已知sin(4分之π-α)=13分之5且α大于0小于4分之π,求cos(4分之π+α)分之cos2α的值
问题描述：
已知sin(4分之π-α)=13分之5且α大于0小于4分之π,求cos(4分之π+α)分之cos2α的值
答
cos(π/4+α)=cos[π/2-(π/4-α)]=sin(π/4-α)=5/13
0所以sin(π/4+α)=cos(π/4-α)=12/13
cos2α=cos[(π/4+α)-(π/4-α)]=cos(π/4+α)cos(π/4-α)+sin(π/4+α)sin(π/4-α)=5/13*12/13+12/13*5/13=120/169
cos(π/4+α)/cos2α=5/13/(120/169)=13/24