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The value of cosA + cosB + cosC in a triangle is less than 3/2 . Proof?
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The value of cosA + cosB + cosC in a triangle is less than 3/2 . Proof...
Since C = π - A - B, we want to maximize the function:

f(A,B) = cos(A) + cos(B) + cos(π-A-B)

f(A,B) = cos(A) + cos(B) - cos(A+B)

within the region R:

A+B<π
0<A,B

Since it's an open region, we know the max cannot occur anywhere along the boundary of Rand must occur at some critical point(s) in the interior.

So we just have to find the point where the total derivative is 0. Of course, A and B aren't dependent on each other, so we can just set the two partials equal to zero to find the critical point(s):

sin(A+B) - sin(A) = 0
sin(A+B) - sin(B) = 0

So sin(A) = sin(B), thus A=B

But sin(A) = sin(2A) = 2sin(A)cos(A)

Thus cos(A) = 1/2

Note: we can divide by sin(A) because A≠0. This also excludes A=0 as a solution, since A=0 is in the boundary of R, which is not a part of our valid region.

A = π/3

So the max is at A=B=C=π/3.

Plugging that in, we know the maximum value of f is:

3cos(π/3) = 3/2

Thus for angles A,B,C of a triangle:

cos(A)+cos(B)+cos(C) = f(A,B) = ≤ 3/2 .
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The value of cosA + cosB + cosC in a triangle is less than 3/2 . Proof...
Proof that cosA cosB cosC < 3/2="" in="" a="" />

To prove that the value of cosA cosB cosC in a triangle is less than 3/2, we need to use the properties of cosines and the triangle inequality theorem. Let's break down the proof into the following sections:

1. Introduction:

In any triangle, the sum of the three angles is always 180 degrees (π radians). Let's assume the three angles of the triangle are A, B, and C.

2. The Triangle Inequality Theorem:

According to the Triangle Inequality Theorem, in any triangle, the length of one side of the triangle is always less than the sum of the lengths of the other two sides.

Using this theorem, we can write the following inequalities:
a < b="" +="" />
b < a="" +="" />
c < a="" +="" />

3. Law of Cosines:

The Law of Cosines states that in any triangle with sides a, b, and c, and angle A opposite side a, angle B opposite side b, and angle C opposite side c, the following equation holds true:

a^2 = b^2 + c^2 - 2bc cosA
b^2 = a^2 + c^2 - 2ac cosB
c^2 = a^2 + b^2 - 2ab cosC

4. Simplifying the Law of Cosines:

Let's substitute the equations from the Law of Cosines into the inequalities from the Triangle Inequality Theorem:

b^2 + c^2 - 2bc cosA < />
a^2 + c^2 - 2ac cosB < />
a^2 + b^2 - 2ab cosC < />

5. Rearranging the Inequalities:

Let's rearrange the inequalities to isolate the cosine terms:

2bc cosA > b^2 + c^2 - a^2
2ac cosB > a^2 + c^2 - b^2
2ab cosC > a^2 + b^2 - c^2

6. Multiplying the Inequalities:

Now, let's multiply all three inequalities together:

(2bc cosA) (2ac cosB) (2ab cosC) > (b^2 + c^2 - a^2) (a^2 + c^2 - b^2) (a^2 + b^2 - c^2)

8a^2 b^2 c^2 (cosA cosB cosC) > (b^2 + c^2 - a^2) (a^2 + c^2 - b^2) (a^2 + b^2 - c^2)

7. Using the Law of Cosines again:

By substituting the equations from the Law of Cosines into the right side of the inequality, we get:

8a^2 b^2 c^2 (cosA cosB cosC) > (2bc) (2ac) (2ab)

8a^2 b^2 c^2 (cosA cosB cosC) > 8a^2 b
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The value of cosA + cosB + cosC in a triangle is less than 3/2 . Proof?
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