Date: 11/19/2014 at 23:45:38 From: Mary Subject: trig I'm trying to use trigonometric identities such as double and half angle sums to find sin(100). So far, none with 30, 45, or 60 degrees has gotten me to 100 degrees. Of course, I can find sin(100) on the calculator, but I want an EXACT value using the identities.
Date: 11/20/2014 at 15:41:46 From: Doctor Carter Subject: Re: trig Dear Mary - I'm unsure of your background, so I will keep this answer as short as I can. I'll begin by deriving the triple angle formula for sine: sin(3x) = sin(2x + x) = sin(2x) cos(x) + cos(2x) sin(x) = 2 sin(x) (cos(x))^2 + (cos(x))^2 sin(x) - (sin(x))^3 = 3 (cos(x))^2 sin(x) - (sin(x))^3 = 3 sin(x) - 4 (sin(x))^3 It follows that 4 (sin(x))^3 - 3 sin(x) + sin(3x) = 0 If x = 100 deg, then 3x = 300 deg = -60 deg. So sin(3x) = -sqrt(3)/2. Writing s for sin(100 deg), we conclude that s is a root of the cubic equation 4 s^3 - 3 s - sqrt(3)/2 = 0. This equation has one irrational coefficient. With just a little more work, (which I'll omit in this response), it can be shown that sin(100 deg) is also a root of the sixth degree equation with rational coefficients 16 s^6 - 24 s^4 + 9 s^2 - 3/4 = 0 It can further be shown (but this requires some advanced techniques) that sin(100 deg) is not a root of any polynomial of smaller degree that has rational coefficients. Because the degree of this polynomial (6) is not a power of 2, sin(100 deg) is not a constructible length, and therefore there is no ruler and compass construction of a 100 degree angle. Also, because the degree of this polynomial is not a power of 2, it follows that sin(100 deg) cannot be expressed using just rational numbers and square roots of rational numbers. So we can't expect sin(100 deg) to be expressed simply in terms of sin(15 deg), sin(30 deg), sin(45 deg), sin(60 deg), sin(75 deg), etc. Bottom line: sin(100 deg) is one of the roots of the irreducible 6th degree equation with rational coefficients given above. It is the unique positive root of the equation 4 s^3 - 3 s - sqrt(3)/2 = 0. This root is approximately 0.984807753012208 - Doctor Carter, The Math Forum http://mathforum.org/dr.math/
Date: 11/21/2014 at 00:18:14 From: Mary Subject: Thank you (trig) Thank you so much. This answers my question, which came up when I was helping a fellow student. At first, I thought it must be a typo. After trying, I was pretty sure sin(100) was impossible, but didn't have a way to prove it. I really liked your derivation of a triple angle. I will keep that and it might come in handy another time. I wonder why they don't teach us that. Thanks again. Your answer confirms my suspicions.
Date: 11/21/2014 at 06:29:57 From: Doctor Carter Subject: Re: Thank you (trig) Glad I could help! - Doctor Carter, The Math Forum http://mathforum.org/dr.math/
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