Sum, difference and multiple-angle formulas

Once we know $\sin\theta$ and $\cos\theta$ for individual angles, we want to compute them for sums, differences, and multiples. The addition formulas , the most important identities in trigonometry , answer this question.

The two master formulas

$\boxed{\cos(A - B) = \cos A \cos B + \sin A \sin B.}$

From this, every other addition formula follows:

• Replace $B$ with $-B$: $\cos(A + B) = \cos A \cos B - \sin A \sin B$.
• Replace $A$ with $\dfrac{\pi}{2} - A$: $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and $\sin(A - B) = \sin A \cos B - \cos A \sin B$.
• Divide to get $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$.

Derivation of $\cos(A - B) = \cos A \cos B + \sin A \sin B$

Consider two points on the unit circle: $P_1 = (\cos A, \sin A)$ and $P_2 = (\cos B, \sin B)$. The angle between them at the origin is $A - B$, so by the distance formula: $|P_1 P_2|^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2.$ Expanding, $= \cos^2 A - 2\cos A \cos B + \cos^2 B + \sin^2 A - 2\sin A \sin B + \sin^2 B$ $= 2 - 2(\cos A \cos B + \sin A \sin B).$

Alternatively, by the law of cosines on the triangle with two sides of length $1$ enclosing angle $A - B$: $|P_1 P_2|^2 = 1 + 1 - 2\cos(A - B) = 2 - 2\cos(A - B).$ Equating: $\cos(A - B) = \cos A \cos B + \sin A \sin B. \qed$

Multiple-angle (double-angle) formulas

Putting $A = B = \theta$ in the addition formulas:

$\sin 2\theta = 2 \sin\theta \cos\theta.$

$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta.$

$\tan 2\theta = \frac{2 \tan\theta}{1 - \tan^2\theta}.$

Rearranging $\cos 2\theta$: $\sin^2\theta = \frac{1 - \cos 2\theta}{2},\qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2}.$

These power-reducing formulas are vital for integrating $\sin^2$ and $\cos^2$ in Class XII.

Half-angle formulas

Replace $\theta$ by $\theta/2$: $\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}, \qquad \cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}, \qquad \tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}.$

Triple-angle formulas

$\sin 3\theta = 3\sin\theta - 4\sin^3\theta.$ $\cos 3\theta = 4\cos^3\theta - 3\cos\theta.$

Proof of the first: $\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta = 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta = 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta$.

The "harmonic combination" formula

Any expression $a\sin\theta + b\cos\theta$ can be written as $a\sin\theta + b\cos\theta = R\sin(\theta + \phi), \quad R = \sqrt{a^2 + b^2}, \quad \tan\phi = \frac{b}{a},$ or equivalently as $R\cos(\theta - \psi)$ for some $\psi$. Hence the max and min of $a\sin\theta + b\cos\theta$ are $\pm \sqrt{a^2 + b^2}$.

Worked examples

Example 1. Find $\sin 75^\circ$.

$75^\circ = 45^\circ + 30^\circ$. So $\sin 75^\circ = \sin 45 \cos 30 + \cos 45 \sin 30 = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Example 2. If $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{12}{13}$, both $A, B$ in quadrant I, find $\sin(A + B)$.

$\cos A = \dfrac{4}{5}$, $\sin B = \dfrac{5}{13}$. $\sin(A + B) = \dfrac{3}{5} \cdot \dfrac{12}{13} + \dfrac{4}{5} \cdot \dfrac{5}{13} = \dfrac{36 + 20}{65} = \dfrac{56}{65}$.

Example 3. Prove $\sin 2\theta = \dfrac{2\tan\theta}{1 + \tan^2\theta}$.

$\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \dfrac{\sin\theta}{\cos\theta} \cdot \cos^2\theta = 2\tan\theta \cdot \dfrac{1}{\sec^2\theta} = \dfrac{2\tan\theta}{1 + \tan^2\theta}$. $\qed$

Example 4. Maximum value of $5\sin\theta + 12\cos\theta$.

$R = \sqrt{25 + 144} = 13$. Max value $= 13$.

Example 5 (harder). If $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$, prove $\cos 3A + \cos 3B + \cos 3C = 3\cos(A + B + C)$.

Set $z_k = \cos\theta_k + i \sin\theta_k$ for $k = 1, 2, 3$ (using Chapter 4's notation). The hypothesis says $z_1 + z_2 + z_3 = 0$. Each $|z_k| = 1$, and the identity $z_1^3 + z_2^3 + z_3^3 - 3 z_1 z_2 z_3 = (z_1 + z_2 + z_3)(z_1^2 + z_2^2 + z_3^2 - z_1 z_2 - z_2 z_3 - z_3 z_1)$ gives $z_1^3 + z_2^3 + z_3^3 = 3 z_1 z_2 z_3$. Taking real parts: $\cos 3A + \cos 3B + \cos 3C = 3\cos(A + B + C)$. $\qed$

(This example previews Chapter 4. A purely trigonometric proof exists but is longer.)

Try it yourself

1. Compute $\cos 15^\circ$, $\sin 105^\circ$, $\tan 75^\circ$.
2. If $\sin\theta = \dfrac{4}{5}$, $\theta$ in quadrant I, find $\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$.
3. Prove $\dfrac{\sin 2A}{1 + \cos 2A} = \tan A$.
4. Find $\sin(A - B)$ if $\cos A = \dfrac{3}{5}$, $\sin B = \dfrac{12}{13}$ (both in quadrant I).
5. Prove $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.
6. Express $\sin 3\theta$ in terms of $\sin\theta$.
7. Find max and min of $f(x) = 7\sin x - 24\cos x$.
8. If $\tan A = \dfrac{1}{2}, \tan B = \dfrac{1}{3}$, find $\tan(A + B)$.
9. Prove $\sin\!\left(\dfrac{\pi}{4} + \theta\right) - \sin\!\left(\dfrac{\pi}{4} - \theta\right) = \sqrt{2}\sin\theta$.
10. Prove $\sin^2\!\dfrac{\theta}{2} = \dfrac{1 - \cos\theta}{2}$.
11. Find $\tan\!\dfrac{\pi}{8}$.
12. Prove $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ when $A + B + C = \pi$.

Pitfalls / Tricks

• $\cos(A + B) \ne \cos A + \cos B$. Always use the formula.
• $\sin 2\theta \ne 2 \sin\theta$. The factor $\cos\theta$ matters.
• For triple-angle, $\sin 3\theta$ has $3\sin - 4\sin^3$ (mind the sign), while $\cos 3\theta = 4\cos^3 - 3\cos$.
• Insight. All the "$R$-sine" computations work because $(\cos\phi, \sin\phi)$ traces the unit circle. Recognising $a\sin + b\cos$ as a single sinusoid is the most-used trick in JEE trigonometry.