Question 42653
<pre><font size = 5><b>Find the exact value of sin 2&#952; if
cos&#952; = -(&#8730;5)/3 and 180° < &#952; < 270°. 
Thank You 

We need to use the formula

sin(2&#952;) = 2sin&#952;·cos&#952;

We know cos&#952; but not sin&#952;.  We have to find sin&#952;.

We are given 180° < &#952; < 270° so we know &#952; is in 
quadrant III.  So we draw the picture of &#952; in the
third quadrant

    _ |
  -&#8730;5 |
-------------
  |  /|
  | /3|
  |/  |

Since the cosine is x/r or adjacent/hypotenuse, we
                       _             _ 
put the numerator of -&#8730;5/3, namely -&#8730;5 the on the x-side
(or adjacent side) and 3 on the r, the radius vector (or
hypotenuse).

We use the Pythagorean theorem to find the y-side (or the
opposite side).
      _______     ____________      ___     _
y = ±&#8730;r² - x² = ±&#8730;(3)² - (-&#8730;5)² = ±&#8730;9-5 = ±&#8730;4 = ±2

We know to take the negative sign since y goes down
from the x-axis, so we have

    _ |
  -&#8730;5 |
-------------
  |  /|
-2| /3|
  |/  |
                        
Now we know sin&#952; = -2/3

So
sin(2&#952;) = 2sin&#952;·cos&#952; 
                     _ 
sin(2&#952;) = 2(-2/3)·(-&#8730;5/3)
                     _ 
sin(2&#952;) = 2(-2/3)·(-&#8730;5/3)
            _ 
sin(2&#952;) = 4&#8730;5/9

Edwin
AnlytcPhil@aol.com</pre>