GucciDoTheDishes
  • GucciDoTheDishes
**medal and fan Simplify: (sin Θ − cos Θ)2 + (sin Θ + cos Θ)2
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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GucciDoTheDishes
  • GucciDoTheDishes
answer choices A) 1 B) 2 C)sin^2Θ D)cos^2
Nnesha
  • Nnesha
let sin theta = x cos theta = y so you have \[\huge\rm (x-y)^2 +(x +y)^2\]
Nnesha
  • Nnesha
(x-y)^2 is same as (x-y)(x-y) apply foil method

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GucciDoTheDishes
  • GucciDoTheDishes
so it cancels out ?
Nnesha
  • Nnesha
no there is a plus sign
GucciDoTheDishes
  • GucciDoTheDishes
im not really familiar with the foil method
GucciDoTheDishes
  • GucciDoTheDishes
do I just multiply (x-y)(x-y) ??
Nnesha
  • Nnesha
ohh okay |dw:1438028749837:dw| multiply first and 2nd term of first parentheses by 2nd parentheses
Nnesha
  • Nnesha
yees right
GucciDoTheDishes
  • GucciDoTheDishes
I know part of it is X^2 so x &y multiplied its just x*y ?
Nnesha
  • Nnesha
yes right but don't forget the negative sign \[\huge\rm (x-y)(x-y)\] |dw:1438029051942:dw|
Nnesha
  • Nnesha
x times -y = -xy
GucciDoTheDishes
  • GucciDoTheDishes
okay but now whats the next step to the problem ?
Nnesha
  • Nnesha
now multiply 2nd parentheses by 2nd term of 1st parentheses|dw:1438029177493:dw|
GucciDoTheDishes
  • GucciDoTheDishes
-yx - y^2
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
well no little mistake -y times -y = ?
GucciDoTheDishes
  • GucciDoTheDishes
positive y ?
Nnesha
  • Nnesha
positive y^2
GucciDoTheDishes
  • GucciDoTheDishes
oh sorry
Nnesha
  • 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
GucciDoTheDishes
  • GucciDoTheDishes
-xy and -xy are like terms do they cancel out ?
Nnesha
  • 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
GucciDoTheDishes
  • GucciDoTheDishes
but I would add them together correct ?
Nnesha
  • Nnesha
yep
GucciDoTheDishes
  • GucciDoTheDishes
So i would get -2xy ?
Nnesha
  • Nnesha
yep right now multiply (x+y)^2
Nnesha
  • Nnesha
(x+y)^2 is same as (X+y)(x+y)
GucciDoTheDishes
  • GucciDoTheDishes
okay so now i ditribute x-y^2 to x^2 + (x+y)^2+y^2?
freckles
  • freckles
Have you expanded (x+y)^2 ? \[\huge\rm \color{Red}{x^2-xy-xy+y^2} +(x+y)^2\]
GucciDoTheDishes
  • GucciDoTheDishes
no ?
freckles
  • freckles
(x+y)^2 is the same as (x+y)(x+y) multiply this out
freckles
  • freckles
x(x+y)+y(x+y) use distributive property twice
GucciDoTheDishes
  • GucciDoTheDishes
so Im going to distribute the x to whats in parentheses and then do the same with the y , correct ?
freckles
  • freckles
yes that is the distributive property it says a(b+c)=ab+ac
GucciDoTheDishes
  • GucciDoTheDishes
okay after distributing I got x^2+xy + yx+y^2
freckles
  • freckles
yes and xy+xy=?
GucciDoTheDishes
  • GucciDoTheDishes
xy^2 ?
freckles
  • freckles
if have an apple and find another apple do you have a square amount of apples or just 2 apples?
GucciDoTheDishes
  • GucciDoTheDishes
so 2xy ?
freckles
  • freckles
yes
freckles
  • freckles
(x+y)^2=x^2+2xy+y^2 and (x-y)^2=x^2-2xy+y^2
freckles
  • freckles
also recall sin^2(u)+cos^2(u)=1
freckles
  • freckles
anyways let us know if you need further help on this one
GucciDoTheDishes
  • GucciDoTheDishes
but I still dont get it lol sorry , what would the next step be ?
freckles
  • 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?
GucciDoTheDishes
  • GucciDoTheDishes
im not sure what it equals
freckles
  • freckles
which one?
freckles
  • freckles
sin^2(theta)+cos^2(theta)? or the same thing - the same thing?
freckles
  • 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
GucciDoTheDishes
  • GucciDoTheDishes
okay thank you

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