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psk981

  • 2 years ago

L{f(t)}. f(t)= cos^2(t)

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  1. sthop456
    • 2 years ago
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    Don't know if this is right, but if you are looking for the limit of cos^2(x), cos(x) oscillates between -1 and 1, so squaring it would still oscillate it between 0 and 1, so the limit would not exist, DNE (does not exist). Not sure, though.

  2. UnkleRhaukus
    • 2 years ago
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    \[\cos^2(\theta)=\frac12\left(\cos(2\theta)-1\right)\]

  3. UnkleRhaukus
    • 2 years ago
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    \[\mathcal L\Big\{ \cos^2\theta\Big\}=\frac12\left(\mathcal L\Big\{ \cos2\theta\Big\}-\mathcal L\Big\{1\Big\}\right)\]

  4. UnkleRhaukus
    • 2 years ago
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    tell me what you get @psk981

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