怎么解三角函数Sin的二次方13度+Cos的平方17度- Sin13Cos17

2025-05-06 14:34:13
推荐回答(2个)
回答1:

(sin13)^2 + (cos17)^2 - sin13 cos 17 = (1 - cos 26) / 2 + (1 + cos 34) / 2 - sin 13 cos 17
= 1 +( cos (30 + 4) - cos (30 - 4) ) / 2 - 1/2 ( sin (13+17) - sin(17 - 13) )
= 1 + ( - 2 sin 30 sin 4 ) / 2 - ( sin 30 - sin 4 ) / 2
= 1 - sin 4 / 2 - 1 / 4 + sin 4 / 2
= 3 / 4

回答2:

  sin²13°+cos²17°-sin13°cos17°
  = sin²13°+cos²17°-sin13°cos17°
  =(1-cos26°)/2+(1+cos26°)/2-sin13°cos17°
  =1+(cos26°-cos34°)/2-sin13°cos17°
  =1-sin[(26°+34°)/2]sin[(26°-34°)/2]-[sin(13°+17°)+sin(13°-17°)]/2(积化和差、和差化积) =1-sin30°sin4°-(sin30°-sin4°)/2
  =1-(1/2)sin4°-1/4+(1/2)sin4°
  =3/4