Question

sin(四分之π-θ)=十三分之五（0＜θ＜四分之π） 求cos2θ，cos（四分之π+θ）的值

要过程喔~~~

Answer 1

0<π/4-θ<π/4,所以cos(π/4-θ)=12/13
cos(π/4+θ)=cos[π/2-(π/4-θ)]=0*cos(π/4-θ)+1*sin(π/4-θ)=5/13
π/4<π/4+θ<π/2,所以 sin(π/4+θ)=12/13
cos2θ=cos[(π/4+θ)-(π/4-θ)]=cos(π/4+θ)*cos(π/4-θ)+sin(π/4+θ)*sin(π/4-θ)=120/169