Deriving the Sum and Difference Formulas

Trigonometry · Proof

Deriving the
Sum & Difference Formulas

Every formula in the family flows from a single geometric argument — the distance formula on the unit circle.

📐 4 Steps · 🔑 Distance Formula + Rotation · ⏱ 8 min read

📌 Strategy

We derive $\cos(\alpha - \beta)$ first — then everything else follows from it:

$\cos(\alpha - \beta)$

→

$\cos(\alpha + \beta)$

→

$\sin(\alpha \pm \beta)$

→

$\tan(\alpha \pm \beta)$

Step 01

Derive $\cos(\alpha - \beta)$

Setup

Place two angles $\alpha$ and $\beta$ on the unit circle. The two points are:

$$P_1 = (\cos\alpha,\ \sin\alpha) \qquad P_2 = (\cos\beta,\ \sin\beta)$$

Key Idea — Two Ways to Compute the Same Distance

Way 1 — Distance formula directly $$|P_1P_2|^2 = (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2$$

Expanding and applying $\sin^2 + \cos^2 = 1$:

$$|P_1P_2|^2 = 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) \quad \cdots (1)$$

Way 2 — Rotate so that P₂ lands on (1, 0)

After rotation, the points become $P_1' = (\cos(\alpha-\beta),\ \sin(\alpha-\beta))$ and $P_2' = (1, 0)$. The angle between the rays is still $\alpha - \beta$, so the distance is unchanged.

$$|P_1'P_2'|^2 = (\cos(\alpha-\beta) - 1)^2 + \sin^2(\alpha-\beta)$$

Expanding and applying $\sin^2 + \cos^2 = 1$:

$$|P_1'P_2'|^2 = 2 - 2\cos(\alpha - \beta) \quad \cdots (2)$$

── Since rotation does not change distance: (1) = (2) ──

$$2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 2 - 2\cos(\alpha - \beta)$$

$$\boxed{\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta}$$

Step 02

Derive $\cos(\alpha + \beta)$

Simply replace $\beta$ with $-\beta$ in the formula from Step 1:

$$\cos(\alpha + \beta) = \cos(\alpha - (-\beta)) = \cos\alpha\cos(-\beta) + \sin\alpha\sin(-\beta)$$

Apply even/odd identities — $\cos(-\beta) = \cos\beta$ and $\sin(-\beta) = -\sin\beta$:

$$\boxed{\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta}$$

Step 03

Derive $\sin(\alpha \pm \beta)$

Use the co-function identity: $\sin\theta = \cos\!\left(\dfrac{\pi}{2} - \theta\right)$

$$\sin(\alpha + \beta) = \cos\!\left(\frac{\pi}{2} - (\alpha + \beta)\right) = \cos\!\left(\left(\frac{\pi}{2} - \alpha\right) - \beta\right)$$

Apply the $\cos(\alpha - \beta)$ formula from Step 1:

$$= \cos\!\left(\frac{\pi}{2} - \alpha\right)\cos\beta + \sin\!\left(\frac{\pi}{2} - \alpha\right)\sin\beta$$

Apply co-function identities — $\cos\!\left(\dfrac{\pi}{2} - \alpha\right) = \sin\alpha$ and $\sin\!\left(\dfrac{\pi}{2} - \alpha\right) = \cos\alpha$:

$$\boxed{\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta}$$

For the minus version, replace $\beta$ with $-\beta$ and apply even/odd identities:

$$\boxed{\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta}$$

Step 04

Derive $\tan(\alpha \pm \beta)$

Use $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and substitute the results from Steps 2 and 3:

$$\tan(\alpha + \beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$$

Divide both the numerator and denominator by $\cos\alpha\cos\beta$:

$$= \frac{\dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\sin\beta}{\cos\beta}}{1 - \dfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}$$

$$\boxed{\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}}$$

Replace $\beta$ with $-\beta$ and apply $\tan(-\beta) = -\tan\beta$:

$$\boxed{\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}}$$

📋 The Complete Derivation Chain

cos(α − β)

distance formula

→ β → −β

cos(α + β)

even/odd identities

→ co-function

sin(α ± β)

co-function identity

→ ÷ cos

tan(α ± β)

quotient identity

💡

Key Takeaway: You do not need to memorize four separate derivations. The entire family of sum and difference formulas flows from just one geometric argument — the distance formula on the unit circle.