高一 数学 三角函数 请详细解答,谢谢! (21 12:29:32)

cos2a=cos²a-sin²a=(cosa-sin)(cosa+sina)

sin(a-π/4)=sinacosπ/4-cosasinπ/4=(-√2/2)(cosa-sina)
所以cos2a/sin(a-π/4)=(cosa+sina)/(-√2/2)=-√2/2
所以sina+cosa=1/2

∵cos(2a)/sin(a-π/4)=-√2/2
∴(cos²a-sin²a)/[sinacos(π/4)-cosasin(π/4)]=-√2/2
==>(cosa+sina)(cosa-sina)/(sina-cosa)=-1/2
==>cosa+sina=1/2
故cosa+sina=1/2。

将左边分母展开为
sin a * cos (pi/4) - cos a * sin (pi/4)
也就是
sqrt(2)/2 (sin a - cos a)
分子分母同时乘以sin a + cos a：

cos 2a * (sin a + cos a)
------------------------------
sqrt(2)/2 * (sin^2 a -cos^2 a)

分母中(sin^2 a -cos^2 a) = -cos 2a，与分子消去剩下-1
所以sin a + cos a = 1/2

cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4)) /(2sin(a-π/4) cos(a-π/4))
cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4)) /sin(2a-π/2)
cos2a/sin(a-π/4)=-(2cos2a*cos(a-π/4)) /cos2a
cos2a/sin(a-π/4)=-2*cos(a-π/4)
-2*cos(a-π/4) =-√2/2
cos(a-π/4) =√2/4
cosa*cos(π/4)+sina*sin(π