# Prove this is an identity? 1-sinx=1-sin²(-x)/1-sin(-x)

Mar 31, 2018

See below.

#### Explanation:

Get rid of $\sin \left(- x\right) .$ Recall that $\sin \left(- x\right) = - \sin x .$

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

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

Now, from the difference of squares, we know that $1 - {\sin}^{2} x = \left(1 + \sin x\right) \left(1 - \sin x\right) :$

$1 - \sin x = \frac{\cancel{1 + \sin x} \left(1 - \sin x\right)}{\cancel{1 + \sin x}}$

$1 - \sin x = 1 - \sin x$

So, this is indeed an identity.

Mar 31, 2018

Look below

#### Explanation:

First, let's look at the numerator on the right side of the equation. $\sin \left(- x\right) = - \sin \left(x\right)$, so ${\sin}^{2} \left(- x\right) = {\left(- \sin \left(x\right)\right)}^{2} = {\sin}^{2} \left(x\right)$, so the numerator is $1 - {\sin}^{2} \left(x\right)$ Looking at the denominator, you'll see that it is equal to $1 + \sin \left(x\right)$. Dividing the numerator and denominator, you'll see that the identity is true.