How many angles in positive integer degrees up to a full circle have sines in radicals?
This problem's question is: How many angles in positive integer degrees up to a full circle have sines in radicals?
By that, the value of the sine of those angles must be expressible using algebraic numbers.
Hints: what angles are constructable as those angles have sines (and cosines, for that matter) that can be expressed in radicals? If a sine of an angle is expressible in radicals, then the cosine of that angle is also.Angles can be bisected but not trisected. Formulas exists for computing the sines and cosines of of sums and differences of angles that if used on algebraic numbers return algebraic numbers. The answer has to be between 1 and 360 inclusive as the problem has been given so those are only permissible integers.
The answer is 120.
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1 solution
See Constructible polygon . The largest integer divisor of 360 whose factors are powers of 2, 3 and 5 as those primes are the factors of 360 is 120. Q.E.D. The sine of 3 degrees can be constructed by computing a quarter of the angle of (72 degrees minus 60 degrees) giving 3 degrees.
The sines and cosines of the respective angles:
3 6 9 1 2 1 5 1 8 2 1 2 4 2 7 3 0 ⎩ ⎨ ⎧ 1 6 1 ( 2 ( 3 + 1 ) ( 5 − 1 ) − 2 ( 3 − 1 ) 5 + 5 ) , 2 1 2 1 5 + 6 ( 5 + 5 ) + 7 + 2 ⎭ ⎬ ⎫ { 8 1 ( − 5 + 3 0 − 6 5 − 1 ) , 4 1 5 + 6 ( 5 + 5 ) + 7 } { 4 1 8 − 2 2 ( 5 + 5 ) , 2 1 2 1 ( 5 + 5 ) + 2 } { 4 1 − 5 − 3 0 − 6 5 + 7 , 8 1 ( 5 + 6 ( 5 + 5 ) − 1 ) } { 2 2 − 3 , 2 3 + 2 } { 4 1 ( 5 − 1 ) , 2 1 2 1 ( 5 + 5 ) } ⎩ ⎨ ⎧ 2 − − 5 + 3 0 − 6 5 + 7 − 4 2 1 , 2 1 2 1 − 5 + 6 ( 5 − 5 ) + 7 + 2 ⎭ ⎬ ⎫ { 4 1 5 − 6 ( 5 + 5 ) + 7 , 8 1 ( 5 + 3 0 − 6 5 + 1 ) } { 2 1 2 − 2 1 ( 5 − 5 ) , 2 1 2 1 ( 5 − 5 ) + 2 } { 2 1 , 2 3 }