if-a-is-a-rational-number-with-a-1-prove-that-cos-n-cos-1-a-is-also-a-rational-number-n-N-

Question Number 79062 by mr W last updated on 22/Jan/20

i f a i s a r a t i o n a l n u m b e r w i t h ∣ a ∣⩽ 1,

p r o v e t h a t \cos (n \cos^{- 1} (a)) i s a l s o a

r a t i o n a l n u m b e r . (n \in N)

Answered by mind is power last updated on 22/Jan/20

l e t U_{n} = c o s ({n c o s}^{-} (a))

U_{0} = 1 \in Q

U_{1} = c o s ({c o s}^{-} (a)) = a \in Q

s u p p o s e, \forall n ⩾ 1 U_{n}, U_{n - 1} \in Q, l e t s S h o w U_{n + 1} \in Q

w e W i l l u s e F o r t r e c u r s i o n

U_{n + 1} + U_{n - 1} = c o s ((n + 1) {c o s}^{-} (a)) + c o s ((n - 1) {c o s}^{-} (a))

= c o s ({n c o s}^{-} (a) + {c o s}^{-} (a)) + c o s ({n c o s}^{-} (a) - {c o s}^{-} (a))

= 2 a c o s ({n c o s}^{-} (a)) = 2 {a U}_{n}

\Rightarrow U_{n + 1} = 2 {a U}_{n} - U_{n - 1}

s i n c e U_{n}, U_{n - 1}, a \in Q

\Rightarrow U_{n + 1} = 2 {a U}_{n} - U_{n - 1} \in Q

Commented by mr W last updated on 22/Jan/20

t h a n k y o u s i r! v e r y c l e a r!

Commented by mind is power last updated on 22/Jan/20

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