let k be Nat; :: thesis: sin (#) cos is PI * (k + 1) -periodic
defpred S1[ Nat] means sin (#) cos is PI * (\$1 + 1) -periodic ;
A1: S1[ 0 ] by Lm24;
A2: for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A3: sin (#) cos is PI * (k + 1) -periodic ; :: thesis: S1[k + 1]
sin (#) cos is PI * ((k + 1) + 1) -periodic
proof
for x being Real st x in dom () holds
( x + (PI * ((k + 1) + 1)) in dom () & x - (PI * ((k + 1) + 1)) in dom () & () . x = () . (x + (PI * ((k + 1) + 1))) )
proof
let x be Real; :: thesis: ( x in dom () implies ( x + (PI * ((k + 1) + 1)) in dom () & x - (PI * ((k + 1) + 1)) in dom () & () . x = () . (x + (PI * ((k + 1) + 1))) ) )
assume A4: x in dom () ; :: thesis: ( x + (PI * ((k + 1) + 1)) in dom () & x - (PI * ((k + 1) + 1)) in dom () & () . x = () . (x + (PI * ((k + 1) + 1))) )
then A5: ( x + (PI * (k + 1)) in dom () & x - (PI * (k + 1)) in dom () ) by ;
A6: ( x + (PI * ((k + 1) + 1)) in () /\ () & x - (PI * ((k + 1) + 1)) in () /\ () ) by ;
then ( x + (PI * ((k + 1) + 1)) in dom () & x - (PI * ((k + 1) + 1)) in dom () ) by VALUED_1:def 4;
then () . (x + (PI * ((k + 1) + 1))) = (sin . ((x + (PI * (k + 1))) + PI)) * (cos . ((x + (PI * (k + 1))) + PI)) by VALUED_1:def 4
.= (- (sin . (x + (PI * (k + 1))))) * (cos . ((x + (PI * (k + 1))) + PI)) by SIN_COS:78
.= (- (sin . (x + (PI * (k + 1))))) * (- (cos . (x + (PI * (k + 1))))) by SIN_COS:78
.= (sin . (x + (PI * (k + 1)))) * (cos . (x + (PI * (k + 1))))
.= () . (x + (PI * (k + 1))) by ;
hence ( x + (PI * ((k + 1) + 1)) in dom () & x - (PI * ((k + 1) + 1)) in dom () & () . x = () . (x + (PI * ((k + 1) + 1))) ) by ; :: thesis: verum
end;
hence sin (#) cos is PI * ((k + 1) + 1) -periodic by Th1; :: thesis: verum
end;
hence S1[k + 1] ; :: thesis: verum
end;
for n being Nat holds S1[n] from NAT_1:sch 2(A1, A2);
hence sin (#) cos is PI * (k + 1) -periodic ; :: thesis: verum