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Mathematics 25 Online
OpenStudy (chaotic_butterflies):

(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...

OpenStudy (chaotic_butterflies):

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

Nnesha (nnesha):

show ur work please :-)

Nnesha (nnesha):

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

OpenStudy (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 (nnesha):

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

OpenStudy (chaotic_butterflies):

a^2 - b^2?

Nnesha (nnesha):

nope...

Nnesha (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 (nnesha):

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

Nnesha (nnesha):

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

OpenStudy (chaotic_butterflies):

Okay, I can at least do that...

Nnesha (nnesha):

ohh nice :-)

Nnesha (nnesha):

let me know what you get :-)

OpenStudy (chaotic_butterflies):

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

OpenStudy (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 (nnesha):

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

Nnesha (nnesha):

no no you can't cancel anything yet

Nnesha (nnesha):

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

OpenStudy (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

OpenStudy (chaotic_butterflies):

*y^4

Nnesha (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 (nnesha):

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

OpenStudy (chaotic_butterflies):

Did I not already do that?

Nnesha (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

OpenStudy (chaotic_butterflies):

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

Nnesha (nnesha):

alright so combine like terms meaning add coefficient of same bases

OpenStudy (chaotic_butterflies):

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

Nnesha (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

OpenStudy (chaotic_butterflies):

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

OpenStudy (chaotic_butterflies):

That loaded too late^

Nnesha (nnesha):

yes that's right

OpenStudy (chaotic_butterflies):

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

Nnesha (nnesha):

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

Nnesha (nnesha):

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

OpenStudy (chaotic_butterflies):

Ooof... forgot about that!

OpenStudy (chaotic_butterflies):

I honestly don't remember.

Nnesha (nnesha):

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

Nnesha (nnesha):

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

OpenStudy (chaotic_butterflies):

So it's just 2*1 = 2?

Nnesha (nnesha):

yes right :-) 2!!

OpenStudy (chaotic_butterflies):

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

Nnesha (nnesha):

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

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