1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    PeterO

    User Avatar
    Homework Helper

    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

    dynamicsolo

    User Avatar
    Homework Helper

    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

    PeterO

    User Avatar
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof sin pi/6
  1. Proof for Pi (Replies: 8)

  2. Sin pi/12 (Replies: 4)

Loading...