# cosA+cosB+cosC<=2/3, in a scalene triangle

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### 2 Answers

This statement is not true, we can disprove it with an example.

Let the angles of our triangle be equal to 90, 89, and 1 degrees.

Cos(1) + cos(90) + cos(89) = .9998 + 0 + .0174 = 1.0173.

1.0173 is greater than 2/3, so this statement is not true.

You need to remember that the sum of angles in triangle, of any type, is `pi.`

`A + B + C = pi => A+B = pi - C => (A+B)/2 = pi/2 - C/2`

`cos (A+B) = cos(pi - C) = -cos C`

`cos (A+B)/2 = cos (pi/2 - C/2) = sin C/2`

You need to transform the sum cos A + cos B in a product, such that:

`cos A + cos B = 2cos (A+B)/2*cos (A - B)/2`

Replacing sin C/2 for c`os (A+B)/2` in `2cos (A+B)/2*cos (A - B)/2` , yields:

`2sin C/2*cos (A - B)/2 + cos C <= 2/3`

You need to consider cos C as `cos 2*(C/2),` hence, using the formula of cosine of double angle, yields:

`cos 2*(C/2) = 1 - 2sin^2(C/2)`

`2sin C/2*cos (A - B)/2 +1 - 2sin^2(C/2) <= 2/3`

Subtracting 1 both sides, yields:

`2sin C/2*cos (A - B)/2 - 2sin^2(C/2) <= 2/3 - 1`

You need to factor out `2sin(C/2),` yields:

`2sin(C/2) (cos (A - B)/2 - sin(C/2)) <= -1/3`

`sin(C/2) (cos (A - B)/2 - cos (A+B)/2) <= -1/6`

You need to transform the difference `cos (A - B)/2 - cos (A+B)/2` in a product, such that:

`-sin(C/2) (2 sin (A-B+A+B)/4 sin (A-B-A-B)/4) <= -1/6`

`2sin(A/2)sin(B/2)sin(C/2) <= -1/6`

`sin(A/2)sin(B/2)sin(C/2) <= -1/12`

Since c`os A cos B + cos C <= 1 + 1 + 1 = 3` , yields:

`( cos A cos B + cos C ) <=3`

But `cos A cos B + cos C = sin(A/2)sin(B/2)sin(C/2) <= -1/12` , hence, replacing `sin(A/2)sin(B/2)sin(C/2) <= -1/12 ` for c`os A cos B + cos C` , yields:

`sin(A/2)sin(B/2)sin(C/2) <= -1/12 <= 3`

**Hence, evaluating if the statement cos A cos B + cos C <= 2/3 holds, yields** **`sin(A/2)sin(B/2)sin(C/2) <= -1/12 <= 3` , that is a tvalid inequality.**