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For what angles can the exact value of all trigonometrig ratios be found?

  1. Jan 29, 2012 #1
    I have learned a lot of formulas for converting trigonometric values, but when I looked up sin(pi/10) I got the exact answer (1/4)(√5 +1). I tried to arive at this using formulas, but I couldn't. How is this found? What other angles can be found exactly.

    Ps. I already know about how taylor series and how they can be used.
     
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  3. Jan 29, 2012 #2

    Integral

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    Not sure where or how you got the exact answer, but it is not right.

    For small angles sinx is very close to x. ∏/10 ≈ .3

    your exact is .8.

    Interesting note your exact answer appears to be sin(∏/10) + .5
     
  4. Jan 29, 2012 #3
    Oops sorry it should be (1/4)(√5 - 1)
    and I got it from wolfram alpha
     
  5. Jan 29, 2012 #4

    Integral

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    Ok, that is better.

    Now just what have you tried?
     
  6. Jan 29, 2012 #5
    Addition formulas (ex: sin(a+b) = sina cosb + cosa sinb )
    Power reducing formulas (sin(a)^2 = (1-cos 2a)/2 )
    Half and double angle formulas (derived from the above two formulas)
    Sum to product formulas ( sina + sinb = 2sin((a+b)/2)cos((a-b)/2) )
    and product to sum formulas ( cosa * sinb = (sin(a+b)-sin(a-b))/2 )
     
  7. Jan 29, 2012 #6
    I can use half-angle formulas to get it to √((1-cos(pi/5))/2)
     
  8. Jan 29, 2012 #7

    Deveno

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    this is actually an interesting question, but space prevents me from giving a complete answer here. the answer has to do with "constructible numbers", which is kind of an advanced algebraic topic, and one that relates to which polynomials can be solved "by taking roots".

    perhaps you might wonder how this particular value was found. i can shed some light on this.

    we start with a (not often used) trig identity:

    [tex]\sin5x = 16\sin^5x - 20\sin^3x+5\sin x[/tex]

    now if [itex]x = \frac{\pi}{10}[/itex] the left-hand side is 1. therefore, [itex]\sin(\frac{\pi}{10})[/itex] satisfies the 5-th degree polynomial:

    [tex]16x^5 - 20x^3 + 5x - 1 = 0[/tex]

    the rational root test shows that x = 1 is a root, so we can factor out x - 1 to obtain:

    [tex](x - 1)(16x^4 + 16x^3 - 4x^2 - 4x + 1) = 0[/tex]

    we're not interested in the root 1 (since we know that's not what the sine is), so we are now just interested in a 4-th degree polynomial (well, it's an improvement) :S.

    but luck is with us! the quartic polynomial is, in fact, a perfect square:

    [tex]16x^4 + 16x^3 - 4x^2 - 4x + 1 = (4x^2 + 2x - 1)^2[/tex]

    we know the sine is positive (it's in the first quadrant), so we're only interested in the positive square root, so we have:

    [tex]4x^2 + 2x - 1 = 0[/tex]

    from here, the quadratic formula gives us the possible solutions:

    [tex]x = \frac{-1 \pm \sqrt{5}}{4}[/tex]

    since only one of these is positive, that leaves us with:

    [tex]\sin(\frac{\pi}{10}) = \frac{\sqrt{5} - 1}{4}[/tex]
     
    Last edited: Jan 29, 2012
  9. Jan 29, 2012 #8
    Awesome! Cool formula and cool solving meathod.
     
  10. Jan 29, 2012 #9

    chiro

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    Think of all the trig identities that exist in terms of sums, differences and so on.

    Also with Eulers identity you can find direct relationships betweens sums and powers of inside expressions of trig functions (like relating sin(5x) to sin^5(x) as is done with the example in an above post).

    Eulers identity (and further exponentiation) allows us to generate all kinds of relationships between inside expressions and powers and you can effectively use these to get all kinds of results.
     
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