Solve for Θ: Find Value of Θ When pi ≤ Θ ≤ 2pi, cos Θ = cos 1

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Homework Statement


If pi ≤ Θ ≤ 2pi and cos Θ = cos 1, what is the value of Θ? Round to nearest hundredth.


Homework Equations





The Attempt at a Solution


cos 1 = ~.54 then I didn't really understand how to interpret the "If pi ≤ Θ ≤ 2pi" Thanks for your help.
 
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Well since [itex]cos(\theta)=cos(1)[/itex] then this would instinctively mean [itex]\theta=1[/itex] but the restriction is [itex]\pi \leq \theta \leq 2\pi[/itex] or [itex]3.14 \leq \theta \leq 6.28[/itex] (approx). Obviously [itex]1<\pi[/itex] so we can't use the instinctive solution.

What other values of [itex]\theta[/itex] make the same value cos(1)? There are infinite values of [itex]\theta[/itex] that do this. Take a look at the cosine graph and find where [itex]\theta=1[/itex] (or x=1). Now the y-value at that point is cos(1). Where else does the same y value occur between [itex]\pi < \theta< 2\pi[/itex]?
 
180 deg <= Θ <= 360 deg

cos Θ = cos 1

cos 359 = cos 1

Particularly,
Θ= 2*pi*n - 1
Θ=2*180*n - 1 = 360*n - 1
Θ= 359, 719, 1079, 1439, etc degrees
180 < 359 < 360 for this case
 
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Think about this.. the cos graph in the region [itex]0 \leq \pi \leq 2\pi[/itex] is symmetrical about [itex]\pi[/itex].

At [itex]x=1[/itex] you have [itex]y=cos(1)[/itex], and you need to find the other point, i.e. at a different x value, where you get the same y value.

Perhaps if you knew the difference from the lower limit to the first x value, you could say this was the same as from the upper limit to the value you require? :wink: if that makes sense, look at the graph again. :smile:

Also, you should be able to see that [itex]cos(1)\neq cos(359)[/itex]
 
heh. Ok enlighten us, for which value of 180<x<360 is cos(x)=cos(1) ?

And don't tell me its cos 5.283185 = cos 1. Nobody said we doing this in radians
 
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.. hah yeah, what's happened there, is that I've done some bad maths! :blushing: How strange! :redface:

Hopefully my method made some sense though. I've been working in radians not degrees as that's what was stated in the question. So what I did was:

[tex]cos(1)=0.54=cos(2\pi -1)=cos(5.28)[/tex]

So [itex]\theta=5.28 radians=302.7 degrees[/itex]

.. Perhaps I need a refresher in Trig! :wink: ..never mind..