anonymous
  • anonymous
Verify the identity. cos(4u) = cos^2(2u) - sin^2(2u)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\cos \left( 4u\right) = \cos ^{2}\left( 2u \right) - \sin ^{2}\left( 2u \right)\]
anonymous
  • anonymous
@Luigi0210
anonymous
  • anonymous
@Hero

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anonymous
  • anonymous
@arindameducationusc
anonymous
  • anonymous
@kaitlyn_nicole
anonymous
  • anonymous
@kaitlyn_nicole
anonymous
  • anonymous
@Michele_Laino
arindameducationusc
  • arindameducationusc
Do you know Cos2theta= cos^2theta-sin^2theta?
anonymous
  • anonymous
yes, double angle identity
Michele_Laino
  • Michele_Laino
hint: we can rewrite the left side as below: \[\Large \cos \left( {4u} \right) = \cos \left( {2u + 2u} \right)\]
Michele_Laino
  • Michele_Laino
now you can apply the formula of addition for cosine function
Michele_Laino
  • Michele_Laino
\[\Large \cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\]
anonymous
  • anonymous
cos ( 2u + 2u ) = cos (2u) cos (2u) - sin (2u) sin (2u) ?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
what do i do from there?
Michele_Laino
  • Michele_Laino
it is simple, since: \[\Large \cos \left( {2u} \right)\cos \left( {2u} \right) = {\left( {\cos \left( {2u} \right)} \right)^2}\]
Michele_Laino
  • Michele_Laino
similarly for sin(2u)*sin(2u)
anonymous
  • anonymous
sin(2u) sin(2u) = (sin(2u))^2 ?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
okay, so since im trying to verify this identity, what else do i do to get the answer to look like cos(4u) = cos(4u) ?
Michele_Laino
  • Michele_Laino
since we have proven that left side is equal to right side, then we are done here
anonymous
  • anonymous
so is the final answer cos ( 2u + 2u ) = cos (2u) cos (2u) - sin (2u) sin (2u) or cos(2u) cos(2u) = (cos(2u))^2 ?
Michele_Laino
  • Michele_Laino
the answer is the entire procedure which shows how to get the right side, starting from left side
anonymous
  • anonymous
ahhhhhh, i see, i see. thank you soooooo much for your help!
Michele_Laino
  • Michele_Laino
:)

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