<PRE><B>Question 690928</B>
If sin (t) = 5/6, and theta is in quadrant 2, 
find the exact value of:
a) cos(t)
Since sin(t) = y/r, y = 5 and r = 6
Then x = sqrt[r^2-y^2] = sqrt(11) 
------
Note: Remember; cos is negative in the QII
------------
So cos(t)= x/r = -sqrt(11)/6
=====================

b) sin(t + pi/6) = sin(t)*cos(pi/6) + cos(t)*sin(pi/6)
----
= (5/6)(sqrt(3)/2) + (-sqrt(11)/6)(1/2)
----
= (5sqrt(3) - sqrt(11))/12
================================
Cheers,
Stan H.
</PRE>