Law of cosines — proof and applications: Visual walkthrough

Step 1 — Draw the triangle and name its parts

WHAT. A triangle is three corners joined by three straight edges. We label the corners with capital letters $A$ , $B$ , $C$ , and each edge with the small letter of the corner it does not touch. So side $a$ sits opposite corner $A$ , side $b$ opposite $B$ , side $c$ opposite $C$ .

WHY. This "opposite" pairing is the whole grammar of the formula. The result will always tie one lone side to the single angle facing it. If we name loosely now, everything downstream mismatches (that is the classic error the parent note warns about).

PICTURE. Corner $C$ is where two sides, $a$ and $b$ , meet — the "hinge". The angle at that hinge is also called $C$ . The side that bridges the gap across from the hinge is $c$ .

Figure — Law of cosines — proof and applications

Step 2 — Slide the triangle onto graph paper

WHAT. We place corner $C$ exactly at the origin $(0, 0)$ of an $x$ – $y$ grid, and lay arm $b$ flat along the positive $x$ -axis.

WHY. A bare triangle floating in space has no numbers. Graph paper gives every corner an address — a pair $(x, y)$ of horizontal and vertical distances. Once corners have addresses, we can compute the length $c$ with arithmetic instead of guesswork. Choosing $C$ at the origin and $b$ along the axis is just picking the simplest possible addresses; the triangle's shape is untouched by sliding and rotating it.

PICTURE. Arm $b$ runs straight right, so corner $A$ lands at horizontal distance $b$ , zero height: $A = (b, 0)$ . Corner $C$ is the origin: $C = (0, 0)$ .

Step 3 — Find the address of the third corner $B$

WHAT. Corner $B$ is arm $a$ away from the origin, but tilted up by the hinge angle $C$ . Its address is $B = (a cos C, a sin C)$ .

WHY THIS TOOL — cosine and sine. We need to turn "distance $a$ , direction $C$ " into " $x$ across, $y$ up". That translation is what $cos$ and $sin$ do, and no other pair of functions does it. Picture a clock hand of length $a$ starting flat and swinging up by angle $C$ :

Its shadow on the floor (horizontal reach) shrinks as it tilts up — that shadow is $a cos C$ . Here $cos C = \frac{across}{hand length}$ .
Its height grows as it tilts up — that height is $a sin C$ , where $sin C = \frac{up}{hand length}$ .

We pick $cos$ / $sin$ (not $tan$ ) because we want the horizontal and vertical lengths separately, not their ratio. This is exactly the polar→cartesian idea from the Unit Circle and Cosine Values.

PICTURE. Drop a dashed vertical from $B$ to the $x$ -axis. That builds a right triangle whose bottom (across) is $a cos C$ and whose upright (up) is $a sin C$ .

Step 4 — Measure the gap $c$ with the distance rule

WHAT. $c$ is the straight-line distance between corners $A = (b, 0)$ and $B = (a cos C, a sin C)$ . The distance rule (Pythagoras applied to the gap's own right triangle) gives

$c^{2} = (run)^{2} + (rise)^{2} = (a cos C - b)^{2} + (a sin C - 0)^{2} .$

WHY THIS TOOL — the distance formula. Two known addresses, one unknown length between them: this is precisely what Pythagoras Theorem answers. The run is the horizontal difference of the two addresses; the rise is the vertical difference. Square each, add, and $c^{2}$ is the hypotenuse-squared of the little right triangle formed by the run and rise.

PICTURE. Draw the run (a horizontal segment from under $B$ across to $A$ ) and the rise (the vertical drop from $B$ ). $c$ is the slanted line closing that right triangle.

Step 5 — Multiply out and let the Pythagorean identity collapse it

WHAT. Expand both squared brackets, then gather the pieces.

WHY. Inside the mess a hidden simplification is waiting: the terms $a^{2} cos^{2} C$ and $a^{2} sin^{2} C$ can merge, because $cos^{2} C + sin^{2} C = 1$ always (that identity is just Pythagoras applied to the unit circle — see Unit Circle and Cosine Values). Merging them is the step that makes the formula short and memorable.

PICTURE. Think of the terms sorted into three bins: the $a^{2}$ -bin (which will shrink to a single $a^{2}$ ), the lone $b^{2}$ , and the cross-term $- 2 ab cos C$ .

