ANSWERS TO MATH 75 CALCULUS HOMEWORK #2, by WARREN SMITH -------------------------------------------------------- PROBLEM 1. To demonstrate your mastery of manipulating fraction formulas, a/b + c/d polynomial simplify --------- to ----------------, i.e. to "rational" form. e/f - g/h other polynomial Here by "polynomial" I mean an expression involving +,- and multiplication, but not division. (a d + c b) f h f h a d + f h c b 2 ANSWERS: --------------- = ----------------- (e h - g f) b d e h b d - g f b d Method: the top is (we put them both over a common denominator bd by multiplying by d/d=1 and by b/b=1 respectively, then add). a d + c b a/b + c/d = --------- b d and the bottom is e h - g f e/f - g/h = --------- f h and now divide these. [Sanity check: try some numerical values in both the answer expression and the problem expression, and you ought to get the same numerical answer.] PROBLEM 2. Consider these 3 equations in 3 unknowns x,y,z: x + 5y + 7z = 10 3x + 2y - 3z = 3 4x - y = 2 What is the only possible value for z (as an exact fraction, such as 99/872, please)? ANSWER: z = 73/140 (and x=109/140, y=39/35=156/140; you can do a sanity check by substituting these values for x,y,z into our equations and checking that they work). SOME OTHER PEOPLE'S METHODS: Cramer's rule, involving determinants of matrices, is one method some may know. Elena S. Dosawat thought of using it: 1 5 10 1 5 7 z = det 3 2 3 / det 3 2 -3 = (-73) / (-140)= 73/140. 4 -1 2 4 -1 0 Some people may know other matrix methods, most of which are really just shorthand forms of the method I will now describe. [For example Zack Bates did matrix row operations to simplify and solve our set of 3 equations. That is actually a better way to do the computation, in the sense that you use less paper, than the method I will now describe, but you have to know the correspondence between the matrix rows and the equations.] MY METHOD: The method I like best is to "eliminate" one unknown at a time, until you are left with a single equation in a single unknown (namely the one we want, namely z). So. Use the last equation 4x - y = 2 to solve for y=4x-2 in terms of the others ("others" meaning x and z, but here, luckily, only x is needed), then substitute this expression for y into the other 2 equations, thus "eliminating" y: x + 5(4x-2) + 7z = 10 3x + 2(4x-2) - 3z = 3 i.e. (simplifying these) 21x + 7z = 20 11x - 3z = 7 now solve the first of these for x: x = (20-7z)/21 and substitute that expression for x into the 2nd equation (thus "eliminating" x): 11(20-7z)/21 - 3z = 7 which is, as planned, a single equation in a single unknown (namely z). since x and y have both been eliminated. We solve it for z to get z = 73/140. PROBLEM 3. The "addition formula" for cos is cos(a+b) = cos(a) cos(b) - sin(a) sin(b). Write down the "addition formula" for sin(a+b) in terms of sin(a), sin(b), cos(a), cos(b). Use these two to derive a formula for tan(a+b) in terms of tan(a) and tan(b). (Remember definition of tan(x) = sin(x)/cos(x). You might want to try some sanity checking...) ANSWER: The addition formula for sin (which I did in class, and also is in the book - the "condensed summary" starting page 582 in the book is about the only part of the book I like much...) is sin(a+b) = sin(a) cos(b) + cos(a) sin(b) now since the definition of tan(x) = sin(x)/cos(x) we have sin(a+b) sin(a) cos(b) + cos(a) sin(b) tan(a+b) = -------- = ----------------------------- cos(a+b) cos(a) cos(b) - sin(a) sin(b) and now the question is, how to rewrite this so it is only in terms of tan(a) and tan(b)? Multiply both top and bottom by 1/(cos(a)cos(b)), in an effort (which works) to convert all the sin's into tan's, and to get rid of (by cancellation) all unwanted cos's. tan(a) + tan(b) ANSWER: tan(a+b) = ----------------- 1 - tan(a) tan(b) Incidentally, if we change b to -x everywhere in the above formula and use tan(-x) = -tan(x), we get tan(a) - tan(x) tan(a-x) = ----------------- 1 + tan(a) tan(x) which is also true. PROBLEM 4. Use the addition formula for sin(a+b) to derive a formula for sin(q/2) in terms of sin(q). ANSWER: sin(a+b) = sin(a) cos(b) + cos(a) sin(b) so use b=a to get sin(2a) = 2 sin(a) cos(a) the sin "double angle" formula. Now let q=2a to get sin(q) = 2 sin(q/2) cos(q/2) (*) now use (from Pythagoras & the defn of sin and cos) 2 2 sin(x) + cos(x) = 1 2 to see that cos(q/2) = sqrt( 1 - sin(q/2) ) (at