====== Proof of the Law of Cosines ====== {{:en:law-of-cosines1-fs8.png?nolink&}} ===== 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 ===== 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|}} ==== Step 1: Identify the lengths ==== Using trigonometry in the smaller right triangle: * 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|}} ==== 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) ===== The proof works by: * 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 The extra term $-2bc\cos(\theta)$ accounts for the fact that the triangle is not necessarily a right triangle. ----- ===== Summary ===== * 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