Math Core

You know $\cos\dfrac{\pi}{6}$ and $\cos\dfrac{\pi}{4}$ exactly, but what about $\cos\dfrac{\pi}{12}$ ? Trig functions don't distribute: $\cos(A - B)$ is not $\cos A - \cos B$ . The sum and difference formulas tell you what $\cos(A \pm B)$ really equals, and setting $A = B$ produces the double-angle formulas, which show up constantly in calculus and physics.

Where the formulas come from

Put angles $A$ and $B$ in standard position. Their terminal sides meet the unit circle at $P = (\cos A, \sin A)$ and $Q = (\cos B, \sin B)$ , and the angle between the two rays is $A - B$ . Now compute the squared distance $PQ^2$ two ways.

With the distance formula:

PQ^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 2 - 2(\cos A\cos B + \sin A\sin B),

using $\cos^2 + \sin^2 = 1$ twice. Now rotate the whole picture so that $Q$ lands on $(1, 0)$ . Then $P$ lands on $(\cos(A - B), \sin(A - B))$ , and the same computation gives $PQ^2 = 2 - 2\cos(A - B)$ . Rotation doesn't change distance, so the two expressions are equal:

\cos(A - B) = \cos A\cos B + \sin A\sin B.

Everything else follows. Replace $B$ with $-B$ and use even/odd symmetry to get $\cos(A + B)$ . Use cofunctions, $\sin(A + B) = \cos\left(\tfrac{\pi}{2} - A - B\right)$ , to get the sine formulas. Divide sine by cosine to get tangent.

Sum and difference formulas

\begin{aligned} \sin(A \pm B) &= \sin A\cos B \pm \cos A\sin B \\ \cos(A \pm B) &= \cos A\cos B \mp \sin A\sin B \\ \tan(A \pm B) &= \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \end{aligned}

Notice the signs. Sine keeps the sign ( $+$ goes with $+$ ) and mixes the functions ( $\sin\cos$ , $\cos\sin$ ). Cosine flips the sign and keeps the functions paired ( $\cos\cos$ , $\sin\sin$ ).

Common mistake

The most common mistake is the sign in the cosine formula. $\cos(A + B)$ has a minus in the middle: $\cos A\cos B - \sin A\sin B$ . Check yourself with $A = B = \dfrac{\pi}{2}$ : $\cos\pi = -1$ , and $0\cdot 0 - 1\cdot 1 = -1$ . It works.

Exact values of new angles

Any angle you can write as a sum or difference of $\dfrac{\pi}{6}$ , $\dfrac{\pi}{4}$ , $\dfrac{\pi}{3}$ and their relatives now has an exact value. For example, $\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$ and $\dfrac{5\pi}{12} = \dfrac{\pi}{6} + \dfrac{\pi}{4}$ .

Worked example: An exact value

Find the exact value of $\cos 15^\circ$ .

Solution. Write $15^\circ = 45^\circ - 30^\circ$ :

\begin{aligned} \cos 15^\circ &= \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ \\ &= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6 + \sqrt2}{4}. \end{aligned}

As a check, $\dfrac{\sqrt6 + \sqrt2}{4} \approx 0.966$ , just under $1$ , which is right for a small angle.

The formulas also work with angles you can't name, as long as you know their trig values. Draw or reason out the quadrant for each angle first.

Worked example: Combining two unknown angles

Suppose $\sin\alpha = \dfrac{3}{5}$ with $\alpha$ in Quadrant I, and $\cos\beta = -\dfrac{12}{13}$ with $\beta$ in Quadrant II. Find $\cos(\alpha - \beta)$ and $\tan(\alpha + \beta)$ .

Solution. First fill in the missing values. $\cos\alpha = \dfrac{4}{5}$ (positive in QI) and $\sin\beta = \dfrac{5}{13}$ (positive in QII). Then

\cos(\alpha - \beta) = \frac{4}{5}\cdot\left(-\frac{12}{13}\right) + \frac{3}{5}\cdot\frac{5}{13} = \frac{-48 + 15}{65} = -\frac{33}{65}.

