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 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 http://mathforum.org/dr.math/
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. Regards. Rattanjeet
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 problems. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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 theorem. 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 http://mathforum.org/dr.math/pre
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