Question

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

Answer 1

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))

Answer 2

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
\[y(t)= \frac{3}{5}\cos 2t-\frac{4}{5} \sin 2t\]

Answer 3

now this is a triangle
so cos x= 3/5
and sin x= 4/5

Answer 4

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
\[y(t)= 5 \cos (2t+53)\]

Answer 5

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

Answer 6

yeah it's simple :)