# 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.**

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**