Question 857618
<pre>
To prove:  cos(20°) + cos(100°) +cos(140°) = 0

Write the angles in the first two terms in terms 
of their average, which is 60°, a special angle,
and use a double angle identity on each:

 cos(20°) = cos(60°-40°) = cos(60°)cos(40°)+sin(60°)sin(40°)

cos(100°) = cos(60°+40°) = cos(60°)cos(40°)-sin(60°)sin(40°)
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  SUM OF FIRST TWO TERMS = 2cos(60°)cos(40°) = 2{{{(1/2)}}}cos(40°) = cos(40°) 

Write the angle in the third term in terms of 40° and use
use a double angle identity: 

cos(140°) = cos(180°-40°) = cos(180°)cos(40°)+sin(180°)sin(40°)
                          = (-1)cos(40°) + 0(sin(40°)
                          = -cos(40°)

So we have:

cos(20°) + cos(100°) + cos(140°) = cos(40°) - cos(40°) = 0

Edwin</pre>