Proof of the Law of Cosines

For a triangle with sides $a$, $b$, and $c$, and angle $\theta$ between sides $b$ and $c$, the Law of Cosines states:

$$ a^2 = b^2 + c^2 - 2bc\cos(\theta) $$

We construct the proof by splitting the triangle into two right-angled triangles.

Drop a perpendicular from the vertex opposite side $a$ onto side $c$.

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:

$$ c - b \cos(\theta) $$

Consider the right triangle formed with:

• Vertical side: $b \sin(\theta)$
• Horizontal side: $c - b \cos(\theta)$
• Hypotenuse: $a$

By the Pythagorean Theorem:

$$ a^2 = (b \sin(\theta))^2 + (c - b \cos(\theta))^2 $$

$$ \begin{aligned} a^2 &= b^2 \sin^2(\theta) + (c - b \cos(\theta))^2 \\ \\ &= b^2 \sin^2(\theta) + c^2 - 2bc\cos(\theta) + b^2 \cos^2(\theta) \end{aligned} $$

Rearrange terms:

$$ a^2 = b^2 \sin^2(\theta) + b^2 \cos^2(\theta) + c^2 - 2bc\cos(\theta) $$

Recall the identity:

$$ \sin^2(\theta) + \cos^2(\theta) = 1 $$

Substitute:

$$ a^2 = b^2(1) + c^2 - 2bc\cos(\theta) $$

$$ a^2 = b^2 + c^2 - 2bc\cos(\theta) $$

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.

• 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