# Pruebalo (sinx-cosx)^2=1-1/csc^2x?

Oct 24, 2016

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

#### Explanation:

The question appears to be in Spanish and I am answering in English. I hope answer will satisfy you.

${\left(\sin x - \cos x\right)}^{2}$ - using ${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 a b$, we get

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

= $1 - \sin 2 x$ - (as ${\sin}^{2} x + {\cos}^{2} x = 1$ and $\sin 2 x = 2 \sin x \cos x$)

= $1 - \frac{1}{\csc} \left(2 x\right)$ - as $\sin A = \frac{1}{\csc} A$

Note ${\left(\sin x - \cos x\right)}^{2} = 1 - \frac{1}{\csc} \left(2 x\right)$ and not $1 - \frac{1}{\csc} ^ 2 x$ as

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