Question 871093
NOTE: I am rushing to get done before I have to go to my day job, so check for mistakes.
 
In this problem "cos^-1" means the inverse function of cosine,
and "tan^-1" means the inverse function of tangent.
We are looking for the exact value of
{{{sin(A-B)}}} knowing that {{{cos(A)=2/3}}} and {{{tan(B)=1/4}}},
and that those inverse functions are defined so that
{{{A}}} will be in quadrants I or II, and
{{{B}}} will be in quadrants I or IV.
 
To calculate {{{sin(A-B)}}} we can use the trigonometric identity
{{{sin(A-B)=sin(A)*cos(B)-sin(B)*cos(A)}}}
So we need to find {{{sin(A)}}} , {{{sin(B)}}} , and {{{cos(B)}}} .
 
In quadrants I and I {{{sin(A)>0}}}, so
{{{sin(A)=sqrt(1-(cos(A))^2)=sqrt(1-(2/3)^2)=sqrt(1-4/9)=sqrt(5/9)=sqrt(5)/9}}}
 
Since {{{tan(B)=1/4>0}}} only hapens in quadrants I and III,
and the definition of the inverse tangent function says that {{{B}}} is in I or IV,
{{{B}}} is definitely in quadrant I, where sine and cosine are positive.
This is how angle {{{B}}} would look in a right triangle.
{{{drawing(300,100,-0.25,4.25,-0.25,1.25,
triangle(0,0,4,0,4,1),
locate(-0.2,0.1,B),locate(1.8,0,4),locate(4.05,0.7,1)
)}}} The hypotenuse of that right triangle measures {{{sqrt(4^2+1^2)=sqrt(16+1)=sqrt(17)}}}
So, {{{sin(B)=1/sqrt(17)}}} and {{{cos(B)=4/sqrt(17)}}}
Substituting all the values found into {{{sin(A-B)=sin(A)*cos(B)-sin(B)*cos(A)}}} ,
{{{sin(A-B)=(sqrt(5)/9)*(4/sqrt(17))-(1/sqrt(17))*(2/3)=4sqrt(5)/(9sqrt(17))-2/(3sqrt(17))=4sqrt(5)/(9sqrt(17))-6/(9sqrt(17))=(4sqrt(5)-6)/(9sqrt(17))}}}
SInce that does not look elegant, and teachers do not like square roots in denominators, we multiply numerator and denominator times {{{sqrt(17)}}} to get
{{{(4sqrt(5)sqrt(17)-6sqrt(17))/(9*17)=(4sqrt(85)-6sqrt(17))/153}}}