Compound, Multiple and Sum-to-Product Formulas

The for­mu­las you will use every day#

Let us clear up the most com­mon mis­take first: $\cos(x + y)$ is not $\cos x + \cos y$. The cor­rect for­mu­las, for sums and dif­fer­ences of angles, for dou­ble and triple angles, and for turn­ing sums of sines and cosines into prod­ucts, are the every­day tools of this chap­ter. You will use them again and again, in cal­cu­lus as well. We will state them, see where the impor­tant ones come from, and put them to work find­ing exact val­ues and prov­ing iden­ti­ties.

Sum and dif­fer­ence for­mu­las#

$$\cos(x + y) = \cos x \cos y - \sin x \sin y, \qquad \cos(x - y) = \cos x \cos y + \sin x \sin y$$
$$\sin(x + y) = \sin x \cos y + \cos x \sin y, \qquad \sin(x - y) = \sin x \cos y - \cos x \sin y$$
$$\displaystyle \tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}, \qquad \cot(x \pm y) = \frac{\cot x \cot y \mp 1}{\cot y \pm \cot x}$$

You don't need to mem­o­rise six sep­a­rate proofs. Only the cosine dif­fer­ence for­mula needs real work: com­pare the chord join­ing the unit-cir­cle points at angles $x$ and $y$ with the chord from $(1, 0)$ to the point at angle $x - y$. The two chords are the same length, and writ­ing that out gives the for­mula. Every­thing else fol­lows by clever sub­sti­tu­tion ($y \to -y$, $\displaystyle x \to \tfrac{\pi}{2} - x$).

Con­se­quences. The famil­iar allied-angle results drop straight out: $\displaystyle \cos\left(\tfrac{\pi}{2} - x\right) = \sin x$, $\sin(\pi - x) = \sin x$, $\cos(\pi + x) = -\cos x$ and $\displaystyle \sin\left(\tfrac{3\pi}{2} + x\right) = -\cos x$.

Dou­ble and triple angles#

$$\displaystyle \sin 2x = 2\sin x \cos x = \frac{2\tan x}{1 + \tan^2 x}$$
$$\displaystyle \cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$$
$$\displaystyle \tan 2x = \frac{2\tan x}{1 - \tan^2 x}$$
$$\displaystyle \sin 3x = 3\sin x - 4\sin^3 x, \quad \cos 3x = 4\cos^3 x - 3\cos x, \quad \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$

Half angles come from read­ing the $\cos 2x$ for­mu­las back­wards: $\displaystyle \sin^2 \tfrac{x}{2} = \tfrac{1 - \cos x}{2}$ and $\displaystyle \cos^2 \tfrac{x}{2} = \tfrac{1 + \cos x}{2}$. The square root leaves a sign to decide, and the quad­rant of the half angle decides it.

Sum-to-prod­uct and prod­uct-to-sum#

$$\displaystyle \sin x + \sin y = 2\sin\tfrac{x + y}{2}\cos\tfrac{x - y}{2}, \qquad \sin x - \sin y = 2\cos\tfrac{x + y}{2}\sin\tfrac{x - y}{2}$$
$$\displaystyle \cos x + \cos y = 2\cos\tfrac{x + y}{2}\cos\tfrac{x - y}{2}, \qquad \cos x - \cos y = -2\sin\tfrac{x + y}{2}\sin\tfrac{x - y}{2}$$

Going the other way, prod­ucts turn back into sums: $2\sin x \cos y = \sin(x + y) + \sin(x - y)$, $2\cos x \cos y = \cos(x + y) + \cos(x - y)$ and $2\sin x \sin y = \cos(x - y) - \cos(x + y)$. When­ever a ques­tion mixes sums of sines or cosines with a frac­tion, try these first.

Prac­tice ques­tions#

1. Find exactly: $\sin 15^\circ$; $\cos 105^\circ$; $\tan 75^\circ$.
2. Prove: $\displaystyle \cos\left(\tfrac{\pi}{4} + x\right) + \cos\left(\tfrac{\pi}{4} - x\right) = \sqrt{2}\cos x$.
3. Prove: $\displaystyle \frac{\tan(\tfrac{\pi}{4} + x)}{\tan(\tfrac{\pi}{4} - x)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2$.
4. $\displaystyle \tan A = \tfrac{1}{2}$, $\displaystyle \tan B = \tfrac{1}{3}$. Show $\displaystyle A + B = \tfrac{\pi}{4}$ (both acute).
5. $\displaystyle \sin x = \tfrac{12}{13}$, $x$ in II. Find $\sin 2x$, $\cos 2x$, $\tan 2x$.
6. $\displaystyle \tan x = \tfrac{3}{4}$, $\displaystyle \pi < x < \tfrac{3\pi}{2}$. Find $\displaystyle \sin \tfrac{x}{2}$, $\displaystyle \cos \tfrac{x}{2}$.
7. Prove: $\displaystyle \frac{\sin 5x - \sin 3x}{\cos 5x + \cos 3x} = \tan x$.
8. Prove: $\cos 4x = 1 - 8\sin^2 x \cos^2 x$.
9. Prove: $\displaystyle \sin 3x + \sin 2x - \sin x = 4\sin x \cos\tfrac{x}{2}\cos\tfrac{3x}{2}$.
10. Find $\sin 18^\circ$ using $\sin 2\theta = \cos 3\theta$ for $\theta = 18^\circ$.

Answers#