Moved - evaluate Θ
- Thread starter JSmith
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You will need two things for this:
i) a calculator (in radian mode) to find \(\displaystyle \theta=\cos^{-1}\left(\dfrac{3}{4} \right)\) (only if you want a decimal approximation).
ii) The identity \(\displaystyle \cos(\theta)=\cos(2\pi-\theta)\) to get the 4th quadrant solution.
Think of the unit circle and the vertical line \(\displaystyle x=\dfrac{3}{4}\). The two points where these intersect correspond to the two solutions on the given interval. Since the inverse cosine function will give you the 1st quadrant solution as its range is \(\displaystyle [0,\pi]\), we need the identity in ii) to give us the 4th quadrant solution.
A further question...are you certain the equation you have been given is not:
\(\displaystyle \cos^2(\theta)=\dfrac{3}{4}\) ?
This would actually make more sense if you are studying the "special angles."
i) a calculator (in radian mode) to find \(\displaystyle \theta=\cos^{-1}\left(\dfrac{3}{4} \right)\) (only if you want a decimal approximation).
ii) The identity \(\displaystyle \cos(\theta)=\cos(2\pi-\theta)\) to get the 4th quadrant solution.
Think of the unit circle and the vertical line \(\displaystyle x=\dfrac{3}{4}\). The two points where these intersect correspond to the two solutions on the given interval. Since the inverse cosine function will give you the 1st quadrant solution as its range is \(\displaystyle [0,\pi]\), we need the identity in ii) to give us the 4th quadrant solution.
A further question...are you certain the equation you have been given is not:
\(\displaystyle \cos^2(\theta)=\dfrac{3}{4}\) ?
This would actually make more sense if you are studying the "special angles."
D
Deleted member 4993
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If, what are the possible values of θ, if 0 ≤ θ ≤ 2π.
Unsure how to go about doing this with a unit circle/in radians... Help?
Please start a new thread with a new problem
If I am not allowed to use calculator - I'll approximate my solution with:
3/4 ~ 1/√2
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HallsofIvy
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Are you asked only for approximate solutions as Subhotosh Kahn says? It would seem very strange to ask for approximate solutions but not allow you to use a calculator which gives very good approximate solutions. And there simply is no formula which will give an exact solution. I would appreciate it if you would give the exact statement of the problem.
To determine the first value, I used the cos-1 function on my calculator, which told me that cos(.7227)=3/4. The first value of theta is 0.7227 radians.
Then, I can use cos(x)= (2pi - x) to get the other point of intersection. This gives us 5.5604 radians.
Therefore, the possible values of theta in the interval 0 ≤ θ ≤ 2π are 0.7227, and 5.5604 radians.
How does that look?
Then, I can use cos(x)= (2pi - x) to get the other point of intersection. This gives us 5.5604 radians.
Therefore, the possible values of theta in the interval 0 ≤ θ ≤ 2π are 0.7227, and 5.5604 radians.
How does that look?
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Your approximation for first quadrant solution is rounded correctly, but the fourth quadrant solution should be rounded to 5.5605.
Unless directed to approximate, I would write the solutions exactly as:
\(\displaystyle \theta=\cos^{-1}\left(\dfrac{3}{4} \right),\,2\pi-\cos^{-1}\left(\dfrac{3}{4} \right)\)
Unless directed to approximate, I would write the solutions exactly as:
\(\displaystyle \theta=\cos^{-1}\left(\dfrac{3}{4} \right),\,2\pi-\cos^{-1}\left(\dfrac{3}{4} \right)\)