anonymous
  • anonymous
Need help with trig!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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)?
jdoe0001
  • jdoe0001
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?
anonymous
  • anonymous
@jdoe0001 I'm not quite sure what does it give you?

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jdoe0001
  • jdoe0001
well, what does \(\large cos^{-1}\) mean? what does it stand for?
anonymous
  • anonymous
@jdoe0001 cot? yeah I have no idea
jdoe0001
  • jdoe0001
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"
jdoe0001
  • jdoe0001
say for example, what's the cosine of 120 degrees?
anonymous
  • anonymous
-1/2
jdoe0001
  • jdoe0001
so, that means that \(\bf cos(120^o) = -\cfrac{1}{2} \quad and \quad cos^{-1}\left(-\frac{1}{2}\right) = 120^o\)
jdoe0001
  • jdoe0001
see the domain and range correlation?
anonymous
  • anonymous
Yeah so how would you figure out cos^-1 (cos(4t))
jdoe0001
  • jdoe0001
so, what's the cos(4t)? well, we dunno, let's call it, "some value" so, \(\bf cos(4t) = \textit{some value} \quad and \quad cos^{-1}(\textit{some value}) = \quad ?\)
jdoe0001
  • jdoe0001
one's domain, is the other's range so, what do you think it's?
anonymous
  • anonymous
90 degrees
jdoe0001
  • jdoe0001
90 degrees? well.... not exactly
anonymous
  • anonymous
but does cos^-1(0) = 90?
jdoe0001
  • jdoe0001
recall, we dunno what "some value" actually is
jdoe0001
  • jdoe0001
ohh for for 0, yes, you're correct, cosine of 0 , is 90 degrees you also get 0 at 270 degrees
jdoe0001
  • jdoe0001
what about the \(\bf cos^{-1}(cos(4t))\)
anonymous
  • anonymous
that I'm still confused about what to do
jdoe0001
  • jdoe0001
erk, lemme correct myself \(\bf cos^{-1}(0) = 90^o\) and also \(\bf cos^{-1}(0) = 270^o\)
jdoe0001
  • jdoe0001
notice, NOT cosine, but arcCosine so-called or \(\bf cos^{-1}\)
jdoe0001
  • jdoe0001
\(\bf cos(\color{red}{120^o}) = -\cfrac{1}{2} \quad and \quad cos^{-1}\left(-\frac{1}{2}\right) = \color{red}{120^o}\\ \quad \\ cos(\color{red}{4t}) = \textit{some value} \quad and \quad cos^{-1}(\textit{some value}) = \quad ?\)
anonymous
  • anonymous
4t
jdoe0001
  • jdoe0001
yeap, \(\large cos^{-1}(cos(\theta)) = \theta\)
anonymous
  • anonymous
ok so what are the exact values
jdoe0001
  • jdoe0001
\(\large {P = 4 cos (4t) \implies 0 = 4 cos (4t) \implies 0 = cos(4t)\\ \textit{taking }cos^{-1} \textit{to both sides}\\ cos^{-1}(0) = cos^{-1}(cos(4t))\\ 90^o \ and\ 270^o = 4t \implies 4t = \begin{cases}90^o \implies& 4t = 90\\ 270^o \implies &4t = 270 \end{cases}}\) if you solve for "t", what values do you get?
anonymous
  • anonymous
22.5 and 67.5
jdoe0001
  • jdoe0001
yeap, that's "t" between \(\bf 0 \le t < \pi\)
anonymous
  • anonymous
ok thank you so much but i also have one more question
jdoe0001
  • jdoe0001
.... ohhh wait a second... I meant 2 pi.. but you only need it up \(\pi\)
anonymous
  • anonymous
how would i sketch one cycle of P (t) with labeled x axis
jdoe0001
  • jdoe0001
so. I guess the 270 is ... no necessary, 270 is greater than 180 degrees,
anonymous
  • anonymous
oh ok good to know
jdoe0001
  • jdoe0001
t = 22.5 ONLY then
anonymous
  • anonymous
and a sketch?
jdoe0001
  • jdoe0001
.... can you tell what the AMPLITUDE for "4cos(4t)" is?
anonymous
  • anonymous
actually nevermind i can probably figure it out but thank you
jdoe0001
  • jdoe0001
yw

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