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anonymous

  • 4 years ago

Rewrite y(t)=3cos2t-4sin2t in terms of y(t)=Acos(2pi(ft)+phi))

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  1. anonymous
    • 4 years ago
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    I have worked out y(t)=Acos(2pi(ft)+phi) y(t)=Acos(phi)cos(2pi(ft))-Asin(phi)sin(2pi(ft)) a=Acos(phi) b=Asin(phi) y(t)=acos(2pi(ft))-bsin(2pi(ft))

  2. ash2326
    • 4 years ago
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    we have \[y(t)= 3\cos 2t-4 \sin 2t\] let's find \[\sqrt{3^2+4^2}\] which is 5 now in y(t), divide and multiply by 5 \[y(t)= \frac{3}{5}\cos 2t-\frac{4}{5} \sin 2t\]

  3. ash2326
    • 4 years ago
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    |dw:1328156984416:dw|now this is a triangle so cos x= 3/5 and sin x= 4/5

  4. ash2326
    • 4 years ago
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    y(t) can be written , sorry I missed 5 in numerator, it'll be there so \[y(t)= 5(\frac{3}{5}* \cos 2t- \frac{4}{5}* \sin 2t)\] now substitutin cos x and sin x in place of 3/5 and 4/5 \[y(t)=5(\cos x* \cos 2t-\sin x* \sin 2t)\] now cos A cos B- sin A sin B= cos (A+B) so \[y(t)=5(\cos (2t+x)\] where \[x =(\cos^{-1} \frac {3}{5})\] or x =53 degrees so \[y(t)= 5 \cos (2t+53)\]

  5. anonymous
    • 4 years ago
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    This looks right, but it seems too simple. Maybe, I am over-complicating things yet again.

  6. ash2326
    • 4 years ago
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    yeah it's simple :)

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