Question

y=5cos(3*x-(pi/4))

can someone help me find the interval that covers one period, i know the left hand side end point is Pi/12 but what is the right end point?

I thought it was 25Pi/12, but my book says 3Pi/4??

Answer 1

the period of cosine is \(2\pi\) which means it does all its business on any interval of length \(2\pi\)
one such interval would be \([0,2\pi]\) so we can solve this problem by setting
\[3x-\frac{\pi}{4}=0\] and solving for \(x\), which gives what you said
\[x=\frac{\pi}{12}\]

Answer 2

the next step would be to set
\[3x-\frac{\pi}{4}=2\pi\] and solve for \(x\)

Answer 3

hmm, i factored out a 3 from the argument.....but i guess we are not suppose to that

Answer 4

you get
\[3x=2\pi+\frac{\pi}{4}\]
\[3x=\frac{9\pi}{4}\]
\[x=\frac{3\pi}{4}\]

Answer 5

if you "factor it out" you would have to write
\[3x-\frac{\pi}{4}=3(x-\frac{\pi}{12})\] and i guess this would work too, but it seems like extra bother

Answer 6

your right :)

Answer 7

\(\spadesuit\)

Answer 8

\spadesuit does it

Answer 9

for \( y=5\cos\left(3x-\frac{\pi}{4}\right)\)
you can match this to
\[ A \cos\left(\frac{2\pi}{T}x + \phi\right) \]
A is the amplitude, T is the period, and ϕ is the phase.
matching\(\frac{2\pi}{T} \)= 3 we find T=\(\frac{2\pi}{3} \)
any interval of this size covers one period.
If you start at \(\frac{\pi}{12} \) (you don't really have to ) add T=\(\frac{2\pi}{3} \) to get the start of the next period:
\(\frac{\pi}{12} + \frac{2\pi}{3}=\frac{9\pi}{12}=\frac{3\pi}{4}\)