Prove that cosA + cosB + cosC is always positive in triangle ABC.

Solution:

Let's understand how we arrived at the proof.

Trigonometry is the branch of maths that deals with the relationships between the sides and angles of different types of triangles. There are six trigonometric ratios and six inverse ratios.

We use various properties of trigonometry to solve this question.

At the LHS, we have cosA + cosB + cosC.

Therefore, LHS = cosA + cosB + cosC

Now, we apply the formula for (cos A + cos B) using the standard trigonometric identity:

⇒ { 2 cos [(A + B) / 2] · cos [(A - B) / 2] } + cos C

Now, we know that A + B + C = π (angle sum property of triangle). Hence, A + B = π - C

⇒ { 2 cos [(π/2) – (C/2)] · cos [(A - B) / 2] } + cos C

Next, we use the trigonometric identity: cos(π/2 - A) = sin A

⇒ 2sin(C/2) · cos [(A - B) / 2] } + cos C

Now, we use trigonometric identity: cos2A = 1 – 2sin²(A), to expand cosC

⇒ 2sin(C/2) · cos [(A - B) / 2] } + 1 – 2sin²(C/2)

Taking 2sin(C/2) common in the above equation:

⇒ 1 + 2 sin (C/2) · { cos [(A - B) / 2] – sin (C/2) }

⇒ 1 + 2 sin (C/2) · { cos [(A - B) / 2] – sin [(π/2) – ( (A + B) / 2 )] } (angle sum property)

⇒ 1 + 2 sin (C/2) · { cos [(A - B) / 2] – cos [(A + B) / 2] }

Using the identity cosA – cos B = 2sin [(A + B) / 2] · sin [(A - B) / 2]:

⇒ 1 + 2 sin (C/2) · 2 sin (A/2) · sin(B/2)

⇒ 1 + 4 sin(A/2) sin(B/2) sin(C/2)

We know that the sine function is always positive in the range [0, π/2].

Hence, the above expression is always positive for 0 < A, B, C < π.

Thus, it is proved that cosA + cosB + cosC is always positive in any triangle ABC.

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Prove that cosA + cosB + cosC is always positive in triangle ABC.

Summary:

It is proved that cosA + cosB + cosC is always positive in any triangle ABC.