Question

求下列三角函数的值：（1）sin（-67π/4）(3)cos398°13'

Answer 1

（1）sin（-67π/4）=sin(-67π/4+16π)=sin(-3π/4)=-sin(3π/4)=-√2/2

(3)cos398°13' =cos(398°13'-360° )=cos(38°13‘)=cos(38°+13/60)=cos38.217°=0.7857