chaotic_butterflies
  • 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...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
chaotic_butterflies
  • 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 \[\huge\rm (a+b)^2 = a^2 +2ab+b^2\] \[\huge\rm (a-b)^2 = a^2 -2ab+b^2\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

chaotic_butterflies
  • 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 \[\huge\rm (a-b)^2 =???\]
chaotic_butterflies
  • 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 :-)
chaotic_butterflies
  • chaotic_butterflies
Okay, I can at least do that...
Nnesha
  • Nnesha
ohh nice :-)
Nnesha
  • Nnesha
let me know what you get :-)
chaotic_butterflies
  • chaotic_butterflies
\[(\sin \theta - \cos \theta) (\sin \theta - \cos \theta) + (\sin \theta + \cos \theta) (\sin \theta + \cos \theta)\]
chaotic_butterflies
  • 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 ) :-)
chaotic_butterflies
  • 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
  • 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
chaotic_butterflies
  • 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
chaotic_butterflies
  • 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
chaotic_butterflies
  • 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
chaotic_butterflies
  • chaotic_butterflies
Oh wait it would just be 2x^2 + 2y^2 wouldn't it?
chaotic_butterflies
  • chaotic_butterflies
That loaded too late^
Nnesha
  • Nnesha
yes that's right
chaotic_butterflies
  • 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 ?
chaotic_butterflies
  • chaotic_butterflies
Ooof... forgot about that!
chaotic_butterflies
  • 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 :-)
chaotic_butterflies
  • chaotic_butterflies
So it's just 2*1 = 2?
Nnesha
  • Nnesha
yes right :-) 2!!
chaotic_butterflies
  • chaotic_butterflies
Yay! Thank you for being so detailed, it really helped
Nnesha
  • Nnesha
my pleasure :-) gO_od job!! :=)

Looking for something else?

Not the answer you are looking for? Search for more explanations.