#[1]" "arcsin[cos((10pi)/9)]#
Convert #cos((10pi)/9)# to #sin# using the co-function identity.
#[2]" "=arcsin[sin(pi/2-(10pi)/9)]#
Simplify.
#[3]" "=arcsin[sin((9pi)/18-(20pi)/18)]#
#[4]" "=arcsin[sin((-11pi)/18)]#
The restricted domain for sine is #[-pi/2,pi/2]# since this involves an #arcsin#. We must look for an angle in this restricted domain whose sine has an equal value to #sin((-11pi)/18)#. That angle is #(-7pi)/18# because it has the same reference angle and sign as #(-11pi)/18#.
#[5]" "=arcsin[sin((-7pi)/18)]#
Arcsin and sin will cancel because of the definition of inverse functions.
#[6]" "color(blue)(=(-7pi)/18)#
