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

show ur work please :-)

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

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

a^2 - b^2?

nope...

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

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

Okay, I can at least do that...

ohh nice :-)

let me know what you get :-)

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

no no you can't cancel anything yet

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

*y^4

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

Did I not already do that?

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

alright so combine like terms meaning add coefficient of same bases

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

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

That loaded too late^

yes that's right

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

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

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

Ooof... forgot about that!

I honestly don't remember.

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

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

So it's just 2*1 = 2?

yes right :-) 2!!

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

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