Question

Need help with trig!!!

Answer 1

A mass hangs from a spring which oscillates up and down. The position P (in feet) of the mass at time t (in seconds) is given by P = 4 cos (4t). For what values of t, 0 <= t < pi, will the position be at equilibrium (position is 0)?

Answer 2

so when the position is at equilibrium, P = 0, that is
\(\bf P = 4 cos (4t) \implies 0 = 4 cos (4t) \implies 0 = cos(4t)\\
\textit{taking }cos^{-1} \textit{to both sides}\\
cos^{-1}(0) = cos^{-1}(cos(4t))\)

what does that give you?

Answer 3

@jdoe0001 I'm not quite sure what does it give you?

Answer 4

well, what does \(\large cos^{-1}\) mean? what does it stand for?

Answer 5

@jdoe0001 cot? yeah I have no idea

Answer 6

in trig, \(\bf cos^{-1}(\textit{some value here)}\\
sin^{-1}(\textit{some value here)}\\
tan^{-1}(\textit{some value here)}\)

means, what is the ANGLE, whose cos, sin or tan, is "some value"

Answer 7

say for example, what's the cosine of 120 degrees?

Answer 8

-1/2

Answer 9

so, that means that \(\bf cos(120^o) = -\cfrac{1}{2} \quad and \quad cos^{-1}\left(-\frac{1}{2}\right) = 120^o\)

Answer 10

see the domain and range correlation?

Answer 11

Yeah so how would you figure out cos^-1 (cos(4t))

Answer 12

so, what's the cos(4t)? well, we dunno, let's call it, "some value"
so, \(\bf cos(4t) = \textit{some value} \quad and \quad cos^{-1}(\textit{some value}) = \quad ?\)

Answer 13

one's domain, is the other's range
so, what do you think it's?

Answer 14

90 degrees

Answer 15

90 degrees? well.... not exactly

Answer 16

but does cos^-1(0) = 90?

Answer 17

recall, we dunno what "some value" actually is

Answer 18

ohh for for 0, yes, you're correct, cosine of 0 , is 90 degrees
you also get 0 at 270 degrees

Answer 19

what about the \(\bf cos^{-1}(cos(4t))\)

Answer 20

that I'm still confused about what to do

Answer 21

erk, lemme correct myself \(\bf cos^{-1}(0) = 90^o\)
and also \(\bf cos^{-1}(0) = 270^o\)

Answer 22

notice, NOT cosine, but arcCosine so-called or \(\bf cos^{-1}\)

Answer 23

\(\bf cos(\color{red}{120^o}) = -\cfrac{1}{2} \quad and \quad cos^{-1}\left(-\frac{1}{2}\right) = \color{red}{120^o}\\ \quad \\
cos(\color{red}{4t}) = \textit{some value} \quad and \quad cos^{-1}(\textit{some value}) = \quad ?\)

Answer 24

4t

Answer 25

yeap, \(\large cos^{-1}(cos(\theta)) = \theta\)

Answer 26

ok so what are the exact values

Answer 27

\(\large {P = 4 cos (4t) \implies 0 = 4 cos (4t) \implies 0 = cos(4t)\\
\textit{taking }cos^{-1} \textit{to both sides}\\
cos^{-1}(0) = cos^{-1}(cos(4t))\\
90^o \ and\ 270^o = 4t \implies 4t =
\begin{cases}90^o \implies& 4t = 90\\
270^o \implies &4t = 270
\end{cases}}\)

if you solve for "t", what values do you get?

Answer 28

22.5 and 67.5

Answer 29

yeap, that's "t" between \(\bf 0 \le t < \pi\)

Answer 30

ok thank you so much but i also have one more question

Answer 31

.... ohhh wait a second... I meant 2 pi.. but you only need it up \(\pi\)

Answer 32

how would i sketch one cycle of P (t) with labeled x axis

Answer 33

so. I guess the 270 is ... no necessary, 270 is greater than 180 degrees,

Answer 34

oh ok good to know

Answer 35

t = 22.5 ONLY then

Answer 36

and a sketch?

Answer 37

.... can you tell what the AMPLITUDE for "4cos(4t)" is?

Answer 38

actually nevermind i can probably figure it out but thank you

Answer 39

yw