anonymous
  • anonymous
find 3 coeff. b1, b2, b3 E of R3 such that: sin(t)cos(2t) = b1sin(t) + b2sin(2t) + b3sin(3t). i just need a clue of the formula i need to use. thx alot.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
It looks like you'd need to check out one of the several double angle formulas for cos(2t) and it should become pretty simple at that point.
anonymous
  • anonymous
that's what i thought, but it's a linear algebra class so I think it's a little tougher than that.
anonymous
  • anonymous
waaaa never mind i think it's that simple.

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anonymous
  • anonymous
So, I'm assuming you're being asked to prove that sin(t)cos(2t) is in the span of {sin(t), sin(2t), sin(3t)} or something similar? Yeah, I think in this case it really is going to be simple. Sometimes with Linear Algebra problems are almost so simple that they look hard! :)
anonymous
  • anonymous
Of course the real problem is that then there is a problem that looks really simple, but ends up being really difficult and it's not always clear which is which. :)
anonymous
  • anonymous
(sin t)(cos (2t)) = (sin t)(1 – 2(sin t)^2) = sin t – 2 (sin t)^3. sin(3t) =sin (2t + t) = sin (2t) cos (t) + cos (2t) sin(t) = … = 3sin t – 4(sin t)^3. sin(2t) = 2(sin t)(cos t). sin (t) = sin (t). So now notice the Left Hand Side has no cosines. This suggests letting b2 = 0. Notice that the Left Hand Side has coefficient -2 on the cubic term. This suggests letting b3 = ½. But if you make b3 equal to 1/2, then it contributes 3*(1/2) to the coefficient of sin t. To compensate, we should let b1 be -1/2. So (-1/2, 0, 1/2) is the solution.
anonymous
  • anonymous
Or is a solution, I'm not sure that it's unique.
anonymous
  • anonymous
thanks a lot i verified my answer and it's for sure correct.
anonymous
  • anonymous
cool

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