Verify the Identity (1+sin alpha)(1-sin alpha)=cos^2 alpha

Question

zepdrix · Answer

Hey Sarah :) Welcome to OpenStudy!

zepdrix · Answer

What we have here are often referred to as `conjugates`:
$\Large\rm (a-b)(a+b)$
When you multiply them out, they give us `the difference of squares`.
$\Large\rm a^2-b^2$

Alternatively, you can FOIL it out the long way if you don't want to remember this.

anonymous · Answer

Hey! Glad to be here!

zepdrix · Answer

$$\Large\rm (1-\sin \alpha)(1+\sin \alpha)=1^2-\sin^2 \alpha$$The square on the 1 isn't necessary so we'll ignore it.$$\Large\rm =1-\sin^2 \alpha$$

Do you remember your `Pythagorean Identity for Sine and Cosine` ?

anonymous · Answer

yeah

anonymous · Answer

sin^u+cos^2=1 ; 1+tan^2u=sec^2u ; 1+cot^2u=csc^2u

zepdrix · Answer

$$\Large\rm \sin^2u+\cos^2u=1$$From this identity, we'll subtract sin^2u from each side,$$\Large\rm \cos^2u=1-\sin^2u$$

zepdrix · Answer

So do you see how we can make our jump from 1-sin^2(alpha) to cos^2(alpha) by using that identity? :O

anonymous · Answer

I tend to get confused when I get to this part.