已知cos(x+兀/6)=1/4,求cos(5兀/6-x)+cos2(兀/3-x)的值
问题描述：
已知cos(x+兀/6)=1/4,求cos(5兀/6-x)+cos2(兀/3-x)的值
答
cos(5兀/6-x)=cos(兀-（x+π/6）=-cos(x+兀/6)=-1/4①
cos2(兀/3-x)=1-2sin^2（兀/3-x）②
又因为sin兀/3-x=sinπ/2-(x+兀/6)=cos(x+兀/6)=1/4
故②=7/8
故①+②=5/8
答
原式=(1+√14)/4∵cos(x+π/6)=1/4∴sin(x+π/6)=√15/4∵cos(5π/6-x)=cos[π-(x+π/6)]=cos(x+π/6)∴cos(5π/6-x)=1/4∵cos2(π/3-x)=cos^2 (π/3-x)-sin^2 (π/3-x)cos(π/3-x)=cos[π/2-(x+π/6)]=sin(x+π/6)=...