#f(pi/8)=2(pi/8)sin(pi/8)+pi/8cot^2(2(pi/8)) =pi/4sin(pi/8)+pi/8cot^2(pi/4)#
By the half angle formulae,
#sin(x/2)=+-sqrt((1-cos(x))/2)#
#sin(pi/8)=+-sqrt((1-cos(pi/4))/2)=+-sqrt((1-sqrt2/2)/2)=+-sqrt(2-sqrt2)/2#
Using the CAST rule, we can deduce that #sin(pi/8)=+sqrt(2-sqrt2)/2#
Hence #f(pi/8)=pi/4sqrt(2-sqrt2)/2+pi/8(1)^2=pi/8(sqrt(2-sqrt2)+1)#
#f(theta)=2thetasin(theta)+thetacot^2(2theta)#
#f'(theta)=2sintheta+2thetacostheta+cot^2(2theta)+theta(2cot(2theta)*(-csc^2(2theta))*2)#
#therefore f'(pi/8)=2sin(pi/8)+pi/4cos(pi/8)+cot^2(pi/4)+pi/2cot(pi/4)*(-csc^2(pi/4))=2sin(pi/8)+pi/4cos(pi/8)+1-pi/2csc^2(pi/4)#
By the half angle formulae,
#cos(x/2)=+-sqrt((cos(x)+1)/2)#
#cos(pi/8)=+-sqrt((cos(pi/4)+1)/2)=+-sqrt((sqrt2/2+1)/2)=+-sqrt(2+sqrt2)/2#
Using the CAST rule, we can deduce that #cos(pi/8)=+sqrt(2+sqrt2)/2#
#because csc(x) = 1/sin(x)#
#therefore csc(pi/8)=1/(sqrt(2-sqrt2)/2)=2/(sqrt(2-sqrt2)#
#therefore csc^2(pi/8)=(2/(sqrt(2-sqrt2)))^2=4/(2-sqrt2)#
#therefore f'(pi/8)=2(sqrt(2-sqrt2)/2)+pi/4sqrt(2+sqrt2)/2+1-pi/2(4/(2-sqrt2))=sqrt(2-sqrt2)+(pisqrt(2+sqrt2))/8+1-(2pi)/(2-sqrt2)#
#therefore (y-y_1)/(x-x_1)=y'#
#(y-f(pi/8))/(x-pi/8)=f'(pi/8)#
#y=f'(pi/8)x-pi/8f'(pi/8)+f(pi/8)#
#therefore y=-0.65013x+0.94875approx-0.650x+0.949#
