$\textcolor{w h i t e}{L H S} = 1 - \frac{1}{\cos} \left(x\right) \cdot {\cos}^{3} \left(x\right)$

$\textcolor{w h i t e}{L H S} = 1 - {\cos}^{3} \frac{x}{\cos} \left(x\right)$

$\textcolor{w h i t e}{L H S} = 1 - {\textcolor{red}{\cancel{\textcolor{b l a c k}{{\cos}^{3} \left(x\right)}}}}^{{\cos}^{2} \left(x\right)} / \textcolor{red}{\cancel{\textcolor{b l a c k}{\cos}}} \left(x\right)$

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

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

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

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

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

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

That is the proof. Hope this helped!