Cannotfind7 Posted November 9, 2010 Share Posted November 9, 2010 Ok so to start out, I was given a puzzle to solve under the conditions that i could not solve it by using a scientific calculator. So hitting Sin(x) is not an option. I was started out with a circle as a base. The circle has a Diameter of 6.5 meters and an idea of putting 80 points along the edge of the circle with the same distance between each of them. I need to find the distance that is between these points, Not the arch curve but the straight line between them. best way i thought of how to do that was to make a triangle, using the radius for 2 of the sides and finding the link the third and final side through law of sines and law of cosines. Side a=3.25 Side b=3.25 Side c= X Angle A= 87.75 Angle B= 87.75 Angle C= 4.5 Sin(a)/A = Sin©/C Sin©= (C * Sin(a))/A Sin(x)= (4.5 * Sin(3.25)) /87.75 Now that i have so far, I could solve it if I knew how to solve " Sin(3.25) " by hand I would be set. But I was only taught on how to plug it in to the calculator. How do you guys solve For Sin(3.25) by hand. Diagrams or links would be helpful so I can better understand how to solve for sin by hand. Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted November 9, 2010 Share Posted November 9, 2010 You could do it by modeling it as an 80-sided polygon rather than as a circle. I believe there are equations for determining the side length of an arbitrarily large regular polygon, aren't there? Link to comment Share on other sites More sharing options...

Bignose Posted November 9, 2010 Share Posted November 9, 2010 (edited) It is pretty much the equivalent of a calculator, but using a chart of values of sin would not technically be a calculator. This is essentially what's been done since the discovery of the trig functions -- tabulate their values and look up and interpolate as needed. You may be able to use the double angle or half angle formulas to get to known values, too. Also, depending on the accuracy needed, you could approximate sin by its series representation: [math]\sin x \approx x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - ... [/math] though, solving a polynomial may not be significantly easier. Edited November 9, 2010 by Bignose Link to comment Share on other sites More sharing options...

imatfaal Posted November 9, 2010 Share Posted November 9, 2010 Cannotfind First off - your Sine Law calcs were getting a bit crazy, I didnt check them but you were taking sin(x) and x is a side length and not taking sin of 87.75 which is an angle (you might want to look through to make sure you are happy with sine law stuff) Secondly - the problem is more simple than sine law. you have an isoceles triangle with two long equal sides of length R and an angle of theta in between two long sides. Drop a vertical splitting theta and making perpendicular with short side; you now have rightangled triangle with known angle (theta/2) and known hypoteneuse (radius). Sin(theta/2) = o/h = Unknown/Radius Radius.sin(theta/2) = Unknown the length of your desired line is of course twice the unknown in the right-angled triangle Answer = 2.radius.Sin(theta/2) Thirdly - I think you can use small angle approximation for a problem like this. the small angle approximation states that for small enough angles in radians Sin(theta) tends to theta. Your angle is theta 2.pi/(80.2) = pi/80. This is still a pretty vile long division but quite doable with a pen,paper and patience. I make the error through using small angle approximation about 2 parts in 100,000 Link to comment Share on other sites More sharing options...

John Cuthber Posted November 9, 2010 Share Posted November 9, 2010 Dividing pi by 80 would be a bit tedious. Multiplying it by 125 then moving the decimal point to divide by 1000 is a lot easier. Or you could halve it 3 times then shift the decimal point one across. (not a lot less effort- but you only need to know your two times table.) In the days before calculators, people were expected to know about that sort of thing and there were tables of reciprocals alongside the log tables etc. As far as I can see this "I think you can use small angle approximation for a problem like this. the small angle approximation states that for small enough angles in radians Sin(theta) tends to theta." is the same as saying that the curve round the circle is the same length as the straight line. That's not allowed because of this requirement " Not the arch curve but the straight line between them." So you need to use at least the first few terms in the polynomial expansion otherwise you are cheating. Don't forget to convert the angle into radians (which brings us back to what's pi/80?). Link to comment Share on other sites More sharing options...

imatfaal Posted November 10, 2010 Share Posted November 10, 2010 John - of course you are right that question said "Not the arch curve but the straight line between them." Even now I tend to read the question too quickly - but I still struggle to see how it could easily be done otherwise. The Taylor Series would be accurate to 1 in 10^9 by the second term because of the smallness of the angle. But even restricting to two terms cubing pi/80 (to 7 dp or more) is an awful job. There has to be a trick that I am missing. I hope the OP posts the intended solution method Link to comment Share on other sites More sharing options...

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