cos²﹙﹣π/12﹚－sin²π/12

cos²﹙﹣π/12﹚－sin²π/12 sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求：(1)sin(2 cos*π/12—sin*π/12 计算cosπ/12*sinπ/12 求值：sin(π/12)+cos(π/12) 已知tan﹙π－α﹚＝2,求sin²α－2sinαcosα－cos²α分之4cos²α－3 根号3×cosπ／12－sinπ／12=? （cosπ/12-sinπ/12)*(COSπ/12+sinπ/12)等于 若﹙cosα﹣2cosβ﹚²﹢﹙sinα﹣2sinβ﹚²＝3,且0＜α＜π／2,0＜β＜π／2,则α 证明恒等式∶cosα﹙cosα－cosβ﹚＋sinα﹙sinα－sinβ﹚＝2sin²×α－β／2 cos²12分之π-sin²12分之π等于多少? y=cos²(x-π/12)+sin²(x+π/12)-1