Question

1)已知cos(x+π/6)=1/4 求cos(5π/6-x)+cos(π/3-x)^2的值（2）已知tanα=2，求1/sinα^2-sinαcosα-3...

1)已知cos(x+π/6)=1/4 求cos(5π/6-x)+cos(π/3-x)^2的值（2）已知tanα=2，求1/sinα^2-sinαcosα-3cos^2 为什么没答案。求给个答案。。

Answer 1

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

Answer 2

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

Answer 3

no