least when cos(q/2) is nonnegative, i.e. when q is in [-180, 180] degrees) so that we may re-express (*) totally in terms of sin: 2 sin(q) = 2 sin(q/2) sqrt( 1 - sin(q/2) ) when q is in [-180, 180] degrees. Square both sides to get rid of the square root: 2 2 2 sin(q) = 4 sin(q/2) (1 - sin(q/2) ) 2 Note this is a quadratic equation for sin(q/2) . To save writing let s = sin(q/2). Then the equation is 2 2 2 sin(q) = 4 s (1 - s ) 2 and let t = s to save even more writing, so this equation is 2 sin(q) = 4 t (1-t) which is the quadratic equation for t 2 2 4 t - 4 t + sin(q) = 0 which we may solve for t using the quadratic formula to get: 2 1 +- sqrt( 1-sin(q) ) t = ------------------------ 2 and now remember that sin(q/2) = s = sqrt(t) if sin(q/2) is nonnegative so... the answer is 2 1 +- sqrt( 1-sin(q) ) sin(q/2) = sqrt( ---------------------- ) 2 As a sanity check, when q=0 we want sin(q)=sin(q/2)=0 so the +- sign MUST be a -: ANSWER: 2 1 - sqrt( 1-sin(q) ) sin(q/2) = sqrt( --------------------- ) 2 which works if q in [0,90] degrees. (A sin angle-halving formula.) [Incidentally, the reason these only work in certain angle ranges (0 to 90 works) is that really, some of these square roots could have been negated, and the other sign choice is the right one in some other angle ranges.] As another sanity check try q=90 degrees: sin(q)=1, sin(q/2)=sin(45 degrees)=1/sqrt(2); it works. It is also possible to rewrite this in these equivalent forms EQUIVALENT ANSWERS: 2 sin(q) / 2 sin(q/2) = sqrt( --------------------- ) 2 1 + sqrt( 1-sin(q) ) 2 sin(q/2) = sin(q) / sqrt( 2 + 2 sqrt( 1-sin(q) ) (*) Note: it often is convenient, if you have some expression of the form A-sqrt(B), to multiply and divide it by A+sqrt(B) which shows that 2 A - B A - sqrt(B) = ----------- A + sqrt(B). In this way you can convert the - sign into a + sign or vice versa (the sqrt is maybe on the bottom rather than top of a fraction now, though). That is how I got these two equivalent formulas. Actually that was a little bit longer that I thought it was going to be, but every step in the process is something you should be able to do. OK, now, we will USE this formula: If the area A of a regular N-sided polygon, all of whose side lengths are equal, and all of whose vertices lie on the perimeter of the unit circle, is A=F(N), then what is F(N), as a formula involving N and trig functions? (For example, F(4) is the area of a square, all 4 of whose corners lie on the unit circle, i.e. all of these corners are at distance=1 from (0,0).) The polygon is the union of N isoceles triangles of angle 360/N degrees (or 2pi/N in radians) each, and legs 1,1. Such a triangle has base 1 and height sin(2pi/N) and hence area=sin(2pi/N)/2. So F(N) = N TriangleArea is ANSWER: N 2 pi F(N) = --- sin(----) 2 N Sanity check: try N=4, get 4 triangles, each of angle 90 degrees and legs 1,1, for TriangleArea=1/2 = sin(90 degrees)/2 and TotalArea = 4/2 = 2 = F(4), which works. Second sanity check: Try N=2 for a flat 2-sided "polygon" of area=0, and sure enough sin(pi)=0 so this sanity check also works. Now we can use our halving formula for sin, above, to see (substitute: sin(2 pi / N) = 2 F(N) / N sin(pi / N) = F(2N) / N ) 2 2 1 - sqrt( 1 - (2/N) F(N) ) F(2N) = N sqrt( --------------------------- ) 2 And I prefer the last equivalent form of the sin-halving formula (labeled *) the best, which gives 2 2 F(2N) = 2 F(N) / sqrt( 2 + 2 sqrt( 1 - (2/N) F(N) ). Of course you could have used the other equivalent answers to get other but equivalent formulas here, I just like this form the best since it has fewer subtractions, which helps reduce roundoff errors. Here is the result of starting with F(4)=2 and doubling: F(8) = sqrt(8) = 2.828427124 F(16) = 3.1214452 = 8 / sqrt(4+sqrt(14)) 32 3.1365485 64 3.1403312 128 3.1412773 256 3.1415138 512 3.1415729 1024 3.1415877 which is getting closer and closer to pi=3.14159... as the polygon gets more and more sides, more closely approximating a perfect circle. A regular polygon with 1024 sides looks very near circular. Note each time we double N to 2N we can compute F(2N) from F(N) and N using ONLY +, -, multiply, divide, and square roots; no trig functions are needed. So we can get arbitrarily good approximations to pi in this way. Archimedes found the area of a regular 92-gon by doubling 3,6,12,24,48,92. Looks like some of those ancient Greeks were not so dumb!