Is $\sin(\alpha + \beta)$ the same as $\sin\alpha + \sin\beta$? Try $\alpha = \beta = 45^\circ$: the left side is $\sin 90^\circ = 1$, but the right side is $\sqrt{2}$. So trig functions do not distribute over sums, and we need the real rule. It comes from distances on the unit circle.
Key idea
$$\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$$ $$\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta$$ $$\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}$$
Why it works
Put $A = (\cos\alpha, \sin\alpha)$ and $B = (\cos\beta, \sin\beta)$ on the unit circle. By the distance formula, $$AB^2 = (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2 = 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta).$$
Now rotate the picture by $-\beta$. Distances do not change, and the points become $(\cos(\alpha - \beta), \sin(\alpha - \beta))$ and $(1, 0)$, whose squared distance is $2 - 2\cos(\alpha - \beta)$. Setting the two equal gives $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$. Replacing $\beta$ with $-\beta$ gives the sum formula, and using $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ gives the sine formulas. Dividing the sine formula by the cosine formula gives tangent.
Now rotate the picture by $-\beta$. Distances do not change, and the points become $(\cos(\alpha - \beta), \sin(\alpha - \beta))$ and $(1, 0)$, whose squared distance is $2 - 2\cos(\alpha - \beta)$. Setting the two equal gives $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$. Replacing $\beta$ with $-\beta$ gives the sum formula, and using $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ gives the sine formulas. Dividing the sine formula by the cosine formula gives tangent.
Worked examples
Worked example: An exact value
Find $\sin 75^\circ$ exactly.
- $75^\circ = 45^\circ + 30^\circ$.
- $$\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}.$$
- $= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Answer: $\frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966$
Worked example: A difference
Find $\tan 15^\circ$ exactly.
- $15^\circ = 45^\circ - 30^\circ$: $\tan 15^\circ = \dfrac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}} = \dfrac{3 - \sqrt{3}}{3 + \sqrt{3}}$.
- Multiply top and bottom by $3 - \sqrt{3}$: $\dfrac{(3 - \sqrt{3})^2}{9 - 3} = \dfrac{12 - 6\sqrt{3}}{6}$.
- $= 2 - \sqrt{3}$.
Answer: $2 - \sqrt{3} \approx 0.268$
Worked example: Without knowing the angles
$\sin\alpha = \frac{3}{5}$ and $\cos\beta = \frac{5}{13}$, both angles in Quadrant I. Find $\sin(\alpha + \beta)$.
- Fill in the missing values with the Pythagorean identity: $\cos\alpha = \frac{4}{5}$, $\sin\beta = \frac{12}{13}$.
- $$\sin(\alpha + \beta) = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{15}{65} + \frac{48}{65}.$$
Answer: $\frac{63}{65}$
Common mistake
Writing $\sin(\alpha + \beta) = \sin\alpha + \sin\beta$. Test with $\alpha = \beta = 45^\circ$: $\sin 90^\circ = 1$, but $\sin 45^\circ + \sin 45^\circ \approx 1.41$. Also watch the sign in the cosine formula: the sum has a minus, $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$.