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

2 Answers
Mar 31, 2018

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.

Mar 31, 2018

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.