Let a = cos 1^{circ} and b = sin 1^{circ}. We | vedclass

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(C) We know that $\cos(n	heta) + i \sin(n	heta) = (\cos 	heta + i \sin 	heta)^n$. For $	heta = 1^{\circ}$ and $n = 90$,we have $\cos 90^{\circ} +

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both $a$ and $b$ are algebraic

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(C) We know that $\cos(n	heta) + i \sin(n	heta) = (\cos 	heta + i \sin 	heta)^n$. For $	heta = 1^{\circ}$ and $n = 90$,we have $\cos 90^{\circ} + i \sin 90^{\circ} = (\cos 1^{\circ} + i \sin 1^{\circ})^{90} = i$. This implies that $\cos 1^{\circ} + i \sin 1^{\circ}$ is a root of the polynomial $z^{90} - i = 0$. Since $a + ib$ is a root of $z^{90} - i = 0$,it is an algebraic number. Any complex number $z = a + ib$ is algebraic if and only if both its real part $a$ and imaginary part $b$ are algebraic. Since $\cos 1^{\circ} + i \sin 1^{\circ}$ is a root of $z^{90} = i$,then $z^{180} = -1$,or $z^{180} + 1 = 0$. Thus,$\cos 1^{\circ} + i \sin 1^{\circ}$ is an algebraic number. It is a known result that if $\alpha$ is an algebraic number,then $\cos \alpha$ and $\sin \alpha$ are algebraic for rational multiples of $\pi$. Since $1^{\circ} = \frac{\pi}{180}$ radians,both $\cos 1^{\circ}$ and $\sin 1^{\circ}$ are algebraic numbers.

Let $a = \cos 1^{\circ}$ and $b = \sin 1^{\circ}$. We say that a real number is algebraic if it is a root of a polynomial with integer coefficients. Then,

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