Question

the law of cosines
follow these steps to derive the law of cosines.

1. in △abd, the trigonometric function \\(\cos(a) = \frac{x}{c}\\)
2. multiply both sides of the equation in step 5 by \\(c\\) to get \\(x = c\cos(a)\\).
3. substitute \\(c\cos(a)\\) for the variable \\(x\\) in the equation \\(a^2 = b^2 - 2bx + c^2\\) to produce \\(a^2 = b^2 - 2bc\cos(a) + c^2\\).
4. the law of cosines is obtained by rearranging/solving the previous equation. \\(a^2 = b^2 + c^2 - 2bc\cos\dots\\)

image of triangle (abc) with (a), (d), (c) on the base, (b) at the top, height (h), segments (x), (b-x), sides (c), (a), base (b) labeled, and a right angle at (d).

Explanation:

Step1: Analyze the equation transformation

We have the equation \(a^{2}=b^{2}-2bx + c^{2}\) and after substitution we get \(a^{2}=b^{2}-2bc\cos(A)+c^{2}\). Now we need to get to the form \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\). This involves rearranging the terms (since we are just re - ordering the terms \(b^{2}\), \(c^{2}\) and \(- 2bc\cos(A)\) without solving for a variable in the traditional sense of isolating a variable. Solving would imply finding a value for a variable, but here we are just re - arranging the terms of the equation).

Step2: Determine the operation

To get from \(a^{2}=b^{2}-2bc\cos(A)+c^{2}\) to \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), we are rearranging the terms. The term \(-2bc\cos(A)\) and \(c^{2}\) (along with \(b^{2}\)) are re - ordered. So the operation is rearranging.

Answer:

rearranging