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?