Answer to Question #312753 in Algebra for deepu

Question

prove that sin (degree(m)) is an algebraic number of any integer m.

Answer 1

180^o= $\pi$ radian

m^o= m$\pi$/180

e^m$\pi$^i/180 = cos ( m$\pi$/180)+ i sin (m$\pi$ /180)

(e^m$\pi$^i/180)¹⁸⁰ = cos (m$\pi$) + i sin (m$\pi$)

(-1)^m =+1= -1

cos (m$\pi$/180) + i sin (m$\pi$/180)

cos (m$\pi$/180) - i sin (m$\pi$/180)

2cos (m$\pi$/180) is an algebraic integer

Hence, cos (m$\pi$/180) becomes an algebraic integer

i sin (m$\pi$/180) is an algebraic integer

sin (m$\pi$/180) also becomes an algebraic integer

$\therefore$ sin (degree(m)) is an algebraic number