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Prove that cosa*cosb<=0 if a-b=pi

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greynose | Student, Undergraduate | Honors

Posted July 5, 2011 at 11:22 PM via web

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Prove that cosa*cosb<=0 if a-b=pi

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giorgiana1976 | College Teacher | Valedictorian

Posted July 5, 2011 at 11:26 PM (Answer #1)

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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.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 5, 2011 at 11:32 PM (Answer #2)

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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

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