We use the identity #sinx=cos(pi/2-x)=cos(x-pi/2)#
Hence #sin(picosa)=cos(pisina)# can be written as
#cos(picosa-pi/2)=cos(pisina)#
Hence #picosa-pi/2=2npi+-pisina#, where #n# is an integer.
or #cosa=2n+1/2+-sina=(4n+1)/2+-sina#
This leads to two situations
(1) #cosa=(4n+1)/2+sina#
or #cosa-sina=(4n+1)/2#
or #sqrt2cos(a+pi/4)=(4n+1)/2#
or #cos(a+pi/4)=(4n+1)/(2sqrt2)#
Now as #|cosa|<=1#, we can have #n=0# only and then
#cos(a+pi/4)=1/(2sqrt2)#
and #a+pi/4=2mpi+-cos^(-1)(1/(2sqrt2))#
and #a=2mpi-pi/4+-cos^(-1)(1/(2sqrt2))#
(2) #cosa=(4n+1)/2-sina#
or #cosa+sina=(4n+1)/2#
or #sqrt2cos(a-pi/4)=(4n+1)/2#
or #cos(a-pi/4)=1/(2sqrt2)# - because of similar restrictions as mentioned above.
and #a-pi/4=2mpi+-cos^(-1)(1/(2sqrt2))#
or #a=2mpi+pi/4+-cos^(-1)(1/(2sqrt2))#
Here #m# is an integer.