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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. ThenA. a is algebraic but b is not algebraicB. b is algebraic but a is not algebraicC. both a and b are algebraicD. neither a nor b is algebraic |
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Answer» Correct Answer - C `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` |
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