Question

помогите пожалуйста
очень срочно

Answer 1

1) 0°<α<90°
a)sin(270° -α) = - cosα <0 .
б) cos(180° - α) = -cosα <0.
в) tq(270 -α) = ctqα  >0.
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2) tq100° =tq(90° +10°) = -ctq10° ;  tq110° =tq(90° +20°) = -ctq20° ;
ctq10° >ctq20⇒ -ctq10° < -ctq20 ⇔ tq100° < tq110°  .
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3)  ((tqπ/3+cosπ/6)*2tqπ/4)/(cosπ +sin1,5π) =
( (√3 +(√3)/2)*2*1 ) / (-1 -1)= 3√3)/(-2) =-1,5√3 .
4) (5/6)*π =5π/6 =(5*180°)/6 =5*30° =150 °.