Question 1054631
.
use the given information to find:
sin(s+t)
tan(s+t)
the quadrant of s+t

sin s= 1/7 QII
sin t= -6/7 QIV
~~~~~~~~~~~~~~~~~~


<pre>
Stap by step:

1.  sin(s)= 1/7 QII  --->  cos(s) = {{{-sqrt(1-sin^2(s))}}} = {{{-sqrt(1-(1/7)^2)}}} = {{{-sqrt(48)/7}}} = {{{-(4*sqrt(3))/7}}}.
           The sign is "-" at sqrt since cosine is negative in QII.


2.  sin(t)= -6/7 QIV  --->  cos(t) = {{{sqrt(1-sin^2(t))}}} = {{{sqrt(1-(-6/7)^2)}}} = {{{sqrt(13)/7}}} = {{{(sqrt(13))/7}}}.
           The sign is "+" at sqrt since cosine is positive in QIV.


3.   Now  sin(s+t) = sin(s)*cos(t) + cos(s)*sin(t) = {{{(1/7)*(sqrt(13)/7) + (-4*sqrt(3)/7)*(-6/7)}}} =  {{{(sqrt(13) + 24*sqrt(3))/49}}}.


4.   Next cos(s+t) = cos(s)*cos(t) - sin(s)*sin(t) = {{{(-4*sqrt(3)/7)*(sqrt(13)/7) - (1/7)*(-6/7)}}} =  {{{(-4*sqrt(3)*sqrt(13) + 6)/49}}} = {{{(6-4*sqrt(39))/7}}}.


5.  Finally,  tan(s+t) = {{{sin(s+t)/cos(s+t)}}} = {{{(sqrt(13) + 24*sqrt(3))/(6-4*sqrt(39))}}}.


6.  From (3), sin(s+t) is positive;  from (4), cos(s+t) is negative.  Hence, s+t lies in QII.
</pre>

All questions are answered.



For detailed solution of similar problem see the lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Calculating-trigonometric-functions-of-angles.lesson>Calculating trigonometric functions of angles</A>, Problems 5 and 7, 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Selected-problems-from-the-archive-on-calculating-trig-functions-of-angles.lesson>Advanced problems on calculating trigonometric functions of angles</A>, Problem 1 

in this site.


Also, you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Trigonometry: Solved problems</U>".