Question

АЛГЕБРА! СРОЧНО! Решите пожалуйста. Номер 416

Answer 1

1)sin(π+α)=-sinα
  cos(π/2+α)=-sinα
  cos(2π-α)=cosα
 sin(3π/2-α)=-cosα
(-sinα-sinα)²+(cosα-(-cosα))²=(-2sinα)²+(2cosα)²=4sin²α+4cos²α=4(sin²α+cos²α)=4
2)tg(π/2-α)=sin(π/2-α)/cos(π/2-α)=cosα/sinα=ctgα
ctg(π/2+α)=cos(π/2+α)/sin(π/2+α)=-sinα/cosα=-tgα
ctg(π+α)=cos(π+α)/sin(π+α)=-cosα/(-sinα)=ctgα
ctg(3π/2+α)=cos(3π/2+α)/sin(3π/2+α)=sinα/(-cosα)=-tgα
(ctgα+tgα)²-(ctgα-tgα)²=(ctgα+tgα-ctgα+tgα)(ctgα+tgα+ctgα-tgα)=2tgα×2ctgα=4tgαctgα=4
3)sin160=sin(180-20)=sin20
cos110=cos(90+20)=-sin20
sin250=sin(270-20)=-cos20
cos340=cos(360-20)=cos20
tg110=sin110/cos110=sin(90+20)/cos(90+20)=cos20/(-sin20)=-ctg20
tg340=sin340/cos340=sin(360-20)/cos(360-20)=-sin20/cos20=-tg20
sin20×(-sin20)+(-cos20)cos20+(-ctg20)(-tg20)=-sin²20-cos²20+1=-(sin²20+cos²20)+1=-1+1=0
4)tg18=sin18/cos18=sin(90-72)/cos(90-72)=cos72/sin72=ctg72
tg288=sin288/cos288=sin(360-72)/cos(360-72)=-sin72/cos72=-tg72
sin32=sin(90-58)=cos58
sin148=sin(90+58)=cos58
sin302=sin(360-58)=-sin58
sin122=sin(180-58)=sin58
ctg72×(-tg72)+cos58cos58-(-sin58)sin58=-1+cos²58+sin²58=-1+1=0