Question 1060448
.
Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sin (u + v)
~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
I am going to use the formula

sin(u+v) = sin(u)*cos(v) + cos(u)*sin(v).     (*)


For it, in addition to the given values  sin(u) = 5/13 and cos(v) = -3/5  I need to know  cos(u)  and  sin(v).


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

    The sign "-" is at the sqrt since cosine is negative in QII.


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

    The sign "+" is at the sqrt since sine is positive in QII.


3.  Now you have everything to use the formula (*). Substitute all given and found values into (*). You will get

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


<U>Answer</U>.  sin(u+v) = {{{-63/65}}}.  
</pre>