Question

请问这道三角函数题怎么做

尽量详细点。。谢谢了

Answer 1

原式=cos(a-π/4)cos(a+π/4）=√2/6
cos(a+π/4-π/2）cos(a+π/4）=√2/6
sin(a+π/4）cos(a+π/4)=√2/6
2 ×sin(a+π/4）cos(a+π/4)=2×√2/6
sin2(a+π/4)=2√2/6
sin(2a+π/2)=2√2/6
cos2a=2√2/6
∵0＜a＜π/2
∴0＜2a＜π
∴sin2a=√1-cos²2a=√7/3 所以选D。

Answer 2

原式=(cosπ/4cosa+sinπ/4sina)(cosπ/4cosa-sinπ/4sina)==√2/6
(cosa+sina)(cosa-sina)==√2/6
cosa^2-sina^2=√2/6
cos2a=√2/6
sin2a=√34/6

追问

你的第二步错了。答案是D