anonymous
  • anonymous
(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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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
−4sin(Θ)cos(Θ) 2 sin2 Θ cos2 Θ
dumbcow
  • dumbcow
distribute, then combine terms note : sin^2 + cos^2 = 1
UnkleRhaukus
  • UnkleRhaukus
use difference of squares, a^2 - b^2 = (a+b)(a-b)

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midhun.madhu1987
  • midhun.madhu1987
\[a ^{2} - b ^{2} = (a-b)(a+b)\]
UnkleRhaukus
  • UnkleRhaukus
where a = sin Θ − cos Θ b = sin Θ + cos Θ
UnkleRhaukus
  • UnkleRhaukus
can you do it @gabbimanges17 ?
anonymous
  • anonymous
@UnkleRhaukus is the answer sin^2?
UnkleRhaukus
  • UnkleRhaukus
nope. what do you get for (a-b)? and what do you get for (a+b)?
anonymous
  • anonymous
@UnkleRhaukus im not sure
UnkleRhaukus
  • UnkleRhaukus
\[ (\sin\theta − \cos\theta)^2 − (\sin\theta + \cos\theta)^2\] \[\Big((\sin\theta − \cos\theta)+(\sin\theta + \cos\theta)\Big)\Big((\sin\theta − \cos\theta)-(\sin\theta + \cos\theta)\Big)\]
anonymous
  • anonymous
looks like they cancel each other out
UnkleRhaukus
  • UnkleRhaukus
many of the terms cancel away, but not all of them
anonymous
  • anonymous
@UnkleRhaukus can you specify?
UnkleRhaukus
  • UnkleRhaukus
\[\color{teal}{(\sin\theta − \cos\theta)}^2 − \color{orange}{(\sin\theta + \cos\theta)}^2\] \[=\Big(\color{teal}{(\sin\theta − \cos\theta)}+\color{orange}{(\sin\theta + \cos\theta)}\Big)\Big(\color{teal}{(\sin\theta − \cos\theta)}-\color{orange}{(\sin\theta + \cos\theta)}\Big)\\ =\Big(\sin\theta − \cos\theta+\sin\theta + \cos\theta\Big)\Big(\sin\theta − \cos\theta-\sin\theta - \cos\theta\Big)\\ = \]
UnkleRhaukus
  • UnkleRhaukus
\[=\Big(\sin\theta +\sin\theta + \cos\theta − \cos\theta\Big)\Big(\sin\theta -\sin\theta - \cos\theta− \cos\theta\Big)\\ =\]
anonymous
  • anonymous
0
UnkleRhaukus
  • UnkleRhaukus
work it out properly
anonymous
  • anonymous
im not sure what youre asking.. it looks like the last two cos dont cancel out
UnkleRhaukus
  • UnkleRhaukus
\[=\Big(\sin\theta +\sin\theta + \cos\theta − \cos\theta\Big)\Big(\sin\theta -\sin\theta - \cos\theta− \cos\theta\Big)\\ =\Big((1+1)\sin\theta + (1-1)\cos\theta\Big)\Big((1-1)\sin\theta +(-1-1)\cos\theta\Big)\\ =\]
anonymous
  • anonymous
2?
anonymous
  • anonymous
you have me in a twist
UnkleRhaukus
  • UnkleRhaukus
simplify!
anonymous
  • anonymous
−4sin(Θ)cos(Θ)
UnkleRhaukus
  • UnkleRhaukus
are you sure?
anonymous
  • anonymous
im not sure about anything anymore, please can you explain in plain english
UnkleRhaukus
  • UnkleRhaukus
The expression given is a difference of squares, we used a formula we already knew for the difference of square to rearrange the expression. Then we simplified the terms in this new form, and we got ...
anonymous
  • anonymous
−4sin(Θ)cos(Θ)
UnkleRhaukus
  • UnkleRhaukus
good.
anonymous
  • anonymous
thank you!

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