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Answer 1 · 2016-08-16T07:49:31

If #phi# is the angle of 3rd quadrant then

#cosphi->-ve and sinphi->-ve#

Now #sin2phi=2sinphicosphi#

#=-2sqrt(1-cos^2phi)*cosphi#

#=-2sqrt(1-(-15/17)^2)*(-15/17)#

#=+2sqrt(17^2-15^2)/17*xx15/17#

#=(2*8*15)/17^2=240/289#

#2phi=sin^-1(240/289)#
#~~56.14^@color(red)(->2phi" is in the 1st quadrant")#

Again

#cosphi=-15/17#

#phi=cos^-1(-15/17)=360-151.93=208.07#
#color(red)(("As "phi" is in the 3rd quadrant"))#

#:.2phi=416.14=360+56 .14#

So #2phi# is the angle of 1st quadrant.