Question

Найдите sin2a, tg2a, ctg2a, cos(п÷6-a),cos2a если cos=-0,8, пи<a<3пи÷2​

Answer 1

cosa=-0,8;π<a<3π/2

sina=-√(1-cos²a)=-0,6

sin2a=2sinacosa=2*(-0,8)*(-0,6)=0,96
1+ctg²2a=1/sin²2a
π<a<3π/2
2π<2a<3π

ctg²2a=1/sin²2a-1=

(100²/96²)-1=(100-96)*196/

96²=4*14²/96²
ctg2a=±2*14/96=±7/24

tg2a=1/ctg2a=±24/7