Find theta given known cos and sin...

  • #26
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If i used: sin2x = 2sinx cos x
Sin 2x = ## \frac{ -\sqrt2}{2}##
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?
 
  • #27
SammyS
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If i used: sin2x = 2sinx cos x
Sin 2x = ## \frac{ -\sqrt2}{2}##
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?
You can reduce those fractions. 45/36 = 5/4, etc., and it's pi, not phi.

More to the point for your question here:

In the problem statement the requirement is, 0 ≤ x < 2π . (Actually it has θ, not x. That's not important.)

Therefore, 0 ≤ 2x < 4π. Right?

So find the additional solutions which fit that. (Add 2π to those two answers you have for 2x .)

By The Way:
In the blue strip above the box in which you compose posts, there is a large ∑ character. Click on that to find a large set of characters useful for math/science.
θ π φ ∅ ℝ etc.
 
Last edited:
  • #28
581
20
X (theta) must be on 4 Quadrant, so 2x must on 8 quadrants, 540<2x<=720
X 1=5/4 π
X2 = 7/4 π
Both added 2π to get x at 8 quadrants
X1 = 13/4 π
X2 = 15/4 π

Both satisfied the domain
How to decide which one is the answer?
 
  • #29
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20
I'm confused, why cos 13/4 π is negative, even though 13/4 π on 8 quadrants?
 
  • #30
Ray Vickson
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If i used: sin2x = 2sinx cos x
Sin 2x = ## \frac{ -\sqrt2}{2}##
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?

Because of "circularity", we have:
1st quadrant ↔ 5th quandrant ↔ 9th quandrant, etc.
4th quadrant ↔ 8th quadrant ↔ 12th quadrant, etc.
(Here, ↔ means "geometrically equivalent".)

Get it? The are only 4 physically different quadrants, so the "8th quadrant" occupies the same physical location as the 4th quadrant, and so forth. That is why both ##\theta## and ##2 \theta## can be in the same (4th = 8th = 12th....) actual, geometric quadrant.
 
  • #31
SammyS
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X (theta) must be on 4 Quadrant, so 2x must on 8 quadrants, 540<2x<=720
X 1=5/4 π
X2 = 7/4 π
Both added 2π to get x at 8 quadrants
X1 = 13/4 π
X2 = 15/4 π

Both satisfied the domain
How to decide which one is the answer?
Those actually should written be something like
2x1 = (5/4) π
2x2 = (7/4) π​
Then adding 2π gives (two new possibilities for x):
2x3 = (13/4) π
2x4 = (15/4) π​

Solve for each xk .
 

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