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Prove that cosa*cosb<=0 if a-b=pi
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We'll consider the constraint a - b = pi => a = pi + b
We'll calculate the cosine function of the angle a:
cos a = cos (pi + b)
We'll use the following identity:
cos (x+y) = cosx*cosy - sinx*siny
Comparing, we'll get:
cos (pi + b) = cos pi*cos b - sin pi*sin b
But cos pi = - 1 and sin pi = 0
cos (pi + b) = -cos b
Therefore cos a = - cos b.
If we'll multiply a negative number by a positive number, the result is a negative number, therefore cos a*cos b = - (cos b)^2 =<0.
Posted by giorgiana1976 on July 5, 2011 at 11:26 PM (Answer #1)
We have to prove that (cos a)*(cos b) <= 0 if a - b = pi
(cos a)*(cos b) = (1/2)[cos (a + b) + cos(a - b)]
as a - b = pi
=> (1/2)[cos (a + b) + cos pi]
=> (1/2)[cos (a + b) - 1]
=> (1/2)*cos (a + b) - 1/2
but cos (a + b) can take a maximum value of 1
=> (1/2)*cos (a + b) - 1/2 <= 0
This proves that (cos a)*(cos b) <= 0 if a - b = pi
Posted by justaguide on July 5, 2011 at 11:32 PM (Answer #2)
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