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HELP ME, PLEASE. I REALLY NEED IT
Find x:
[tex]S=\frac{1}{sin2x}+\frac{1}{sin4x}+\frac{1}{sin8x}+...+\frac{1}{sin2^{n}x }[/tex]

Respuesta :

mhanifa

Answer:

  • cotx - cot2ⁿx

Step-by-step explanation:

  • Note: finding the sum not x

First rewrite the first term as:

  • 1/sin2x =
  • sinx / (sinxsin2x)  =
  • sin(2x-x) / (sinxsin2x)  =
  • (sin2xcosx - cosxsinx) / (sinxsin2x) =
  • (sin2xcosx)/(sinxsin2x) - (cos2xsinx)/(sinxsin2x) =
  • cosx/sinx - cos2x/sin2x =
  • cotx - cot2x

Now the sum is:

  • S = cotx - cot2x + cot2x - cot4x +... + cot2ⁿ⁻²x - cot2ⁿx =
  • cotx - cot2ⁿx

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