let th be real number ; ( th in ].0 ,(PI / 2).[ implies cos . th > 0 )
assume that
A1:
th in ].0 ,(PI / 2).[
and
A2:
cos . th <= 0
; contradiction
cos | REAL is continuous
by Th72, FDIFF_1:33;
then A4:
cos | [.0 ,th.] is continuous
by FCONT_1:17;
A5:
0 < th
by A1, XXREAL_1:4;
A6:
th is Real
by XREAL_0:def 1;
( [.(cos . 0 ),(cos . th).] \/ [.(cos . th),(cos . 0 ).] = [.(cos . th),(cos . 0 ).] & 0 in [.(cos . th),(cos . 0 ).] )
by A2, Th33, XXREAL_1:1, XXREAL_1:222;
then
ex th2 being Real st
( th2 in [.0 ,th.] & cos . th2 = 0 )
by A4, A5, A6, Th27, FCONT_2:16;
then consider th2 being Real such that
A9:
th2 in [.0 ,th.]
and
0 < th
and
A10:
cos . th2 = 0
by A5;
A11:
0 <= th2
by A9, XXREAL_1:1;
A12:
th2 <= th
by A9, XXREAL_1:1;
A13:
th < PI / 2
by A1, XXREAL_1:4;
A14:
0 < th2
by A10, A11, Th33;
th2 < PI / 2
by A12, A13, XXREAL_0:2;
then A16:
th2 / 2 < (PI / 2) / 2
by XREAL_1:76;
PI in ].0 ,4.[
by Def32;
then
PI < 4
by XXREAL_1:4;
then
PI / 4 < 4 / 4
by XREAL_1:76;
then A20:
th2 / 2 < 1
by A16, XXREAL_0:2;
0 =
cos . ((th2 / 2) + (th2 / 2))
by A10
.=
((cos . (th2 / 2)) ^2 ) - ((sin . (th2 / 2)) * (sin . (th2 / 2)))
by Th79
.=
((cos . (th2 / 2)) - (sin . (th2 / 2))) * ((cos . (th2 / 2)) + (sin . (th2 / 2)))
;
then A22:
( (cos . (th2 / 2)) - (sin . (th2 / 2)) = 0 or (cos . (th2 / 2)) + (sin . (th2 / 2)) = 0 )
;
A23:
th2 / 2 in ].0 ,1.[
by A14, A20, XXREAL_1:4;
].0 ,1.[ c= [.0 ,1.]
by XXREAL_1:25;
then A25:
( cos . (th2 / 2) > 0 & sin . (th2 / 2) >= - 0 )
by A23, Lm18, Th74;
4 * (th2 / 2) < 4 * 1
by A20, XREAL_1:70;
then A27:
2 * th2 in ].0 ,4.[
by A14, XXREAL_1:4;
(sin . (th2 / 2)) * ((cos . (th2 / 2)) " ) = 1
by A22, A25, XCMPLX_0:def 7;
then
tan . ((2 * th2) / 4) = 1
by A23, Th75, RFUNCT_1:def 4;
then
2 * th2 = PI
by A27, Def32;
hence
contradiction
by A1, A12, XXREAL_1:4; verum