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Exact value of sin 345.5°

  1. Nov 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the exact value of sin 345.5o

    2. Relevant equations
    Trigonometry Identities

    3. The attempt at a solution
    Don't know where to start.

    Tried sin 345.5o = - sin 14.5o but stuck. Also tried multiply 345.5 with positive integer to get sin 2θ or sin 3θ or sin 4θ but also stuck

    Thanks
     
  2. jcsd
  3. Nov 14, 2015 #2

    Ray Vickson

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    You should definitely start with 2*345.5 = 691. Now try to reduce 691 to an angle < 90 degrees, by subtracting suitable multiples of 180 or 90.
     
  4. Nov 29, 2015 #3
    Sorry for taking long time to reply

    sin 691 = sin 331 = - sin 29.

    You mean finding sin 29 through the link you gave in other thread and using double angle formula?

    Thanks
     
  5. Nov 29, 2015 #4
    This is impossible. You can't get an exact value of the sign for any angle in degrees that's not divisible by 3. You won't ever get rid of the factor 3 by halving/doubling adding or subtracting angles.
     
  6. Nov 29, 2015 #5

    SteamKing

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    I think you mean sine. The trig function is called the 'sine'.
     
  7. Nov 29, 2015 #6

    BvU

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  8. Nov 29, 2015 #7

    Ray Vickson

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    No, that is incorrect. A paper giving exact algebraic formulas for the sine of all angles from 1 to 90 degrees, in 1 degree increments, has been published on-line (with proofs included). It was done as a retirement project by an ex-professor of mathematics; for a precise citation, see one of my responses in the previous thread by user 'songoku' on a related topic.

    Note added in edit: I see that BvU has already dealt with this issue, in a post that appeared on my screen only after I pressed the 'enter' key.
     
  9. Nov 29, 2015 #8

    haruspex

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    To go further, I believe it should be possible to find an exact representation of cos(pi/n) only involving square roots if and only if there exists a ruler and compass construction for a regular n-sided polygon. As is well known, that is possible whenever n is the product of a power of 2 and distinct Fermat primes. That is enough to get all multiples of 1 degree as well as pi/17 etc.

    But if we allow other surds then there are more possibilities. Since cos(nx) can be expanded as an nth order polynomial in cos(x), and cos(x)=-cos(pi-x), we can expand cos(4pi/7)=-cos(3pi/7) to obtain a quartic in cos(pi/7).
    I feel there should be a generalization of the Fermat primes that corresponds to roots up to cubic and quartic, but I'm not aware of such.
     
  10. Nov 29, 2015 #9

    SammyS

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    °
    Doesn't that only get you down to multiples of 3° ?
     
  11. Nov 29, 2015 #10

    haruspex

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    Sorry, yes, 3°. To get to 1° you need to use the cos(3x) expansion or similar, so does involve cube roots.
     
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