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

1. anonymous

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

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

|dw:1328156984416:dw|now this is a triangle so cos x= 3/5 and sin x= 4/5

4. ash2326

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

This looks right, but it seems too simple. Maybe, I am over-complicating things yet again.

6. ash2326

yeah it's simple :)