## anonymous one year ago **medal and fan Simplify: (sin Θ − cos Θ)2 + (sin Θ + cos Θ)2

1. anonymous

answer choices A) 1 B) 2 C)sin^2Θ D)cos^2

2. Nnesha

let sin theta = x cos theta = y so you have $\huge\rm (x-y)^2 +(x +y)^2$

3. Nnesha

(x-y)^2 is same as (x-y)(x-y) apply foil method

4. anonymous

so it cancels out ?

5. Nnesha

no there is a plus sign

6. anonymous

im not really familiar with the foil method

7. anonymous

do I just multiply (x-y)(x-y) ??

8. Nnesha

ohh okay |dw:1438028749837:dw| multiply first and 2nd term of first parentheses by 2nd parentheses

9. Nnesha

yees right

10. anonymous

I know part of it is X^2 so x &y multiplied its just x*y ?

11. Nnesha

yes right but don't forget the negative sign $\huge\rm (x-y)(x-y)$ |dw:1438029051942:dw|

12. Nnesha

x times -y = -xy

13. anonymous

okay but now whats the next step to the problem ?

14. Nnesha

now multiply 2nd parentheses by 2nd term of 1st parentheses|dw:1438029177493:dw|

15. anonymous

-yx - y^2

16. Nnesha

yes right

17. Nnesha

well no little mistake -y times -y = ?

18. anonymous

positive y ?

19. Nnesha

positive y^2

20. anonymous

oh sorry

21. Nnesha

$\huge\rm\color{reD}{ (x-y)^2} +(x +y)^2$ $\huge\rm \color{Red}{x^2-xy-xy+y^2} +(x+y)^2$ combine like terms and then foil (x+y)^2

22. anonymous

-xy and -xy are like terms do they cancel out ?

23. Nnesha

no you need to be careful abt signs -xy -xy you would cancel out if there are opposite signs -xy -xy = ? add example -1-1=-2

24. anonymous

but I would add them together correct ?

25. Nnesha

yep

26. anonymous

So i would get -2xy ?

27. Nnesha

yep right now multiply (x+y)^2

28. Nnesha

(x+y)^2 is same as (X+y)(x+y)

29. anonymous

okay so now i ditribute x-y^2 to x^2 + (x+y)^2+y^2?

30. freckles

Have you expanded (x+y)^2 ? $\huge\rm \color{Red}{x^2-xy-xy+y^2} +(x+y)^2$

31. anonymous

no ?

32. freckles

(x+y)^2 is the same as (x+y)(x+y) multiply this out

33. freckles

x(x+y)+y(x+y) use distributive property twice

34. anonymous

so Im going to distribute the x to whats in parentheses and then do the same with the y , correct ?

35. freckles

yes that is the distributive property it says a(b+c)=ab+ac

36. anonymous

okay after distributing I got x^2+xy + yx+y^2

37. freckles

yes and xy+xy=?

38. anonymous

xy^2 ?

39. freckles

if have an apple and find another apple do you have a square amount of apples or just 2 apples?

40. anonymous

so 2xy ?

41. freckles

yes

42. freckles

(x+y)^2=x^2+2xy+y^2 and (x-y)^2=x^2-2xy+y^2

43. freckles

also recall sin^2(u)+cos^2(u)=1

44. freckles

anyways let us know if you need further help on this one

45. anonymous

but I still dont get it lol sorry , what would the next step be ?

46. freckles

recall you started with $(\sin(\theta)+\cos(\theta))^2+(\sin(\theta)-\cos(\theta))^2 \\ \text{ then you expanded} \\ \sin^2(\theta)+2\sin(\theta)\cos(\theta)+\cos^2(\theta)+\sin^2(\theta)-2\sin(\theta)\cos(\theta)+\cos^2(\theta)$ are you not sure what sin^2(theta)+cos^2(theta) equals ? or is it you don't know what 2sin(theta)cos(theta)-2sin(theta)cos(theta) is?

47. anonymous

im not sure what it equals

48. freckles

which one?

49. freckles

sin^2(theta)+cos^2(theta)? or the same thing - the same thing?

50. freckles

the first one is a Pythagorean identity sin^2(u)+cos^2(u)=1 the second one is using the additive inverse property that is u+(-u)=0 or this could be written as u-u=0 which should make since because if you have exactly one apple and so one takes that apple away you have no apples aka 0 apples

51. anonymous

okay thank you