SOLUTION: Find cos(s - t) given that cos s = - 1/6, with s in quadrant III, and cos t=-3/5 with t in quadrant III

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Question 1176584: Find cos(s - t) given that cos s = - 1/6, with s in quadrant III, and cos t=-3/5 with t in quadrant III
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
First you draw a picture of angle s in quadrant III, and draw a perpendicular
from the end of the terminal side up to the x-axis.



You are given cos(s) = -1/6. You know that cosine = adjacent/hypotenuse,
which is x/r, so we label the x value as the numerator of -1/6, which
is -1, and we label the hypotenuse (the terminal side) as the denominator
of -1/6, which is 6. (The hypotenuse or terminal side is always taken
positive.



We use the Pythagorean theorem to find the value of y:

x%5E2%2By%5E2=r%5E2
%28-1%29%5E2%2By%5E2=6%5E2
1%2By%5E2=36
y%5E2=35
y=%22%22%2B-sqrt%2835%29

Since y goes down from the x-axis we give it a negative sign.
We label the opposite side, which is y, as -√35



We will need the sine and cosine of s,

cos%28s%29+=+-1%2F6 which is given, and
sin%28s%29+=+opposite%2Fhypotenuse=y%2Fr=-sqrt%2835%29%2F6




The same way, draw the angle t, and get this graph:



We will need the sine and cosine of t,

cos%28s%29+=+-3%2F5 which is given, and
sin%28s%29+=+opposite%2Fhypotenuse=y%2Fr=-4%2F5

Now we use the formula for 

cos%28s-t%29+=+cos%28s%29cos%28t%29%2Bsin%28s%29sin%28t%29

cos%28s-t%29+=+%28-1%2F6%29%28-3%2F5%29%2B%28-sqrt%2835%29%2F6%29%2A%28-4%2F5%29

cos%28s-t%29+=+3%2F30%2B%284sqrt%2835%29%2F30%29

cos%28s-t%29+=+%283%2B4sqrt%2835%29%29%2F30

Edwin