急求三角函数cos10°cos50°cos70°的值.

代代花红 1年前 已收到3个回答 举报

uaaaa 花朵

共回答了11个问题采纳率:90.9% 举报

用积化和差做
cos10°cos50°cos70°
=0.5*{cos(10°+50°)+cos(10°-50°)}*cos70°
=0.5(cos60°+cos40°)cos70°
=0.5(0.5+cos40°)cos70°
=0.25*cos70°+0.5*cos40°cos70°
=0.25*cos70°+0.5*0.5{cos(40°+70°)+cos(40°-70°)}
=0.25*cos70°+0.25[cos110°+cos30°]
=0.25(cos70°+cos110°+cos30°)
因为cos70°=cos(180°-110°)=-cos110°
所以cos70°+cos110°=0
所以原式=0.25*cos30°=八分之根号三

1年前

3

fox_vitas 幼苗

共回答了6个问题 举报

分别为0.984,0.642,0.342

1年前

2

hx168 幼苗

共回答了9个问题 举报

cos10°cos50°cos70°
=1/2*{cos(10°+50°)+cos(10°-50°)}*cos70°
=1/2(cos60°+cos40°)cos70°
=1/2(1/2+cos40°)cos70°
=1/4*cos70°+1/2*cos40°cos70°
=1/4*cos70°+1/2*1/2{cos(40°+70°)+cos(40°-70...

1年前

1
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