## anonymous one year ago Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2. Options include: -sin^2 Θ -cos^2Θ 0 2

1. anonymous

So what would you get should you foil out both of those?

2. anonymous

Ok I'm going to let sin Θ = x and cos Θ = y (x-y)(x-y) (x^2-xy+y^2) (x+y)(x+y) (x^2 +xy+y^2)

3. anonymous

so (sin Θ^2 - sin Θ times cos Θ + cos Θ^2) + (sin Θ^2 + sin Θ times cos Θ + cos Θ^2) but what is sin theta times cos theta??

4. anonymous

Well, you should have a $$-2\sin(\theta)\cos(\theta)$$ and a $$+2\sin(\theta)\cos(\theta)$$, so those would cancel anyway, we don't need to worry about them.

5. anonymous

ohh I forgot the outside values

6. anonymous

But once you cancel and combine everything, you would have $$2\sin^{2}(\theta) + 2\cos^{2}(\theta) = 2(\sin^{2}(\theta) + \cos^{2}(\theta))$$

7. anonymous

Okay so it should be (sin^2 theta + cos^2 theta) + (sin^2 theta + cos^2 theta)? And the pythagorean identity says that sin^2 theta + cos^2 theta = 1, so 1 + 1 = 2?

8. anonymous

Yep :)

9. anonymous

Thank you so much!

10. anonymous

Yep, no problem :3

11. xapproachesinfinity

i have seen this question here over and over lol did teacher run out of problems or something hehe

12. xapproachesinfinity

teachers*