Problem 64 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use a half-angle formula and the Law of cosines to show that, for any triangle, (a) \(\cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{a b}}\) and (b) \(\sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}\) where \(s=\frac{1}{2}(a+b+c)\).

Step-by-Step Solution

Answer

(a) \( \cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}} \), (b) \( \sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}} \)

Step 1: State the half-angle formulas

The half-angle identities for a given angle \( C \) are \[\cos\frac{C}{2} = \pm \sqrt{\frac{1+\cos C}{2}} \]and\[\sin\frac{C}{2} = \pm \sqrt{\frac{1-\cos C}{2}}\]The signs depend on the quadrant where the half angle lies. Here, angle \( C \) of a triangle is always acute, so the half-angle will lie in the first quadrant where both sine and cosine are positive. Hence we can use the positive square root.

Step 2: State the Law of Cosines

For any triangle with sides \( a, b, \) and \( c, \) and the angle \( C \) opposite side \( c, \) the Law of Cosines states that\[c^2 = a^2 + b^2 - 2ab \cos C\]Rearranging for \( \cos C, \) we get\[\cos C = \frac{a^2 + b^2 - c^2}{2ab}\]

Step 3: Derive \(\cos \frac{C}{2}\) formula

Substitute the expression for \( \cos C \) into the half-angle formula for \( \cos \frac{C}{2} \).\[\cos\frac{C}{2} = \sqrt{\frac{1 + \frac{a^2 + b^2 - c^2}{2ab}}{2}}\]Simplify to get\[\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}\]where \( s \) is the semi-perimeter of the triangle, \( s = \frac{a+b+c}{2} \).

Step 4: Derive \(\sin \frac{C}{2}\)

Substitute the expression for \( \cos C \) into the half-angle formula for \( \sin \frac{C}{2} \).\[\sin\frac{C}{2} =\sqrt{\frac{1- \frac{a^2 + b^2 - c^2}{2ab}}{2}}\]Simplify to get\[\sin\frac{C}{2} =\sqrt{\frac{(s-a)(s-b)}{ab}}.\]

Key Concepts

Law of Cosines

The Law of Cosines is a fundamental relation in trigonometry that extends the Pythagorean Theorem to any triangle, not just right triangles. It helps us find the unknown side of a triangle when we know the other two sides and the angle between them. The formula for the Law of Cosines is:

• \[ c^2 = a^2 + b^2 - 2ab \cos C \]

In this equation, \(c\) is the side opposite the angle \(C\), and \(a\) and \(b\) are the other two sides of the triangle. The Law of Cosines becomes especially useful in non-right triangles where traditional trigonometric identities might not directly apply. By rearranging the equation, we can solve for \(\cos C\):

• \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

This rearrangement is often used with half-angle formulas to derive expressions for the half angles, which are necessary for solving complex triangle problems with non-standard angles.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved angles. Among these identities, the half-angle formulas are particularly useful in simplifying trigonometric expressions. These identities allow us to find the sine or cosine of half of an angle when the full angle's trigonometric value is known:

• Cosine Half-Angle: \[ \cos\frac{C}{2} = \pm \sqrt{\frac{1+\cos C}{2}} \]
• Sine Half-Angle: \[ \sin\frac{C}{2} = \pm \sqrt{\frac{1-\cos C}{2}} \]

In any triangle, the angle \(C\) is typically acute or obtuse, and by evaluating the quadrant of \(\frac{C}{2}\), we determine the correct sign. For triangles, the half-angle typically lies in the first quadrant, making both sine and cosine positive. These identities simplify calculations and are instrumental in solving problems involving half-angles, as seen in deriving expressions involving the Law of Cosines.

Triangle Perimeter

The perimeter of a triangle is a straightforward concept, referring to the total length of its three sides. However, the concept of the semi-perimeter is of special interest in trigonometry and geometry, especially when using trigonometric formulas. The semi-perimeter, denoted as \(s\), is calculated as:

• \[ s = \frac{a + b + c}{2} \]

This value is significant in various formulas, such as Heron's formula for the area of a triangle and in deriving expressions for the half-angle formulas. In the half-angle derivations of the form \(\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}\), \(s\) helps break down the relationships between the triangle's angles and sides into more manageable components. Expressing relationships through \(s\) allows for simpler spellings of complex relationships that involve proportions, thus aiding in understanding the geometric properties of triangles.