Question 1174754

you got it right!


given:

{{{sin(2a)}}}
{{{cos(a)=-4/5}}}

when a terminates in quadrant {{{II}}}, {{{0<= a <=180}}}

use identity:

{{{sin^2(a)=1-cos^2(a)}}}.........substitute {{{cos(a)=-4/5}}}
{{{sin^2(a)=1-(-4/5)^2}}}
{{{sin^2(a)=1-(16/25)}}}
{{{sin^2(a)=25/25-16/25}}}
{{{sin^2(a)=9/25}}}
{{{sin(a)=sqrt(9/25)}}}

then {{{sin(a) = 3/5}}} or {{{sin(a) =- 3/5}}}

use identity:

{{{sin(2a)=2 sin(a) cos(a) }}}..........substitute {{{sin(a)}}}  and{{{ cos(a)}}}

 note: since {{{a}}} is in quadrant {{{II}}}, the info leads to{{{ sin(a) =  3/5}}},so  we have

{{{sin(2a)=2 (3/5)(-4/5)}}}
{{{sin(2a)=-24/25}}}