cos10°的平方+cos50°的平方-sin40°sin80°= ?
答案:2 悬赏:0
解决时间 2021-02-20 03:03
- 提问者网友:心裂
- 2021-02-19 11:18
cos10°的平方+cos50°的平方-sin40°sin80°= ?
最佳答案
- 二级知识专家网友:晚安听书人
- 2021-02-19 11:27
(cos10°)^2+(cos50°)^2-sin40°*sin80°的值
=(cos10°)^2+(cos50°)^2-cos50°*cos10°
=(cos10°)^2+(cos50°)[cos50°-cos10°]
=(cos10°)^2-2(cos50°)[sin30°sin20°]
=(1/2)+[(cos20°)/2]-cos50°cos70°
=(1/2)+[(cos20°)/2]-(1/2)cos120°-(1/2)cos20°
=(1/2)-(1/2)×(-1/2)=3/4
=(cos10°)^2+(cos50°)^2-cos50°*cos10°
=(cos10°)^2+(cos50°)[cos50°-cos10°]
=(cos10°)^2-2(cos50°)[sin30°sin20°]
=(1/2)+[(cos20°)/2]-cos50°cos70°
=(1/2)+[(cos20°)/2]-(1/2)cos120°-(1/2)cos20°
=(1/2)-(1/2)×(-1/2)=3/4
全部回答
- 1楼网友:不服输的倔强
- 2021-02-19 12:10
您好,很高兴为你解答:
(cos10°)^2+(cos50°)^2-sin40°*sin80°
=(cos10°)^2+(cos50°)^2-cos50°*cos10°
=(cos10°)^2+(cos50°)[cos50°-cos10°]
=(cos10°)^2-2(cos50°)[sin30°sin20°]
=(1/2)+[(cos20°)/2]-cos50°cos70°
=(1/2)+[(cos20°)/2]-(1/2)cos120°-(1/2)cos20°
=(1/2)-(1/2)×(-1/2)
=3/4
谢谢,望采纳
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