Question

Question #bab8a

Answer 1

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.