Skoro:
sin(180°–α) = sin(90° + (90°–α)) = cos(90°–α) = sin α
cos(180°–α) = cos(90° + (90°–α)) = –sin(90°–α) = –cos α
tg(180°–α) = tg(90° + (90°–α)) = –ctg(90°–α) = –tg α
ctg(180°–α) = ctg(90° + (90°–α)) = –tg(90°–α) = –ctg α
360° = 2 ⋅ 180°, więc:
sin(360°–α ) = sin(2⋅180°–α) = sin(180° + (180°–α)) = –sin(180°–α) = –sin α
cos(360°–α ) = cos(2⋅180°–α) = cos(180° + (180°–α)) = –cos(180°–α) = cos α
tg(360°–α ) = tg(2⋅180°–α) = tg(180° + (180°–α)) = tg(180°–α) = –tg α
ctg(360°–α ) = ctg(2⋅180°–α) = ctg(180° + (180°–α)) = ctg(180°–α) = –ctg α
W tym zadaniu zauważ, że kąt 360° = 2 ⋅ 180°, więc skorzystaj ze wzorów redukcyjnych dla kątów: (180°–α) i 180° + (180°–α) i wyprowadź wzory dla wszystkich czterech funkcji trygonometrycznych.