Deriving Sines of 30, 45, 60 and 90 DegreesDate: 06/02/99 at 09:41:14 From: Michael Subject: Where does the value of sin(pi/3) come from? To Dr. Math, Where do the values for sin 30, sin 60, etc. come from? We're told to memorize the values and take them for granted, but what is the actual derivation for getting the values? In other words, is there any way to prove that sin 30 = 1/2 or that sin 60 = sqrt(3)/2? Of course, I have the same question about cos, tan, etc, but if you explain sine I can figure those out. Thanks, Mike Date: 06/02/99 at 10:29:23 From: Doctor Rick Subject: Re: Where does the value of sin pi/3 come from? Hi, Mike, thanks for the question! Surely you weren't just told to take these facts for granted - they aren't hard at all to derive. Perhaps you were absent the day you were shown the reason for them. You didn't mention 45 degrees (sin 45 = pi/4). Do you understand why sin(pi/4) = sqrt(2)? I'll go over it, because this will lead into the angles you asked about. Consider a 45-45-90 triangle. Because 2 angles are equal, it is isosceles: both legs have the same length. If you know one leg (a), you know both legs and you can use the Pythagorean Theorem to find the hypotenuse c. /| / | c / | / | a / | / | /____________| a c = sqrt(a^2 + a^2) = sqrt(2*a^2) = sqrt(2)*a So these are the proportions of the sides of a 45-45-90 triangle: B /| / | sqrt(2) / | / | 1 / | / | /____________| A 1 C In other words, the ratio of the hypotenuse to either leg is sqrt(2):1. In terms of trigonometry, sin(45) = sin(A) = BC / AB = 1/sqrt(2) = sqrt(2)/2 Now for the 30-60-90 triangle. It isn't isosceles like the 45-45-90 triangle, but it is half of an equilateral triangle: /|\ / | \ c / | \ c / | \ / a| \ / | \ /______|______\ b b The angles of the equilateral triangle are 60 degrees. The top angle is bisected, giving the 30 degree angle. And since the sides of the equilateral triangle are equal, 2b = c. That's the key - the side opposite the 30 degree angle (b) is half the hypotenuse (c). If we know b, then we know c = 2b, and we can fill in the Pythagorean Theorem: (2b)^2 = a^2 + b^2 a = sqrt((2b)^2 - b^2) = sqrt(3b^2) = sqrt(3)*b Now we have the proportions of the sides of a 30-60-90 triangle: B /| / | / | 2 / | / | sqrt(3) / | /______| A 1 C As before, we can evaluate the sine of angle A, which is 60 degrees: sin(60) = BC / AB = sqrt(3)/2 Likewise, for angle B: sin(30) = AC / AB = 1/2 It's good to memorize those two proportion figures (or, if you're like me, remember how to derive them using Pythagoras). Each figure is a family of facts that will be useful for trigonometry and geometry. I think this is much better than just memorizing the trig facts in isolation; the figures tie them all together, and show how to derive them at the same time. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/