Question

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

Answer 1

See below.

Explanation:

Get rid of `sin(-x).` Recall that `sin(-x)=-sinx.`

`1-sinx=(1-sin^2x)/(1-(-sinx))`

`1-sinx=(1-sin^2x)/(1+sinx)`

Now, from the difference of squares, we know that `1-sin^2x=(1+sinx)(1-sinx):`

`1-sinx=(cancel(1+sinx)(1-sinx))/(cancel(1+sinx))`

`1-sinx=1-sinx`

So, this is indeed an identity.

Answer 2

Look below

Explanation:

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