Proving Sin and Cos Values of pi/6: Analytical Solutions

[leitz]

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:

Homework Statement

Use the properties of sine and cosine to prove that sin pi/6 =1/2. (It must be analytical, not geometrical)

Homework Equations

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

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?

 

[PeterO]

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:

Homework Statement

Use the properties of sine and cosine to prove that sin pi/6 =1/2. (It must be analytical, not geometrical)

Homework Equations

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

The Attempt at a Solution

sin 2x = 2sinxcosx
sin 3x = 3sinx - 4sin³x

=> sin 6x = 2sin3x cosx = 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?

When you did this line:

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

why was it not 2 sin3x.con3x ??

 

[leitz]

sorry 'bout that typo..

 

[dynamicsolo]

... then we are only left with √3/2 and 1/2.

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?

 

[PeterO]

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:

Homework Statement

Use the properties of sine and cosine to prove that sin pi/6 =1/2. (It must be analytical, not geometrical)

Homework Equations

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

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?

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