Let `a=cos 1^(@)` and `b= sin 1^(@)`. We say that a real number is algebraic if is a root of a polynomial with integer coefficients. Then

Options

A. a is algebraic but b is not algebraic
B. b is algebraic but a is not algebraic
C. both a and b are algebraic
D. neither a nor b is algebraic

Answer

C

Explanation

`cos x =(e^(ix)+e^(-ix))/(2)`
`cos 60x =(e^(60ix)+e^(-60ix))/(2)`
`cos 60x =((cos s + i sin x)^(60)+(cos x-i sin x)^(60))/(2)`
`(1)/(2)=((cos^(60)+1^(@)+60c_(2) cos^(58) i sin^(2)1+.......oo)^(60)c_(60)sin^(60)1^(@))/(2)`
Change all `sin 1^(@)` to `cos 1^(@)` using the identity `sin^(2)1^(@)= 1 cos^(2) 1^(@)` equation with root `cos 1^(@)` so it is algebraic . Similarly for `b = sin 1` also algebric .
let a poly. `-x^(2)+1 cos^(2) 1^(@)=0`
`-x^(2)+1= - sin^(2) 1^(@)=0`