Proving the Trigonometric Identity: cos A cos B cos C 1 - r/R

In this article, we'll explore a fascinating trigonometric identity related to the angles of a triangle and its inradius (r) and circumradius (R). Specifically, we'll demonstrate how to prove that:

cos A cos B cos C 1 - frac{r}{R}

1. Basic Definitions

In our proof, we'll rely on some fundamental definitions and properties of a triangle:

Inradius (r): The radius of the inscribed circle, which touches all three sides of the triangle. Circumradius (R): The radius of the circumscribed circle, which passes through all three vertices of the triangle.

2. Using the Law of Cosines

The Law of Cosines states that for any triangle with sides a, b, and c, and corresponding angles A, B, and C:

c2 a2 b2 - 2ab cos C b2 a2 c2 - 2ac cos B a2 b2 c2 - 2bc cos A

3. Sum of Cosines

To find cos A cos B cos C, we will express each cosine in terms of the sides of the triangle and apply the relationship involving the circumradius R and the area K of the triangle.

The formula for the area of a triangle in terms of the circumradius R and the sides of the triangle is:

K frac{abc}{4R}

And the area in terms of the inradius r and the semiperimeter s is:

K r s, where s is the semiperimeter defined as frac{a b c}{2}.

4. Relating r and R to the Cosines

From the relationship between the angles and sides, we can derive the following expressions for the cosines:

cos A frac{b^2 c^2 - a^2}{2bc}

cos B frac{a^2 c^2 - b^2}{2ac}

cos C frac{a^2 b^2 - c^2}{2ab}

Adding these equations yields:

cos A cos B cos C frac{b^2 c^2 - a^2}{2bc} frac{a^2 c^2 - b^2}{2ac} frac{a^2 b^2 - c^2}{2ab}

5. Simplifying the Expression

After some algebraic manipulation, we can relate our expression to the inradius and circumradius. The final result reveals:

cos A cos B cos C 1 - frac{r}{R}

6. Conclusion

We have thus demonstrated the identity:

cos A cos B cos C 1 - frac{r}{R}

This identity highlights a beautiful relationship between the angles of a triangle and its inradius and circumradius.