Sine Function and Unit CircleDate: 08/15/2003 at 10:05:38 From: Sean Subject: Trigonometry I have a problem I am just completely stuck on, and was wondering if you could help me solve it. Solve for x: sin(pi/x) = square root2/2 where x is an integer I have tried the question using formula sin^2 x=cos^2 x-cos2x but am having little luck. In my mind I can see it turning out to be a quadratic, but I'm not sure. Date: 08/15/2003 at 18:59:52 From: Doctor Warren Subject: Re: Trigonometry Hi Sean, Before setting out to algebraically manipulate your equation, you might want to spend some time getting re-acquainted with the sine function. In particular, you should try to recall the following four points: sin(pi/6) = 1/2 sin(pi/4) = sqrt(2)/2 sin(pi/3) = sqrt(3)/2 sin(pi/2) = 1 Remember that if you draw a one-unit-long line at an angle theta from the positive x-axis, measured counterclockwise, the y-coordinate of the endpoint is the sine of the angle. Take a look at this diagram of the unit circle: | | y=1 | sin(pi/2) = 1 sin(2pi/3) = sqrt(3/2) _..--+--.._ sin(pi/3) = sqrt(3/2) .-' | `-. sin(3pi/4) = sqrt(2)/2 ,' | `. sin(pi/4) = sqrt(2)/2 ,' | `. sin(5pi/6) = 1/2 / | \ sin(pi/6): y = 1/2 / | \ . | . sin(pi) = 0 | | | sin(0) = 0 ---+--------------+--------------+--- | | | . | . \ | / sin(7pi/6) = -1/2 \ | / sin(11pi/6) = -1/2 `. | .' sin(5pi/4) = -sqrt(2)/2 `. | .' sin(7pi/4) = -sqrt(2)/2 `-._ | _.-' sin(4pi/3) = -sqrt(3)/2 ''--+--'' sin(7pi/6) = -sqrt(3)/2 y=-1 | sin(3pi/2) = -1 | | In every case, the sine of each angle shown is the value of the y-coordinate at that point on the unit circle. The four values I showed you above get repeated again in each quadrant, and the pattern is really pretty easy to remember. For example, if I want to know the sine of (7 pi/4), I do the following: 1. I reduce the fraction as much as possible. If I'm given 6pi/4, for example, I immediately reduce it to 3pi/2. In this case, I cannot reduce 7pi/4. 2. I think about where 7pi/4 is on the unit circle. It's all the way around on the bottom right of the unit circle, beyond 3pi/2. In fact, 7pi/4 is an angle of pi/4 beyond 3pi/2, which is of course the same as 6pi/4. Just think of this angle as a "quarter-pi" angle, some multiple of pi/4. 3. I know that sines of quarter-pi angles are either sqrt(2)/2 or -sqrt(2)/2. (Except of course where the fraction can be reduced to a multiple of pi or pi/2, which you should have already done at step 1.) 4. If the angle puts me on the bottom half of the circle, I pick the negative choice. If the angle puts me on the top of the circle, I pick the positive one. In this case, the angle 7pi/4 is on the bottom, so I'll pick the negative one. Thus sin(7pi/4) = -sqrt(2)/2. Your question is: Solve sin(pi/x) = sqrt(2)/2 for x. You should recognize the right-hand side, sqrt(2)/2, as one of those convenient quantities that the sine function spits out for quarter-pi angles. Which quarter-pi angles? There are two whose sines are positive sqrt(2)/2: sin(pi/4) sin(3pi/4) Your equation therefore has two solutions. Can you see what they are? The second one is a little tricky, so I'll give you a hint: pi b pi ------- = ------ (a/b) a Let me know if you need any more help! - Doctor Warren, 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/