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sin20^2+cos50^2+sin30*cos70=?
人气:163 ℃ 时间:2020-09-14 05:44:16
解答
原式=sin^2 20°+cos^2 50°+sin(50°-20°)-sin(50°+20°)
用两角和公式
=sin^2 20°+cos^2 50°+(sin50°cos20°-sin20°cos50°)(sin50°cos20°+sin20°cos50°)
=sin^2 50°cos^2 20°-sin^2 20°cos^2 50°+sin^2 20°+cos^2 50°
用降幂公式
=[(1-cos100°)/2 *(1+cos40°)/2]-[(1-cos40°)/2 *(1+cos100°)/2]+(1-cos40°)/2 +(1-cos100°)/2
提出1/4 并整理
=1/4 *[(1-cos100°+cos40°-cos100°cos40°)-(1+cos100°-cos40°-cos100°cos40°)+2+2cos100°+2-2cos40°]
=1/4*4=1
2 sinA sinB = cos(A-B) - cos(A+B)
sin30°sin70°=cos(30°-70°)-cos(30°+70°)=cos40°-cos100°
sin20°的平方=(1-cos2*20°)/2=(1-cos40°)/2
同理cos50°的平方=(1-cos2*50°)/2=(1-cos100°)/2
原式=(1-cos40°)/2+(1-cos100°)/2+(cos40°-cos100°)/2=1
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