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Homework Help: Proof sin pi/6

  1. Aug 18, 2011 #1
    I'm currently reading Apostol's Calculus. I'm now on the part where I need to prove sin and cos values for specific values. Right now I'm working on sin and cos of pi/6. However, I always get at 1/2 and √3/2 for the final value. Here's my solution so far:

    1. The problem statement, all variables and given/known data
    Use the properties of sine and cosine to prove that sin pi/6 =1/2. (It must be analytical, not geometrical)

    2. Relevant equations
    The book said that I could use exercise 4, which was to prove that sin3x= 3 sinx - 4sin3x , cos3x = cos x - 4sin2x cosx.

    3. The attempt at a solution
    sin 2x = 2sinxcosx
    sin 3x = 3sinx - 4sin3x

    => sin 6x = 2sin3x cos3x = 2sin3x√(1-sin23x)

    =>sin 6x = 2(3sinx - 4sin3x)√[1-(3sinx - 4sin3x)2]

    let x = pi/6
    => sin pi = 2(3sin(pi/6) - 4sin3(pi/6))√[1-(3sin(pi/6) - 4sin3(pi/6))2] = 0 (Using the property of sine - sin pi = sin 0 = 0)

    By using the zero factor theorem (Theorem I.11 on Apostol's book), either
    (3sin(pi/6) - 4sin3(pi/6))=0 or √[1-(3sin(pi/6) - 4sin3(pi/6))2]=0

    After simplifying the equations, it will turn out that sin pi/6 is equal to 0, √3/2 , -√3/2 , -1, and 1. But since pi/6 is in the above the real axis and is in the first quadrant, then the value of sin pi/6 MUST be positive, eliminating the negative roots. Furthermore, since it was also established as a property in the book that sin 0 = 0 and sin pi/2 =1, then we are only left with √3/2 and 1/2.

    This is where my problem lies. Any idea how to continue? Also, What would be a nice start for the cosine counter part of it?
    Last edited: Aug 19, 2011
  2. jcsd
  3. Aug 19, 2011 #2


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    When you did this line:

    => sin 6x = 2sin3x cosx = 2sin3x√(1-sin23x)

    why was it not 2 sin3x.con3x ??
  4. Aug 19, 2011 #3
    sorry 'bout that typo..
  5. Aug 19, 2011 #4


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    You still have the equation for, say, cos 3x. Presumably, you also have the Pythagorean Identity, so you know that cos(pi/2) = 0 , since sin(pi/2) = 1 . So we have

    cos(pi/6) [ 1 - 4 sin2(pi/6) ] = 0 .

    If sin(pi/6) = 1/2 , then cos(pi/6) = (√3)/2 , and vice versa [also by Pythagorean Identity]. Which of these possibilities solves the equation?
  6. Aug 19, 2011 #5


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    Read a little further and found that you had already established that sin(pi/2) =1

    Why not use
    sin 3x = 3sinx - 4sin3x

    and let x = pi/6

    You get a cubic which fully factorises and yields your answer of 1/2

    I suspect using the cos3x expression you may get an answer there -especially of you have already proved that cos(pi/2) = 0
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