Expanding: $c^{2} = from run a^{2} cos^{2} C cross term - 2 ab cos C + from run b^{2} + from rise a^{2} sin^{2} C$

Group the two $a^{2}$ pieces and use the identity: $c^{2} = a^{2} = 1 (cos^{2} C + sin^{2} C) + b^{2} - 2 ab cos C = a^{2} + b^{2} - 2 ab cos C ■$

WHAT. The single term $- 2 ab cos C$ decides whether $c$ is shorter, equal, or longer than the Pythagorean value. Its sign is controlled entirely by $cos C$ .

WHY. $a$ , $b$ are lengths so $2 ab > 0$ always. Thus the sign of the whole correction is the opposite of the sign of $cos C$ . We must check all three angle regimes so the reader never meets an unshown case.

PICTURE. Walk the hinge open from narrow to wide and watch $cos C$ on the unit circle:

Hinge angle $C$	$cos C$	Correction $- 2 ab cos C$	Effect on gap $c$
Acute, $0 < C < 90°$	positive	negative	$c$ shrinks below $a^{2} + b^{2}$
Right, $C = 90°$	$0$	zero	$c = a^{2} + b^{2}$ — pure Pythagoras
Obtuse, $90° < C < 180°$	negative	positive	$c$ grows above $a^{2} + b^{2}$

Step 7 — The degenerate limits (fully flat hinges)

WHAT. Push the hinge to its two extremes: $C = 0°$ (arms folded onto each other) and $C = 180°$ (arms stretched into one straight line).

WHY. A rule you trust must survive its boundary cases. These are where "triangle" collapses to a line — a good stress-test that the formula still says something sensible.

PICTURE.

$C = 0°$ : arms point the same way. $cos 0° = 1$ , so $c^{2} = a^{2} + b^{2} - 2 ab = (a - b)^{2}$ , giving $c = ∣ a - b ∣$ — the gap is just the difference of the arm lengths. Correct: overlapping sticks leave a stub of length $∣ a - b ∣$ .
$C = 180°$ : arms point opposite ways. $cos 180° = - 1$ , so $c^{2} = a^{2} + b^{2} + 2 ab = (a + b)^{2}$ , giving $c = a + b$ — the arms lie end to end, gap is the sum.

The one-picture summary

Everything above, in a single frame: hinge $C$ at the origin, arms $a$ and $b$ , the third corner $B$ addressed by $(a cos C, a sin C)$ , the run–rise right triangle giving $c$ , and the final equation with its two halves colour-coded — Pythagoras part plus correction.

Recall Feynman retelling — the whole walkthrough in plain words

I put the sharp corner of my triangle at the middle of graph paper and laid one arm flat. Now that corner sits at $(0, 0)$ and the flat arm's tip sits at $(b, 0)$ — easy addresses. The other arm has length $a$ but leans up by the corner's angle, so its tip is $a$ -across-and-up: $(a cos C, a sin C)$ , because cosine is the flat shadow and sine is the height. The side I want, $c$ , just joins those two tips, and the distance between two points is Pythagoras on the horizontal gap and the vertical gap. When I multiply that out, an $a^{2} cos^{2}$ and an $a^{2} sin^{2}$ appear, and since those always add to $a^{2}$ , they fuse. What's left is $a^{2} + b^{2}$ (plain Pythagoras) minus a leftover $2 ab cos C$ . That leftover is the only thing carrying the angle: a narrow corner makes cosine positive so it shortens $c$ ; a right corner makes cosine zero so it vanishes; a wide corner makes cosine negative so it lengthens $c$ . Fold the arms fully and I get $∣ a - b ∣$ ; stretch them into a line and I get $a + b$ . That's the whole law, seen not memorised.

Recall check

Recall Quick self-test

Why is corner $B$ at $(a cos C, a sin C)$ ? ::: Arm $a$ leaves the origin at angle $C$ ; cosine gives its horizontal reach, sine its height. Which single term carries the angle? ::: The correction $- 2 ab cos C$ ; everything else is $a^{2} + b^{2}$ . What identity collapses the two $a^{2}$ terms? ::: $cos^{2} C + sin^{2} C = 1$ . What does $c$ become at $C = 0°$ and $C = 180°$ ? ::: $∣ a - b ∣$ and $a + b$ respectively. Why does an obtuse hinge give a longer $c$ ? ::: $cos C < 0$ , so $- 2 ab cos C > 0$ adds length.

Connections

Pythagoras Theorem — used twice: in the distance step and inside $cos^{2} + sin^{2} = 1$ .
Unit Circle and Cosine Values — why $B = (a cos C, a sin C)$ and why $cos$ flips sign past $90°$ .
Dot Product — the same $∣ u ∣∣ v ∣ cos θ$ correction, in vector clothing.
Law of Sines · Solving Triangles — where this result plugs into the full toolkit.