prove that sin (degree(1)) is an algebraic number
180o= "\\pi" radian
Let 1 be k
sin ko
ko= k"\\pi"/180
ek"\\pi"i/180 = cos ( k"\\pi"/180)+ isin (k"\\pi" /180)
(ek"\\pi"i/180)180 = cos (k"\\pi") + i sin (k"\\pi")
(-1)k =+1= -1
cos (k"\\pi"/180) + i sin (k"\\pi"/180)
cos (k"\\pi"/180) - i sin (k"\\pi"/180)
2cos (k"\\pi"/180) is an algebraic integer
cos (k"\\pi"/180) becomes an algebraic integer
i sin (k"\\pi"/180) is an algebraic integer
sin (k"\\pi"/180) is also an algebraic integer
"\\therefore" sin (degree(1)) is an algebraic number
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