已知a属于( π/2,π)且sina=3/5.求cos(a- π/4)的值；求sin(a/2)的平方+tan(a+ π/4)的值.

题目

已知a属于( π/2,π)且sina=3/5.求cos(a- π/4)的值；求sin(a/2)的平方+tan(a+ π/4)的值.

答案

因为sina=3/5,a属于( π/2, π)
所以cosa=-4/5
(1)cos(a- π/4)=cosa*cos π/4+sina*sin π/4=-根号2/10；
(2)sin(a/2)的平方+tan(a+ π/4)
=(-1/2)乘[-2sin(a/2)的平方]+(tana +1)/(1-tan a)
=(1/2)-(1/2)乘[1-2sin(a/2)的平方]+[(sina/cosa)+1]/[1-(sina/cosa)]
=(1/2)-(1/2)乘cosa + (sina+cosa)/(cosa -sina)
=(1/2) + (1/2)乘(4/5) + [(3/5)-(4/5)]/[-(4/5)-(3/5)]
=73/70

不知道对不对.