Question Number 79062 by mr W last updated on 22/Jan/20
$${if}\:{a}\:{is}\:{a}\:{rational}\:{number}\:{with}\:\mid{a}\mid\leqslant\mathrm{1}, \\ $$$${prove}\:{that}\:\mathrm{cos}\:\left({n}\:\mathrm{cos}^{−\mathrm{1}} \left({a}\right)\right)\:{is}\:{also}\:{a} \\ $$$${rational}\:{number}.\:\left({n}\in\mathbb{N}\right) \\ $$
Answered by mind is power last updated on 22/Jan/20
$${let}\:{U}_{{n}} ={cos}\left({ncos}^{−} \left({a}\right)\right) \\ $$$${U}_{\mathrm{0}} =\mathrm{1}\in\mathbb{Q} \\ $$$${U}_{\mathrm{1}} ={cos}\left({cos}^{−} \left({a}\right)\right)={a}\in\mathbb{Q} \\ $$$${suppose}\:,\forall{n}\geqslant\mathrm{1}\:\:{U}_{{n}} ,{U}_{{n}−\mathrm{1}} \in\mathbb{Q},{lets}\:{Show}\:{U}_{{n}+\mathrm{1}} \in\mathbb{Q} \\ $$$${we}\:{Will}\:{use}\:{Fort}\:{recursion}\: \\ $$$${U}_{{n}+\mathrm{1}} +{U}_{{n}−\mathrm{1}} ={cos}\left(\left({n}+\mathrm{1}\right){cos}^{−} \left({a}\right)\right)+{cos}\left(\left({n}−\mathrm{1}\right){cos}^{−} \left({a}\right)\right) \\ $$$$={cos}\left({ncos}^{−} \left({a}\right)+{cos}^{−} \left({a}\right)\right)+{cos}\left({ncos}^{−} \left({a}\right)−{cos}^{−} \left({a}\right)\right) \\ $$$$=\mathrm{2}{acos}\left({ncos}^{−} \left({a}\right)\right)=\mathrm{2}{aU}_{{n}} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} =\mathrm{2}{aU}_{{n}} −{U}_{{n}−\mathrm{1}} \\ $$$$\:{since}\:{U}_{{n}} ,{U}_{{n}−\mathrm{1}} ,{a}\in\mathbb{Q} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} =\mathrm{2}{aU}_{{n}} −{U}_{{n}−\mathrm{1}} \in\mathbb{Q} \\ $$
Commented by mr W last updated on 22/Jan/20
$${thank}\:{you}\:{sir}!\:{very}\:{clear}! \\ $$
Commented by mind is power last updated on 22/Jan/20
$${Withe}\:{pleasur}\:{Sir} \\ $$