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Chapter 5: Problem 63

What familiar formula do you obtain when you use the standard form of the Law of cosines \(c^{2}=a^{2}+b^{2}-2 a b \cos C,\) and you let \(C=90^{\circ} ?\) What is the relationship between the Law of cosines and this formula?

Short Answer

Expert verified
When the Law of Cosines is used with \(C=90^{\circ}\), it simplifies to the Pythagorean theorem, \(c^{2} = a^{2} + b^{2}\). This means the Pythagorean theorem is a specific case of the Law of Cosines for right-angled triangles.

Step by step solution

01

Insert the given value

Place the value of \(C=90^{\circ}\) into the formula for the Law of Cosines. Now we have \(c^{2}=a^{2}+b^{2}-2ab\cos90^{\circ}\).
02

Simplify the expression

It is well known that the cosine of 90 degrees, or \(\cos90^{\circ}\), is 0. Therefore, the formula becomes \(c^{2} = a^{2} + b^{2} - 2ab * 0\) which simplifies further to \(c^{2} = a^{2} + b^{2}\).
03

Identify the formula

The resulting formula, \(c^{2} = a^{2} + b^{2}\), is a familiar one. This is the Pythagorean theorem. This theorem works with right triangles, describing the relationship between the lengths of two sides and the hypotenuse.
04

Understanding the relationship

The Law of Cosines simplifies to the Pythagorean theorem when dealing with right triangles, or more specifically when \(\cos C = 0\) or when \(C = 90^{\circ}\). This reveals that the Pythagorean theorem is a special case of the Law of Cosines.

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