Finding Values by HandDate: 08/26/98 at 09:30:41 From: Victor Subject: Exact value Hello there. I have a question here I would like to ask you. How can I find the exact value for the following without using a calculator? 1) sin 18 2) cos pi/16 3) tan^-1 (5/6) + tan^-1 (1/11) - tan^-1 (1/3) 4) sin 20 . sin 40 . sin 60 . sin 80 Thank you. Date: 08/26/98 at 17:32:57 From: Doctor Floor Subject: Re: Exact value Hi Victor, Thank you for sending in your questions! 1) sin(18) I will suppose you work in degrees here. It is easier to compute cos(72) = sin(18), because cos(2*72) = cos(3*72) (check!). Now we know that cos(2x) = 2(cos(x))^2-1 and we can derive that: cos(3x) = cos(x+2x) = cos(x)cos(2x)-sin(x)sin(2x) = cos(x)(2(cos(x))^2-1) - sin(x)2sin(x)cos(x) = cos(x)(2(cos(x))^2-1) - 2cos(x)(1-(cos(x))^2) = 4cos^3(x)-3cos(x) So when you take t = cosx, we can make an equation in t having t = cos(72) as root. Then cos(2x) = cos(3x) becomes: 2t^2 - 1 = 4t^3 - 3t 4t^3 - 2t^2 - 3t + 1 = 0 (t-1)(4t^2 + 2t - 1) = 0 4t^2 + 2t - 1 = 0 [we can factor out t-1 because cos(72) is not 1] t = -0.25 + 0.25*SQRT(5) or t = -0.25 - 0.25*SQRT(5) Since cos(72) is positive, this must mean that sin(18) = cos(72) = -0.25 + 0.25*SQRT(5) 2) cos(pi/16) I will suppose you work in radians here. For the answer to your question we will use that: cos(2x) = 2(cos(x))^2 - 1 2(cos(x))^2 = 1 + cos(2x) (cos(x))^2 = (1 + cos(2x))/2 = 0.5 + 0.5cos(2x) cos(x) = +- SQRT(0.5 + 0.5cos(2x)) [+-: plus or minus] cos(x) = SQRT(0.5 + 0.5cos(2x)) [since we will be working with x such that cos(x) is positive] So we can apply this to: cos(pi/16) = SQRT(0.5 + 0.5cos(pi/8)) = SQRT(0.5 + 0.5SQRT(0.5 + 0.5cos(pi/4))) = SQRT(0.5 + 0.5SQRT(0.5 + 0.25SQRT(2))) where I suppose you know that cos(pi/4) = 0.5SQRT(2), which you can derive from the 45-45-90 triangle. 3) arctan(5/6) + arctan(1/11) - arctan(1/3) Let arctan(5/6) = A, arctan(1/11) = B, arctan(1/3) = C 5/6 = tan(A) 1/11 = tan(B) 1/3 = tan(C) tan(A) + tan(B) 5/6 + 1/11 tan(A+B) = ------------------ = ----------------- 1 - tan(A).tan(B) 1 - (5/6)(1/11) multiply through by 66 55 + 6 61 tan(A+B) = ------ = ---- = 1 66 - 5 61 So A+B = 45 degrees or pi/4 radians. I will work in degrees: tan(45) - tan(C) tan(A+B-C) = tan(45-C) = ------------------ 1 + tan(45).tan(C) 1 - 1/3 3 - 1 2 tan(A+B-C) = --------- = ------- = --- = 1/2 1 + 1/3 3 + 1 4 and so tan(A+B-C) = 1/2 A + B - C = arctan(1/2) and we have: arctan(5/6) + arctan(1/11) - arctan(1/3) = arctan(1/2) This is the exact answer! 4. sin(20)sin(40)sin(60)sin(80) Again I will suppose you work in degrees. As with sin(18) I will work with cosines instead of sines. My goal is to calculate: sin(20)^2sin(40)^2sin(60)^2sin(80)^2 = sin(40)^2sin(80)^2sin(120)^2sin(160)^2 = [ sin(160)=sin(20) and sin(120) = sin(60) ] (1-cos(40)^2)(1-cos(80)^2)(1-cos(120)^2)(1-cos(160)^2) When we have that, the rest is easy. First I will use that for x = 40, 80, 120, 160 the following identity holds: cos(4x) = cos(5x) (Check!) We can work out cos(4x): cos(4x) = 2(cos(2x))^2 - 1 = 2(2(cos(x))^2 - 1)^2 - 1 = 8(cos(x))^4 - 8(cos(x))^2 + 1 We can also work out cos(5x) cos(5x) = cos(x+4x) = cos(x)cos(4x) - sin(x)sin(4x) = cos(x)cos(4x) - sin(x)2sin(2x)cos(2x) = cos(x)cos(4x) - sin(x)4sin(x)cos(x)(2(cos(x))^2-1) = cos(x)(8(cos(x))^4 - 8(cos(x))^2 + 1) - 4(1-(cos(x))^2)cos(x)(2(cos(x))^2-1) = 16(cos(x))^5 - 20(cos(x))^3 + 5cos(x) When we take t = cos(x), then cos(4x) = cos(5x) becomes: 8t^4 - 8t^2 + 1 = 16t^5 - 20 t^3 + 5t 16t^5 - 8t^4 - 20t^3 + 8t^2 + 5t - 1 = 0 This is a polynomial of degree 5, so it has 5 roots. I suggested 4 of them, t = cos(40), cos(80), cos(120) and cos(160). The fifth is very simple, namely t = cos(0) = 1. The fact that t = 1 is a root means that we can factor out t-1, because we don't need this one: 16t^5 - 8t^4 - 20t^3 + 8t^2 + 5t - 1 = 0 (t - 1)(16t^4 + + 8t^3 - 12 t^2 - 4t + 1) = 0 16t^4 + + 8t^3 - 12 t^2 - 4t + 1 = 0 t^4 + (1/2)t^3 - (3/4)t^2 - (1/4)t + 1/16 = 0 So we have a polynomial and we know its roots. But then we also know how to factorize it: (t-cos(40))(t-cos(80))(t-cos(120)(t-cos(160)) = t^4 + (1/2)t^3 - (3/4)t^2 - (1/4)t + 1/16 So when we take t = 1 we have: (1-cos(40))(1-cos(80))(1-cos(120)(1-cos(160)) = 9/16 And if we take t = -1 we have: (-1-cos(40))(-1-cos(80))(-1-cos(120)(-1-cos(160)) = 1/16 But then also: (1+cos(40))(1+cos(80))(1+cos(120)(1+cos(160)) = 1/16 And since (1-cos(40))(1+cos(40)) = (1-cos(40)^2) we have: (1-cos(40)^2)(1-cos(80)^2)(1-cos(120)^2)(1-cos(160)^2) = 9/256 And here we have what I suggested in the beginning. So this means that: sin(20)^2sin(40)^2sin(60)^2sin(80)^2 = 9/256 And because sin(20), sin(40), sin(60) and sin(80) are all positive, this means that: sin(20)sin(40)sin(60)sin(80) = 3/16 Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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