若sin(x+6/π)=1/3,求sin(7π/6+x)+cos2(5π/6-x)

若sin(x+6/π)=1/3,求sin(7π/6+x)+cos2(5π/6-x)
若sin(x+6/π)=1/3,求sin(7π/6+x)+cos2(5π/6-x)

若sin(x+6/π)=1/3,求sin(7π/6+x)+cos2(5π/6-x)
sin(7π/6+x)
=sin(π+π/6+x)
=-sin(π/6+x)
=-1/3
cos²(5π/6-x)
=1-sin²(5π/6-x)
=1-sin²[π-(5π/6-x)]
=1-sin²(π/6+x)
=8/9
所以原式=5/9

是不是假设条件应该是sin(x+π/6)=1/3?
sin(π + （x+ π /6）) + 1-2(sin(5π /6 -x))²
=-sin(x+π /6)+1-2(sin(π -(π /6+x))²
=1/3 + 1 -2*(-sin(π /6+x))²
=4/3 -2*(1/9)
=10/9