求值sin²20º+cos²50º+sin20ºcos50º
人气:258 ℃ 时间:2020-05-08 05:37:37
解答
cos50° = cos(30°+20°) = cos30°cos20°-sin30°sin20° = (√3/2)cos20°-(1/2)sin20° ;
原式
= sin²20°+[(√3/2)cos20°-(1/2)sin20°]²+sin20°[(√3/2)cos20°-(1/2)sin20°]
= sin²20°+[(3/4)cos²20°+(1/4)sin²20°-(√3/2)sin20°cos20°]+(√3/2)sin20°cos20°-(1/2)sin²20°
= (1+1/4-1/2)sin²20°+(3/4)cos²20°+(√3/2-√3/2)sin20°cos20°
= (3/4)sin²20°+(3/4)cos²20°+0
= (3/4)(sin²20°+cos²20°)
= 3/4
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