Відповіді АЛГЕБРА: 11.15

11.15

1) sin(–930°)+ 3cos(–210°) + tg 675° = –sin(900° + 30°) + 3cos(180° + 30°) + tg(720° – 45°) sin30° – 3cos30° – tg45° = 12 – 3 • 32 – 1 = 12 – 32 – 1 = –2.

2) tg15π4 + 2sin (–7π6) – 2cos (–23π6) = tg(4π – π4) – 2sin(π + π6) –

– 2cos(4π – π6) = –tgπ4 + 2sin π6 – 2 cosπ6 = –1 + 2 • 12 – 2 • 32 = –1 +

+ 1 – 3 = –3