anonymous
  • anonymous
Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
(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
anonymous
  • anonymous
please check my answer
anonymous
  • anonymous
@Luigi0210 @KyanTheDoodle @mathmath333 @sleepyjess

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More answers

Nnesha
  • Nnesha
\[\huge\rm (x+y)^2 \cancel= x^2 + y^2\]
Nnesha
  • Nnesha
nope let sin = x cos = y so \[(x -y)^2=(x-y)(x-y)\] now foil
anonymous
  • anonymous
My options: A. −sin2 Θ B. −cos2 Θ C. 0 D. 2
Nnesha
  • Nnesha
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 ?
anonymous
  • anonymous
one moment so i can foil
Nnesha
  • Nnesha
yes right
Pawanyadav
  • Pawanyadav
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)
anonymous
  • anonymous
is it x^2+2xy+y^2
Pawanyadav
  • Pawanyadav
Yes
Nnesha
  • Nnesha
yes right (x+y) = x^2 +2xy +y^2 :-)
Nnesha
  • Nnesha
what about (x-y)^2 ??
anonymous
  • anonymous
the same thing
anonymous
  • anonymous
(x-y)^2 is what i solved for
Nnesha
  • Nnesha
expand you have to foil it (X-y)^2 is same as (x-y)(x-y)
Pawanyadav
  • Pawanyadav
No 2xy is replaced by-2xy
anonymous
  • anonymous
right
Nnesha
  • Nnesha
just like (x+y)(x+y)
Nnesha
  • Nnesha
\[\huge\rm x^2 -2xy +y^2 + x^2 +2xy +y^2\]\] combine like terms
anonymous
  • anonymous
2x^2 + 2y^2
Pawanyadav
  • Pawanyadav
I hope you can add the two square carefully.
Nnesha
  • Nnesha
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 \[\large\rm sin^2 \theta + \cos^2 \theta = 1 \]
anonymous
  • anonymous
=2
Nnesha
  • Nnesha
yes right \[\huge\rm \color{reD}{ 2} \sin^2 +\color{reD}{ 2}\cos^2\] \[\huge\rm \color{reD}{ 2} (\sin^2 \theta +cos^2 \theta)\] \[\huge\rm \color{reD}{ 2} (1) =2\]
anonymous
  • anonymous
Thank you so much for your help @Nnesha i really appreciate it.
Nnesha
  • Nnesha
my pleasure :-)
anonymous
  • anonymous
<3

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