Question 1204731
given:

{{{sin(a) =3/4}}} and {{{cos(b) = -1/3}}}, with {{{a}}} and {{{b }}}both in the interval {{{pi/2}}}, {{{pi }}}=> {{{Q }}}{{{II}}}

ln {{{Q }}}{{{II}}} sine is {{{positive}}} and cosine is {{{negative}}}


{{{sin(a) =3/4=a/c}}}  => {{{a=3 }}}and {{{c=4}}}

{{{cos(a)=b/c=b/4}}}

using the Pythagorean Theorem, the missing side {{{b=sqrt(4^2-3^2)=sqrt(16-9)=sqrt(7)}}}


{{{cos(a)=b/c=sqrt(7)/4}}}

{{{cos(a)=sqrt(7)/4 }}}or {{{cos(a)=-sqrt(7)/4 }}}

we need {{{ cos(a)=-sqrt(7)/4 }}}


and

{{{cos(b) = -1/3}}}, => {{{b=-1}}}, {{{c=3}}}

{{{sin(b)=a/3}}}

{{{a=sqrt(3^2-(-1)^2)=sqrt(9-1)=sqrt(8)=sqrt(2*4)=2sqrt(2)}}}

{{{sin(b)=(2sqrt(2))/3 }}}or {{{sin(b)=-(2sqrt(2))/3}}}


we need {{{sin(b)=(2sqrt(2))/3}}}



find the exact values of


{{{sin(a + b)=cos(b)*sin(a) + cos(a)*sin(b)}}}

{{{sin(a + b)=(-1/3)(3/4)+(-sqrt(7)/4 )((2sqrt(2))/3)}}}

{{{sin(a + b)=-1/4-sqrt(14)/6}}}



and


{{{cos(a -b)=cos(a)*cos(b) + sin(a)*sin(b)}}}

{{{cos(a -b)=(-sqrt(7)/4)(-1/3)+(3/4)((2sqrt(2))/3)}}}

{{{cos(a -b)=sqrt(7)/12+1/sqrt(2) }}}

{{{cos(a -b)=sqrt(7)/12+sqrt(2)/2}}}