Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

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Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

Mathematics
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(sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2 (sin^2-cos^2) + (sin^2+cos^2) (y^2-x^2) + (y^2+x^2) eliminate x^2 (y^2+y^2)= 2sin^2
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Other answers:

\[\huge\rm (x+y)^2 \cancel= x^2 + y^2\]
nope let sin = x cos = y so \[(x -y)^2=(x-y)(x-y)\] now foil
My options: A. −sin2 Θ B. −cos2 Θ C. 0 D. 2
alright so u didn't foil it right \[(x-y)^2 \cancel= x^2 -y^2\] (x-y)^2 is same as (x-y)(x-y ) can you foil these ?
one moment so i can foil
yes right
Open the square Then put sin^2x+cos^x=1 You get 1-2sinxcosx +1+2sinxcosx =1+1=2 (I'm putting theta as x)
is it x^2+2xy+y^2
Yes
yes right (x+y) = x^2 +2xy +y^2 :-)
what about (x-y)^2 ??
the same thing
(x-y)^2 is what i solved for
expand you have to foil it (X-y)^2 is same as (x-y)(x-y)
No 2xy is replaced by-2xy
right
just like (x+y)(x+y)
\[\huge\rm x^2 -2xy +y^2 + x^2 +2xy +y^2\]\] combine like terms
2x^2 + 2y^2
I hope you can add the two square carefully.
yep now replace x and y by sin and cos \[\huge\rm 2 \sin^2 + 2\cos^2\] take out common factor apply this identity \[\large\rm sin^2 \theta + \cos^2 \theta = 1 \]
=2
yes right \[\huge\rm \color{reD}{ 2} \sin^2 +\color{reD}{ 2}\cos^2\] \[\huge\rm \color{reD}{ 2} (\sin^2 \theta +cos^2 \theta)\] \[\huge\rm \color{reD}{ 2} (1) =2\]
Thank you so much for your help @Nnesha i really appreciate it.
my pleasure :-)
<3

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