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En

  • one year ago

PLEASE HELP!!!!!!!!! find the derivative of z=1/2x+1/4 sin2x. i only got 1/2+1/2 cos2x but my book says the answer is cos^2x... pleaaseee explain it to me.

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  1. Loser66
    • one year ago
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    I think IrishBoy explained you already, right? they are the same.

  2. En
    • one year ago
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    i still dont get it.. i tried using the double angle but i only get 2cos^2x :/

  3. Loser66
    • one year ago
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    \(cos (2x) =2cos^2(x) -1\\2cos^2(x) = cos (2x) +1\\cos^2(x) = \dfrac{cos(2x) +1}{2}= \dfrac{1}{2}+\dfrac{cos(2x)}{2}\)

  4. En
    • one year ago
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    so the only equivalent of cos(2x) is can use is 2cos^2x-1? i cant use the \[\cos2x=\cos^2x-\sin^2x\] ?

  5. Loser66
    • one year ago
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    Yes, just take one more step : sin^2 = 1 - cos^2 to get cos (2x) = cos^2 - 1+ cos^2 = 2cos^2 -1

  6. Loser66
    • one year ago
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    it is back to my comment.

  7. En
    • one year ago
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    \[=1/2(1+\cos^2x-\sin^2x) =1/2(\cos^2x+\cos^2x)=1/2(2\cos^2x)=\cos^2x\] is this fine?

  8. En
    • one year ago
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    @Loser66 ?

  9. Loser66
    • one year ago
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    Fine

  10. En
    • one year ago
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    thanks :) sorry for the trouble.

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