Question

1)已知cos(x+π/6)=1/4 求cos(5π/6-x)+cos(π/3-x)^2的值

Answer 1

∵cos(x+π/6)=1/4

∴cos(5π/6+x)=cos[π-（π/6+x)]

=-cos(π/6+x)

=-1/4

cos(π/3-x)^2=cos[π/2-(π/6+x)]^2

=sin(π/6+x)^2

=1-cos(π/6+x)^2

=1-(1/4)^2

=15/16

∴cos(5π/6-x)+cos(π/3-x)^2=-1/4+15/16

=11/16