Question

Trigonometric function help? will medal!!

Answer 1

For the function given, state the starting point for a sample period:

Answer 2

the answers are:

Answer 3

@SolomonZelman Can you help me? sorry for the tag

Answer 4

\[f(t)=a \cdot \cos(b(t-c))\]
here you have phase shift number is c
and you have 2pi/b is the period

so you have a slightly different form:
\[f(t)=acos(bt-cb)\]
just manipulate this form into the form I have notice the number in front of the t

Answer 5

ok

Answer 6

so what is the phase shift number?

Answer 7

i want to say 5pi/6

Answer 8

Here is an example:
\[y=2 \sin(\frac{1}{2}t-\pi) \\ y=2 \sin(\frac{1}{2}t-\frac{2}{2}\pi) \\ y=2 \sin(\frac{1}{2}(t-2\pi))\]
so the phase shift number is 2pi
and we can find the corresponding y value by saying 0(2) since we know this is the sin function and we know how it starts out or you can just plug in if you aren't sure
\[f(2\pi)=2 \sin(\frac{1}{2}(2\pi-2\pi)) \\ f(2\pi)=2 \sin(\frac{1}{2}(0)) \\ f(2\pi)=2\sin(0) \\ f(2\pi)=2(0) \\ f(2\pi)=0\]
so the start point of a one cycle is (2pi,0) for this example

Answer 9

Well you haven't seemed to tried to put your problem into the form I have stated

Answer 10

notice the number in front of your t
factor that number out of both terms inside your cos function

Answer 11

really really big hint:
\[f(t)=3 \cos(\frac{1}{3}t-\frac{1}{3} \frac{5\pi}{2})\]

Answer 12

ok

Answer 13

sorry im kind of slow

Answer 14

so i have to factor 1/3 by what?

Answer 15

you want to write what you have in the form f(t)=acos(b(x-c))

Answer 16

to do that you notice the number in from of your t variable and factor it out of the expression inside your cos function

Answer 17

\[f(t)=3 \cos(\frac{1}{3}(t-\frac{5\pi}{2}))\]

Answer 18

and I already told you you can easily determine the phase shift number once it is in this form

Answer 19

f(t)=acos(b(t-c))
tells us the phase shift number is c

Answer 20

thats the 5pi/6

Answer 21

\[f(t)=3 \cos(\frac{1}{3}(t-\color{ red}{\frac{5\pi}{2}})) \\ f(t)=a \cos(b(t-\color{red}c))\]
what is c ?

Answer 22

would it be a decimal?

Answer 23

ok if that didn't make it obvious to you what c was
...
the only other way I have for you is to find the value of t that makes the inside of your cos function 0
that is solve this for t
\[\frac{1}{3}t-\frac{5\pi}{6}=0\]

Answer 24

that will give you the phase shift number

Answer 25

ok i got 7.85398

Answer 26

could you please leave it in exact form

Answer 27

your answers are in exact form so it makes no sense to write it like that

Answer 28

\[\frac{1}{3} t -\frac{5 \pi}{6}=0 \\ \frac{1}{3} t =\frac{5\pi}{6} \\ \\ \text{ now multiply 3 on both sides } \\ t=\frac{5\pi(3)}{6} \\ t=\frac{3}{6} \cdot 5 \pi\]
reduce 3/6

Answer 29

im using a calculator and it says t= 5pi/2 or its the same as the decimal

Answer 30

that is right
but no it isn't the same as the decimal
the decimal thing is an approximation of 5pi/2

Answer 31

reducing 3/6 is 1/2 right?

Answer 32

\[t=\frac{3}{6} \cdot 5\pi =\frac{1}{2} \cdot 5 \pi =\frac{5\pi}{2}\]

Answer 33

that is the phase shift number

Answer 34

now you can find the y coordinate the easy way or the hard way
and it sounds like you should do it the hard way
\[f(t)=3 \cos(\frac{1}{3}t-\frac{5\pi}{6})\]
plug in 5pi/2

Answer 35

and it isn't too hard to evaluate that
because you know cos(0)=?

Answer 36

multiply that result by 3
since you have a 3 outside the cos( ) part

Answer 37

and you will be done

Answer 38

lol i got zero T-T

Answer 39

cos(0) isn't 0 :(

Answer 40

shoot

Answer 41

wait is it the last one? 5pi/2,3?

Answer 42

yes cos(0) is 1
and 3(1)=3

Answer 43

sweet, because i got 3 lol

Answer 44

\[f(\frac{5\pi}{2})=3 \cos(\frac{1}{3} \cdot \frac{5\pi}{2}-\frac{5\pi}{6}) \\ f(\frac{5\pi}{2})=3 \cos(\frac{5\pi}{6}-\frac{5\pi}{6}) \\ f(\frac{5\pi}{2})=3 \cos(0) \\ f(\frac{5\pi}{2})=3(1) \\ f(\frac{5\pi}{2})=3\]

Answer 45

sorry for being complicated >_< im not good at math, it you didnt notice lol