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--- en:proof-of-the-law-of-cosines [2026/04/01 07:01]
federico [Proof of The Law of Cosines]
+++ en:proof-of-the-law-of-cosines [2026/04/01 07:05] (current)
federico
@@ Line 1: / Line 1: @@
-====== Proof of The Law of Cosines ======
+====== Proof of the Law of Cosines ======
 {{:en:law-of-cosines1-fs8.png?nolink&}}
-**Law of Cosines**: a<sup>2</sup>  = b<sup>2</sup>  + c<sup>2</sup>  - 2bccos(θ)
+===== Statement =====
+For a triangle with sides $a$, $b$, and $c$, and angle $\theta$ between sides $b$ and $c$, the **Law of Cosines** states:
+$$
+a^2 = b^2 + c^2 - 2bc\cos(\theta)
+$$
+-----
 ===== Proof =====
-Divide the triangle into 2 right angle triangles:
+We construct the proof by splitting the triangle into two right-angled triangles.
+Drop a perpendicular from the vertex opposite side $a$ onto side $c$.
 {{:en:law-of-cosines2-fs8.png?nolink|}}
-Using trigonometry, the sides of the green triangle are
+==== Step 1: Identify the lengths ====
-  * Longest side: **a**
-  * Side on the left: **b . sin θ** ; because the sine of θ is opposite divided the hypotenuse, and the opposite is the length of this side. Then, the length of this side is the hypotenuse multiplied by the sine of θ
+Using trigonometry in the smaller right triangle:
-  * Side on the right: ***c - b cos(θ)** ; we must subtract from the length "c", the part of c that is occupied by the red triangle. We know that, in the red triangle, cos θ is the adjacent side divided the hypotenuse. The adjacent is the "base" of the red triangle, and it is cos θ multiplied by the hypotenuse. The hypotenuse is b. Then, the base of the red triangle is (cos θ)b. So, the length of the "base" of the green triangle is c - b.cos θ
+  * The vertical height (altitude) is:
+    $$
+    b \sin(\theta)
+    $$
+  * The horizontal projection of side $b$ onto side $c$ is:
+    $$
+    b \cos(\theta)
+    $$
+  * Therefore, the remaining horizontal segment is:
+    $$
+    c - b \cos(\theta)
+    $$
 {{:en:law-of-cosines3-fs8.png?nolink|}}
-Using the Pythagorean Theorem:
+==== Step 2: Apply the Pythagorean Theorem ====
+Consider the right triangle formed with:
+  * Vertical side: $b \sin(\theta)$
+  * Horizontal side: $c - b \cos(\theta)$
+  * Hypotenuse: $a$
+By the Pythagorean Theorem:
+$$
+a^2 = (b \sin(\theta))^2 + (c - b \cos(\theta))^2
+$$
+==== Step 3: Expand the expression ====
+$$
+\begin{aligned}
+a^2 &= b^2 \sin^2(\theta) + (c - b \cos(\theta))^2 \\
+\\
+&= b^2 \sin^2(\theta) + c^2 - 2bc\cos(\theta) + b^2 \cos^2(\theta)
+\end{aligned}
+$$
+Rearrange terms:
+$$
+a^2 = b^2 \sin^2(\theta) + b^2 \cos^2(\theta) + c^2 - 2bc\cos(\theta)
+$$
+==== Step 4: Use a trigonometric identity ====
+Recall the identity:
+$$
+\sin^2(\theta) + \cos^2(\theta) = 1
+$$
+Substitute:
+$$
+a^2 = b^2(1) + c^2 - 2bc\cos(\theta)
+$$
+-----
+===== Final Result =====
+$$
+a^2 = b^2 + c^2 - 2bc\cos(\theta)
+$$
+-----
+===== Why this works (Intuition) =====
-a<sup>2</sup> = (b sin θ)<sup>2</sup> + (c - b cos(θ))<sup>2</sup>
+The proof works by:
-= b<sup>2</sup> sin <sup>2</sup> θ + c<sup>2</sup> - 2cbcos(θ) + b <sup>2</sup> cos <sup>2</sup> θ
+  * Breaking the triangle into right triangles (where we can use the Pythagorean Theorem)
+  * Expressing unknown lengths using sine and cosine
+  * Reassembling everything into a single equation
-= b<sup>2</sup> (sin <sup>2</sup> θ + cos <sup>2</sup> θ) + c<sup>2</sup> - 2cbcos(θ)
+The extra term $-2bc\cos(\theta)$ accounts for the fact that the triangle is not necessarily a right triangle.
-Knowing that (sin <sup>2</sup> θ + cos <sup>2</sup> θ = 1) ((see https://en.wikibooks.org/wiki/Trigonometry/Sine_Squared_plus_Cosine_Squared))
+-----
-= b<sup>2</sup> * 1 + c<sup>2</sup> - 2cbcos(θ)
+===== Summary =====
-**a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2cbcos(θ)**
+  * The Law of Cosines generalizes the Pythagorean Theorem
+  * It works for **all triangles**, not just right-angled ones
+  * It is essential in geometry, physics, engineering, and computer graphics

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