SOLUTION: 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

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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

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
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
~~~~~~~~~~~~~~~~~~

Stap by step:

1.  sin(s)= 1/7 QII  --->  cos(s) = -sqrt%281-sin%5E2%28s%29%29 = -sqrt%281-%281%2F7%29%5E2%29 = -sqrt%2848%29%2F7 = -%284%2Asqrt%283%29%29%2F7.
           The sign is "-" at sqrt since cosine is negative in QII.


2.  sin(t)= -6/7 QIV  --->  cos(t) = sqrt%281-sin%5E2%28t%29%29 = sqrt%281-%28-6%2F7%29%5E2%29 = sqrt%2813%29%2F7 = %28sqrt%2813%29%29%2F7.
           The sign is "+" at sqrt since cosine is positive in QIV.


3.   Now  sin(s+t) = sin(s)*cos(t) + cos(s)*sin(t) = %281%2F7%29%2A%28sqrt%2813%29%2F7%29+%2B+%28-4%2Asqrt%283%29%2F7%29%2A%28-6%2F7%29 =  %28sqrt%2813%29+%2B+24%2Asqrt%283%29%29%2F49.


4.   Next cos(s+t) = cos(s)*cos(t) - sin(s)*sin(t) = %28-4%2Asqrt%283%29%2F7%29%2A%28sqrt%2813%29%2F7%29+-+%281%2F7%29%2A%28-6%2F7%29 =  %28-4%2Asqrt%283%29%2Asqrt%2813%29+%2B+6%29%2F49 = %286-4%2Asqrt%2839%29%29%2F7.


5.  Finally,  tan(s+t) = sin%28s%2Bt%29%2Fcos%28s%2Bt%29 = %28sqrt%2813%29+%2B+24%2Asqrt%283%29%29%2F%286-4%2Asqrt%2839%29%29.


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

All questions are answered.


For detailed solution of similar problem see the lessons
    - Calculating trigonometric functions of angles, Problems 5 and 7,
    - Advanced problems on calculating trigonometric functions of angles, Problem 1
in this site.

Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

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