1)已知cos(x+π/6)=1/4 求cos(5π/6-x)+cos(π/3-x)^2的值(2)已知tan...
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发布时间:2024-10-01 08:33
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热心网友
时间:2024-11-14 03:36
no
热心网友
时间:2024-11-14 03:32
cos(x+π/6)=1/4
cos(5π/6-x)+cos(π/3-x)^2
=cos(x-5π/6)+cos(x-π/3)^2
=cos(x+π/6-π)+cos(x+π/6-π/2)^2
=-cos(x+π/6)+sin(x+π/6)^2
=-1/4+15/16
=11/16
热心网友
时间:2024-11-14 03:31
1)这一题的关键是,找出x+π/6和5π/6-x、π/3-x的关系,很明显5π/6-x=π-(x+π/6),π/3-x=π/2-(x+π/6)。
cos(5π/6-x)+cos(π/3-x)^2
=cos(π-(x+π/6))+cos(π/2-(x+π/6))^2
=-cos(x+π/6)+sin(x+π/6)^2
=-1/4+15/16
=11/16
2)这一题最安全的方法是把sinα和cosα解出来,等你做多了就直接能看出来 sinα=√5/5,cosα=2√5/5,然后直接代进去算就好了