(sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2
I've worked this out multiple times, and my answer isn't a possible choice...

- chaotic_butterflies

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- schrodinger

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- chaotic_butterflies

Answer choices:
A) 1
B) 2
C) sin^2 theta
D) cos^2 theta

- Nnesha

show ur work please :-)

- Nnesha

remember these two formulas \[\huge\rm (a+b)^2 = a^2 +2ab+b^2\]
\[\huge\rm (a-b)^2 = a^2 -2ab+b^2\]

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## More answers

- chaotic_butterflies

Oh goodness, I came at this a couple of different ways. At first I tried to replace sin and cos theta with regular numbers to see if the setup of the problem was some sort of rule, and that didn't come out right.

- Nnesha

alright so let
sin theta = a
cos theta = b
so \[\huge\rm (a-b)^2 =???\]

- chaotic_butterflies

a^2 - b^2?

- Nnesha

nope...

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
remember these two formulas \[\huge\rm (a+b)^2 = a^2 +2ab+b^2\]
\[\huge\rm (a-b)^2 = a^2 -2ab+b^2\]
\(\color{blue}{\text{End of Quote}}\)

- Nnesha

square of first number and square of 2nd term
thne MULTIPLY BOTH TERMS by 2

- Nnesha

or in other words (a-b)^2 is same as (a-b)(a-b) so you can foil :-)

- chaotic_butterflies

Okay, I can at least do that...

- Nnesha

ohh nice :-)

- Nnesha

let me know what you get :-)

- chaotic_butterflies

\[(\sin \theta - \cos \theta) (\sin \theta - \cos \theta) + (\sin \theta + \cos \theta) (\sin \theta + \cos \theta)\]

- chaotic_butterflies

Well, that's at least the first step... am I wrong or doesn't that look like it should cancel out? @Nnesha

- Nnesha

yes that's right now apply foil method :-)

- Nnesha

no no you can't cancel anything yet

- Nnesha

let sin = x
cos = y
it will be easy like simple algebra :-)
(x-y)(x-y ) :-)

- chaotic_butterflies

Oh phoey... well I'll continue to solve. If you don't mind, I like to simplify by grouping.
(x-y)(x-y)+(x+y)(x+y)
x^2 - xy (-y) (x-y) + (x+y)(x+y)
x^2 - xy - yx + y^2 + (x+ y) (x+y)
x^2 - xy - yx + y^2 + x^2 + xy (y)(x+y)
x^2 - xy - yx + y^2 + x^2 + xy + yx + y^2
x^4 + y^2

- chaotic_butterflies

*y^4

- Nnesha

alright good job thanks!
now x^2 +y^2 +x^2 +y^2 here COMBINE LIKE terms
when you MULTIPLY same bases THEN you should add their exponent

- Nnesha

\[\huge\rm x^2+y^2+x^2+y^2= ??\]
combine like terms

- chaotic_butterflies

Did I not already do that?

- Nnesha

nope last step is wrong
don't add their exponent
here is a different
COMBINE like terms :
x+x = 2x
Multiply
x times x= x^2
there is a plus sign so you can't add their exponents

- chaotic_butterflies

Goodness... I need to fresh up on my basic algebra >.<

- Nnesha

alright so combine like terms meaning add coefficient of same bases

- chaotic_butterflies

So 2x^4 + 2y^4..?

- Nnesha

yeah but exponent should the same
remember exponent rule \[\large\rm x \times x = x^2\] when u MULTIPLY same bases then u should add exponents

- chaotic_butterflies

Oh wait it would just be 2x^2 + 2y^2 wouldn't it?

- chaotic_butterflies

That loaded too late^

- Nnesha

yes that's right

- chaotic_butterflies

So that leaves me with \[\sin \theta ^{2} + \cos \theta ^{2}\]

- Nnesha

now 2 is common factor s take it out \[\huge\rm 2x^2+2y^2\]\[\large\rm 2(x^2+y^2)\]

- Nnesha

yes right change back to sin and cos
so remember special identity sin^2 x + cos^2 x = what ?

- chaotic_butterflies

Ooof... forgot about that!

- chaotic_butterflies

I honestly don't remember.

- Nnesha

hmm that's the only you shouldn't forget \[\huge\rm sin^2 \theta + \cos^2 \theta =1\]

- Nnesha

so replace sin^2 + cos ^2 by 1 :-)

- chaotic_butterflies

So it's just 2*1 = 2?

- Nnesha

yes right :-) 2!!

- chaotic_butterflies

Yay! Thank you for being so detailed, it really helped

- Nnesha

my pleasure :-)
gO_od job!! :=)

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