数学题高一下学期2倍角正，余弦公式cos10*cos20*cos40

显示全部楼层 · 2012-2-12 22:39:18

2倍角正弦：sin2a=3sina*cosa2倍角余弦：cos2a=cosa^2-sina^2=1-2sin^2a=2cos^2a-1还有公式cosa*cosb=1/2[cos(a+b)+cos(a-b)]而此题如果要化简。则如下：原式=cos10*1/2[cos60+cos20]=cos10*1/2[1/2+cos20]=1/4cos10+1/2cos10*cos20=1/4cos10+1/4[cos30+cos10]=1/2cos10+根号3/8（你还是自己对一下上面的公式吧！）...

千问 · 2012-2-12 22:39:18

=8sin10cos20cos40cos10/(8sin10) =4sin20cos40cos20/(8sin10) =2sin40cos40/(8sin10) =sin80/(8sin10) =sin10/(8cos10) =1/8tan10...

千问 · 2012-2-12 22:39:18

2楼的很精辟...