As #thetain"Quadrant"III# and #sintheta=-15/17=-0.882353#

#costheta=-sqrt(1-(15/17)^2)=-sqrt(1-225/289)=-sqrt(64/289)=-8/17=-0.470588#

Then #sin3theta=3sinx-4sin^3x=3(-0.882353)-4(-0.882353)^3#

= #-2.647059+2.747813=0.100753#

and #cos3theta=4cos^3x-3cosx=4(-0.470588)^3-3(-0.470588)#

= #-0.416853+1.411764=0.994911#

Note that as #sin3theta# and #cos3theta# are positive, #3theta# is in #Q1#,

Now #sin6theta=2sin3thetacos3theta=2xx0.100753xx0.994911#

= #0.200481#

and #cos6theta=(0.994911)^2-(0.200481)^2=0.949655#

and #6theta# is in #Q2#

Continuing this way #sina2theta=2sin6thetacos6theta=2xx0.200481xx0.949655#

= #0.380776#

and #cos12theta=(0.949655)^2-(0.200481)^2=0.861652#

Hence #12theta# is in #Q1#.