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Find theta given known cos and sin...

  1. Jun 23, 2017 #1
    1. The problem statement, all variables and given/known data
    15_Mat_B_1.2.png

    2. Relevant equations
    cos 2theta = costheta^2 - sintheta^2

    3. The attempt at a solution
    cos2theta = 1
    2theta = 0, 2phi

    but i get wrong answer.. how is it?
     
  2. jcsd
  3. Jun 23, 2017 #2

    Ray Vickson

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    Well, given your values for ##\sin \theta## and ##\cos \theta## you should NOT get ##\cos 2 \theta = 1##. Check your algebra.
     
  4. Jun 23, 2017 #3
    oh.. cos 2 theta = root 2 / 2
    2theta = 45degrees
    the sin is negative, theta must on quadrant 2 or 3
    225 or 315 degrees.
     
  5. Jun 23, 2017 #4

    phinds

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    oops. never mind
     
  6. Jun 23, 2017 #5
    π
    No, with cos +ve and sin -ve, the angle must be in the 4th quadrant. All silly tom cats. (A calculator will give you different angles.)
    θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 (note the question asked for the answer as a multiple of π, not in degrees).
     
  7. Jun 23, 2017 #6

    phinds

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    Yes, I had already deleted my post before you posted this
     
  8. Jun 23, 2017 #7

    Ray Vickson

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    Or, the angle could be ##-\pi/4##, assuming negative angles are allowed.
     
  9. Jun 23, 2017 #8
    The question said 0 ≤ θ < 2π.
     
  10. Jun 23, 2017 #9
    But the answer also 15/4 pi.. Which is 337.5 degree
     
  11. Jun 23, 2017 #10

    Ray Vickson

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    No, it isn't. ##2 \pi \leftrightarrow 360^o##, and ##15/4> 3> 2##, so ##15/4 \pi ## is a lot bigger than ##360^0##.
     
  12. Jun 23, 2017 #11
    no
    the key answer it is. the 2 theta allowed to be bigger than 2 pi. just theta < 2pi
     
  13. Jun 24, 2017 #12

    SammyS

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    Yes, but you said:
    As Ray said, that's not true. After all, (15/4)π ≠ 337.5° .

    Perhaps you meant that
    one of the answers is 2θ = (15/4)π , which means that θ = 337.5° ,​
    so, of course, θ = (15/8)π < 2π .
     
  14. Jun 25, 2017 #13
    But the restrictions say theta is defined on [0,2pie)?
    Hence, 2theta=-pie/4.

    From our restriction, this -pie/4 should be equal to 7pie/4 (sin is negative and cos is positive/ IV Quadrant).

    Therefore 2theta=7pie/4
    theta=7pie/8
     
  15. Jun 25, 2017 #14

    Mark44

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    You do know that there is a difference between "pie" and "pi" with the latter being the name of a Greek letter, right?
     
  16. Jun 26, 2017 #15
    typing with a new tablet... with auto correct...

    You do know that making a post within a topic, that is not relevant to the discussion is considered trolling? That is a violation of PF rules of conduct.
     
  17. Jun 26, 2017 #16
    The restriction applies to θ, not 2θ. It is also θ that must be in the fourth quadrant, therefore θ = 7π/8 is not acceptable
     
  18. Jun 26, 2017 #17

    ehild

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    The restriction applies to theta, but cos(2θ) is positive and sin(2θ) is negative, therefore 2θ is in the fourth quadrant or 2pi more. And the problem asks 2θ.
     
  19. Jun 26, 2017 #18
    2θ is indeed in the fourth quadrant, but so is θ, as cos θ is positive and sin θ negative, so 2θ must be in the "8th quadrant", as it were.
     
  20. Jun 26, 2017 #19
    @mjc123 why θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 ?
    Why you substract π/8 from 2π?
     
  21. Jun 26, 2017 #20
    @mjc123 2 θ = is 45 degrees, theta = 22.5 degrees
    So quadrant 4 is 360-22.5 = 337.5
    So 2theta = 337.5 x 2 = 675 degrees (15/4)pi
    Why can't i use directly 45degrees, so theta in quadrant 4 is 360-45 = 315 degrees
    Since 315 and 675 have same cos value, why the answer is 675?
     
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