Question 1056959
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let sinA= 3/5 with A in Q2 and sinB= -5/13 with B in quadrant 3. find sin(A+B), cos(a+b), and tan(a+b). where does a and b terminate
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<pre>
Use the formulas 

{{{sin(alpha+beta)}}} = {{{sin(alpha)*cos(beta) + cos(alpha)*sin(beta)}}}    (1)

   and

{{{cos(alpha+beta)}}} = {{{cos(alpha)*cos(beta) - sin(alpha)*sin(beta)}}}.   (2)


Regarding these formulas, see the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A> in this site.


In addition to the given  sin(A) = {{{3/5}}}  and  sin(B) = {{{-5/13}}}, you need to know  cos(A) and cos(B).


1.  cos(A) = {{{sqrt(1-sin^2(A))}}} = {{{-sqrt(1 - (3/5)^2)}}} = {{{-sqrt(1 - 9/25)}}} = {{{-sqrt((25-9)/25)}}} = {{{-sqrt(16/25)}}} = {{{-4/5}}}.

   The sign "-" was chosen for the square root because cosine is negative in Q2.


2.  cos(B) = = {{{-sqrt(1-sin^2(B))}}} = {{{-sqrt(1 - (-5/13)^2)}}} = {{{-sqrt(1 - 25/169)}}} = {{{-sqrt((169-25)/169)}}} = {{{-sqrt(144/169)}}} ={{{-12/13}}} = {{{-12/13}}}.

   The sign "-" was chosen for the square root because cosine is negative in Q3.


Now all you need to do is to substitute everything into the formulas (1) and (2) and make the calculations.


{{{sin(A+B)}}} = {{{(3/5)*(-12/13) + (-4/5)*(5/13)}}} = {{{-36/65 - 20/65}}} = {{{(-36-20)/65}}} = {{{-56/65}}},   and

{{{cos(A+B)}}} = {{{(-4/5)*(-12/13) - (3/5)*(-5/13)}}} = {{{48/65 + 15/65}}} = {{{(48+15)/65}}} = {{{63/65}}}.

Finally, tan(A+B) = {{{sin(A+B)/cos(A+B)}}} = {{{((-56/65))/((63/65))}}} = {{{-56/63}}}.
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The last question "where does a and b terminate" is just answered in your condition.