Exact Values of the Sine and Cosine Functions in Increments of 3 degrees
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1 Exact Values of the Sine and Cosine Functions in Increments of 3 degrees The sine and cosine values for all angle measurements in multiples of 3 degrees can be determined exactly, represented in terms of square-root radicals, and the four common operations of arithmetic. From the degree triangle, set the hypotenuse to 1, and use the Pythagorean formula to determine the legs, which are both 2 2. Using the right-triangle constructions sin= and cos=, we have sin45 = 2 2 and cos45 = 2 2. From the equilateral ( ) triangle with all legs of length 1, we can view half the triangle, forming a triangle, with the hypotenuse measuring 1, the short leg 1/2, and the long leg 3 2. Therefore, we have sin30 =1/2, cos30 = 3 2, sin60 = 3 2 and cos60 =1/2. From the isosceles triangle, let the two long sides be, and the short side 1. Then, bisect one of the 72 angles, extending the ray to intersect the other side of the triangle. This forms two triangles: an isosceles triangle whose short sides are both 1, and a smaller isosceles triangle whose long sides are 1, and short side is 1, as shown in the diagrams below: The two triangles are proportional: The ratio of the long sides :1 is the same proportion as the ratio of the short sides: 1:( 1). Therefore, we solve for by equating two ratios. We have =, which gives =1 after cross-multiplying. Solving for x using the quadratic formula (and ignoring the negative root), we have =(1+ 5) 2. Splitting the triangle in half, we now have a right triangle, a triangle, with hypotenuse 1, long leg /2 and short leg 1 ( 2). Since =(1+ 5) 2, the long leg of the triangle is (1+ 5) 4 and the short leg is
2 Therefore, we can state the following: sin36 = = = ; cos36 = = ( ) =. The remaining exact representations for angles of multiples of 3 can now be found, using sum-difference or half-angle identities. For example, sin6 =2sin3 cos3, and so on. Since different identities may be used at various steps, the exact representations may look different than other possible representations, but can be shown to be identical in value. The following is a table of all exact values for the sine and cosine of angles of multiples of 3, up through 45. All radicals were simplified so that none contained any quotients within them. Angle 0 degrees 0 radians degrees radians degrees radians degrees radians degrees radians degrees radians 2 3 or or degrees radians
3 Angle 21 degrees radians degrees radians degrees radians degrees radians 3 33 degrees radians degrees radians degrees radians degrees radians degrees radians 2 2 For sine and cosine measurements above 45 radians, use the identities cos=sin(90 ) and sin=cos(90 ). For example, sin48 =cos42, and so on. All of these values are algebraic numbers, meaning they are the root of some polynomial with rational coefficients. For example, sin90 is algebraic. It equals 1, which is the root of the polynomial 1. Given an algebraic number in the form of radicals and arithmetic operations such as those listed in the
4 above table, one can build a polynomial for which the given number is a root. For example, if = 2, then squaring both sides and collecting terms to one side, we get =0, and so 2 1 is a polynomial for which 2 is a root. Building such polynomials is easy, but tedious. For example, sin9 = , and thus is a root of the polynomial Interestingly, this polynomial also has cos9 as a root, too. Is that surprising? Or not? The Interesting Case of Sin 10 and the Polynomial += The expression sin(3) can be written as 4sin +3sin by observing that sin(3)=sin(2+) and using the sum identities. If we let =10, we have sin(30 )= 4sin (10 )+3sin(10 ), or = 4sin (10 )+3sin(10 ). Multiplying by 2, we have 1= 8sin (10 )+6sin(10 ). Thus, the polynomial 8 6+1=0 has sin (10 ) as a root, and thus sin (10 ) is algebraic. Graphing this polynomial and using the zero feature, the three roots of this polynomial are (rounded) = ,= and = A calculator also shows that sin(10 )= What are these other values? Since sin (10 ) is a root of 8 6+1=0, we factor this polynomial by dividing by ( sin(10 )) using synthetic division. The first pass through gives 8 6+1=( sin(10 ))8 +(8sin(10 ))+(8sin (10 ) 6). The remaining quadratic can be solved using the quadratic formula: After considerable simplification, we get = 8sin(10 )±(8sin(10 )) 4(8)(8sin (10 ) 6). 2(8) = 1 2 sin(10 )± 3cos(10 ). A calculator verifies that sin(10 )+ 3cos(10 ) , and that sin(10 ) 3cos(10 )) We can also try factoring this cubic directly. This is a depressed cubic (missing its quadratic term). It is known that the closed form solution will consist of cube roots of complex numbers, even though the answer is clearly real. Nevertheless, it s interesting to explore this avenue. The methods here date back to
5 Tartaglia, del Ferro, Cardan and his contemporaries of the th - and 17 th -Centuries. First, let =, where r is a temporary variable and k is some constant to be determined: =0. Expanding, we have Note what happens when = 1 4: ) 8 24( 1 4)+ 24( 1 4 Multiplying through by 8, we have =0. 8 ( 1 4) 6+ 6( 1 4) +1= = =0. This is quadratic if we treat =( ). Using the quadratic formula, we have = 8±8 4(1)(64) 2(64) = 8± =0. = 1± 3. Thus, Since =+, we have = 1± 3 = 1± ± 3. A calculator shows that Interestingly, the value of is also Thus, it is clear (in a manner of speaking) that = sin(10 ) 3cos(10 ).
6 Determining the cube roots of complex numbers requires more wrestling with demoivre s Theorem. Let s move on. What About the Sine of 1 degree? The following construction results in a general way to illustrate the value of sin. Start with an isosceles triangle with two sides of length 1, and the remaining side of length. Let the angle opposite be (2). In the above diagram, continue the construction as follows: Draw segment such that its length is also, then draw segment such that it meets segment at a right angle. We can now label the various lengths as follows: =cos(3) and =sin(3), so therefore, =1 sin(3). The Pythagorean formula gives the length of =1+ 2sin(3). In turn, the length =1 1+ 2sin(3). This is shown in the following figure: Now, drop a perpendicular from to segment, and also a perpendicular from to segment. In doing so, we have split the angle measurement (2) into. Importantly, note that triangles and are proportional. We can now define sin in two ways using the opposite over hypotenuse construction for right angles: sin = 2 and 2sin(3) sin =
7 Relating the two expressions, we have: sin(3) 2 This simplifies to =1 1+ 2sin(3). After squaring away the radical, the equation becomes = sin(3)=0 Since =0 produces a trivial case, we ignore it and divide out by : 3+2sin(3)=0 This cubic polynomial has three roots. Let be the positive root of this polynomial that is closest to 0. Therefore, sin( )=, or =2sin( ). For example, to find sin1, let =1 and we get have 3+2sin(3 )=0. A calculator shows that = is a root of this polynomial. Therefore, = , which is also confirmed via calculator. sin1 =. Prepared by Scott Surgent (surgent@asu.edu) Please report errors to me if you see one. Updated Thanks to Martin Muccarione at Linear Corp. in Carlsbad CA for useful insights, which I have included here.
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