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A Sine, a Double Angle Sine, a Square Root -- and an Imaginative Substitution

Date: 05/19/2011 at 14:32:33
From: Rattanjeet
Subject: Trigonometric Identities

Prove that if A = 2pi/7, then

      sinA + sin2A + sin 4A = sqrt(7/2)

I have expanded the left side as follows:

     sinA + sin2A + sin 4A
   = sinA + 2sinAcosA + 2sin2Acos2A
   = sinA + 2sinAcosA + 2[(2sinAcosA)(cos^2A - sin^2A)]

But I do not understand which one of the three cosine formulas for double
angles to use to get sqrt(7/2). I do not know what to do next.

Date: 05/22/2011 at 14:51:26
From: Doctor Anthony
Subject: Re: Trigonometric Identities

This is a lot more difficult than your usual identity problem. The
technique is very indirect.

We wish to evaluate the following expression:

   sin(2pi/7) + sin(4pi/7) + sin(8pi/7)

Let k = cos(2pi/7) + i.sin(2pi/7). We form the quadratic with these roots:

   k + k^2 + k^4 and k^3 + k^5 + k^6

The sum of these roots is

   (1 + k + k^2 + k^3 + k^4 + k^5 + k^6) - 1

      1 - k^7
   = --------- - 1   
       1 - k
   = -1                                            (since k^7 = 1)

The product of these roots is

   k^4 + k^5 + k^6 + 3k^7 + k^8 + k^9 + k^10

   = k^4 + k^5 + k^6 + 3k^7 + k + k^2 + k^3        and k^7 = 1

   = 1 + k + k^2 + k^3 + k^4 + k^5 + k^6 + 2

   = (1 - k^7)/(1 - k) + 2

   = 2

So our quadratic is

   x^2 + x + 2 = 0

Its solutions are

    -1 +/- sqrt(1 - 8)
   -------------------  =  -1/2 +/- i.sqrt(7)/2

But one root is k + k^2 + k^4  = cos(2pi/7) + i.sin(2pi/7) 
                               + cos(4pi/7) + i.sin(4pi/7)
                               + cos(8pi/7) + i.sin(8pi/7)

From this, we see that 

    sin(2pi/7) + sin(4pi/7) + sin(8pi/7) = sqrt(7)/2   <---------

We take the positive sign because sin(8pi/7) = -sin(pi/7) and this is less
than sin(2pi/7).

- Doctor Anthony, The Math Forum 

Date: 05/22/2011 at 17:16:03
From: Rattanjeet
Subject: Trigonometric Identities

Dr. Anthony:

Thanks a lot for such an elaborate working of the problem. But I do not
understand where you get this from:

   k + k^2 + k^4 and k^3 + k^5 + k^6

Nor do I understand how and why you declare:

   Let k = cos(2pi/7) + i.sin(2pi/7)

I have solved quite a number of questions relating to trigonometric
identities, always using the double angle formulas. But your type of
approach has not been involved in any of those; it is quite different.

Can my problem be solved using double angle formulas?

Please clarify.


Date: 05/22/2011 at 19:26:01
From: Doctor Anthony
Subject: Re: Trigonometric Identities

Double angle formulae and similar manipulations are OK if you want to end up with 
another trig expression. But in this case, we want an actual numerical value, so the 
approach has to be different.

I used de Moivre's theorem with ...

   k = cos(2pi/7) + i.sin(2pi/7) 

... because the k^2 and k^4 are immediately 

   cos(4pi/7) + i.sin(4pi/7)

... and so on.

Similarly, I chose

   1 + k + k^2 + k^3 + .. + k^6

... as this sum is 

   (1 - k^7)/(1 - k) = 0

... thanks to k^7 = 1. We then easily find a quadratic with roots that
include the desired expression and its sum.

While indirect, this method is standard for such specialized types of trig

- Doctor Anthony, The Math Forum 

Date: 05/22/2011 at 21:24:40
From: Doctor Rick
Subject: Re: Trigonometric Identities

Hi, Rattanjeet.

I don't mean to interefere in your discussion with Doctor Anthony; he has
much better ideas on this than I would! I'd just like to emphasize one
point: this is NOT an IDENTITY.

An identity is an equation that is true for ALL values of the variable.
Typically, you are asked to PROVE the identity, which is essentially a

This is an EQUATION. It seems a bit different from the usual equation
because you aren't asked to FIND the solutions, only to CONFIRM that 
A = 2pi/7 is a solution of the equation.

You might be able to use trig identities in the process of confirming the
solution, but you also need to make use of the specific value of A at some
point in your work. 

For example, it might be possible to use the fact that

  sin(4A) = sin(8pi/7) = -sin(6pi/7) = -sin(3A)

Now, this equivalence may ultimately prove to be of no value. It's just
an example of the sort of strategy to consider if you want to try a method
different from Doctor Anthony's.

- Doctor Rick, The Math Forum 
Associated Topics:
High School Trigonometry

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