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教育/科学

(sin10度)^2+(sin50度)^2-2sin10度sin?

2006-03-25 13:39:40x***
(sin10度)^2+(sin50度)^2-2sin10度sin50度cos120度=?(sin10度)^2+(sin50度)^2-2sin10度sin50度cos120度=?(sin10度)^2+(sin50度)^2-2sin10度sin50度cos120度=?(sin10度)^2+(sin50度)^2-2sin10度sin50度c?

最佳回答

  • 省略“度”： (sin10)^+(sin50)^-2sin10sin50cos120 =(sin10)^+(sin50)^-2sin10sin50(-1/2) =(sin10)^+(sin50)^+sin10sin50 =(1/2)[(1-cos20)+(1-cos100)-(cos60-cos40)] =(1/2)[3/2-(cos20+cos100)+cos40] =(1/2)[3/2-2cos60cos40+cos40] =3/4
    2006-03-25 14:46:19
  • 解: (sin10度)^2+(sin50度)^2-2sin10度sin50度cos120度 =(sin10度)^2+(sin50度)^2-2sin10度sin50度cos(180-60度) =(sin10度)^2+(sin50度)^2+sin10度sin50度cos60度 =sin＾10°+sin＾50°+2sin10°sin50°-sin10°sin50° =(sin10°+sin50°)＾+(1/2)(cos60°-cos40°) =4sin＾30°cos＾20°+1/4-cos＾20°+1/2 =3/4
    2006-03-25 14:15:58
    • 很赞哦！ (114)

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