**medal and fan Simplify: (sin Θ − cos Θ)2 + (sin Θ + cos Θ)2

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**medal and fan Simplify: (sin Θ − cos Θ)2 + (sin Θ + cos Θ)2

Mathematics
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answer choices A) 1 B) 2 C)sin^2Θ D)cos^2
let sin theta = x cos theta = y so you have \[\huge\rm (x-y)^2 +(x +y)^2\]
(x-y)^2 is same as (x-y)(x-y) apply foil method

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Other answers:

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

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