How do you use half angle formula to find ##sin 67.5##?

The half of ##135^o## is ##67.5^o##

Using the Half Angle Formula:

##sin (theta/2) = +-sqrt((1-costheta)/2)##

##sin (135^o/2) = +-sqrt((1-cos135^o)/2)##

##cos135^o = cos(90^o +45^o)= cos90^ocos45^o-sin90^oc0s45^o = cancel(0(sqr2/2))-(1)(sqrt2/2) = -sqrt2/2##

plugged in this value in the formula:

##sin67.5^o=sin (135^o/2) = +-sqrt((1-(-sqrt2/2))/2)= +-sqrt((1+sqrt2/2)/2)= sqrt(((2+sqrt2)/2)/2)= sqrt((2+sqrt2)/4)= 1/2sqrt(2+sqrt2) ##