Вычислите sin t и cos t, если:

а) t = 13 П/6

б) t = - 8 П/3

в) t=23 П/6

г) t=-11 П/3

д) t=9 П/4

1) 13π/6=2π+π/6

sin (2π+π/6) = sinπ/6=1/2; cos (2π+π/6) = cosπ/6=√3/2

2) 8π/3=2π+2π/3

sin ( - (2π+2π/3)) = - sin (2π+2π/3) = - sin2π/3=-sin (π-π/3) = - sinπ/3=-√3/2;

cos ( - (2π+2π/3)) = cos (2π+2π/3) = cos2π/3=-cos (π-π/3) = - cosπ/3=-1/2;

3) 23π/6=4π-π/6

sin (4π-π/6) = - sinπ/6=-1/2; cos (4π-π/6) = cosπ/6=√3/2

4) 11π/3=4π-π/3

sin ( - (4π-π/3)) = - sin (4π-π/3) = sinπ/3=√3/2;

cos ( - (4π-π/3)) = cos (4π-π/3) = cosπ/3=1/2;

5) 9π/4=2π+π/4

sin (2π+π/4) = sinπ/4=√2/2; cos (2π+π/4) = cosπ46=√2/2

cosα - функция четная

sinα - функция нечетная