For the tangent, $\tan\alpha = \dfrac{3}{4}$ and $\tan\beta = -\dfrac{5}{12}$ :

\tan(\alpha + \beta) = \frac{\frac{3}{4} - \frac{5}{12}}{1 - \frac{3}{4}\cdot\left(-\frac{5}{12}\right)} = \frac{\frac{4}{12}}{1 + \frac{15}{48}} = \frac{\frac{1}{3}}{\frac{63}{48}} = \frac{16}{63}.

Read backward, the formulas also compress expressions. For example, $\sin 50^\circ\cos 20^\circ - \cos 50^\circ\sin 20^\circ$ matches $\sin(A - B)$ , so it equals $\sin 30^\circ = \dfrac{1}{2}$ .

Double-angle formulas

Set $B = A$ in the sum formulas:

Double-angle formulas

\sin 2A = 2\sin A\cos A, \qquad \tan 2A = \frac{2\tan A}{1 - \tan^2 A}

\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A

The three versions of $\cos 2A$ are all the same thing, rewritten with $\sin^2 A + \cos^2 A = 1$ . Pick the one that matches what you know: if you're given only $\sin A$ , use $1 - 2\sin^2 A$ and you won't need cosine at all.

Worked example: Double angles from one value

If $\tan\theta = -\dfrac{3}{4}$ and $\theta$ is in Quadrant II, find $\sin 2\theta$ , $\cos 2\theta$ and the quadrant of $2\theta$ .

Solution. A reference triangle with legs $3$ and $4$ has hypotenuse $5$ . In Quadrant II, $\sin\theta = \dfrac{3}{5}$ and $\cos\theta = -\dfrac{4}{5}$ .

\sin 2\theta = 2\cdot\frac{3}{5}\cdot\left(-\frac{4}{5}\right) = -\frac{24}{25}, \qquad \cos 2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.

Sine negative and cosine positive puts $2\theta$ in Quadrant IV. That's consistent: $\theta$ is between $\dfrac{\pi}{2}$ and $\pi$ , and since $\tan\theta = -\dfrac{3}{4}$ is closer to $0$ than $-1$ , $\theta$ is actually past $\dfrac{3\pi}{4}$ , so $2\theta$ is between $\dfrac{3\pi}{2}$ and $2\pi$ .

The double-angle formulas are also the key to many identity proofs. To verify $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$ , start on the left and choose the form of $\cos 2x$ that cancels the $1$ :

\frac{1 - (1 - 2\sin^2 x)}{2\sin x\cos x} = \frac{2\sin^2 x}{2\sin x\cos x} = \frac{\sin x}{\cos x} = \tan x.

Had you picked $\cos^2 x - \sin^2 x$ instead, the proof would still work but would take an extra step. Choosing the right version of $\cos 2A$ is a skill worth practicing: use $1 - 2\sin^2 A$ next to a $1 -$ , use $2\cos^2 A - 1$ next to a $1 +$ , and use $\cos^2 A - \sin^2 A$ when you want to factor a difference of squares. You can also apply the sum formula twice to reach higher multiples. For instance, $\sin 3x = \sin(2x + x) = \sin 2x\cos x + \cos 2x\sin x$ , which becomes $3\sin x - 4\sin^3 x$ after you expand and replace $\cos^2 x$ with $1 - \sin^2 x$ .

Power reducing and half angles

Solve the last two forms of $\cos 2A$ for the squares and you get formulas that trade a square for a double angle. Calculus uses these to integrate $\sin^2 x$ and $\cos^2 x$ .

\sin^2 A = \frac{1 - \cos 2A}{2}, \qquad \cos^2 A = \frac{1 + \cos 2A}{2}.

Replace $A$ by $\dfrac{\theta}{2}$ and take square roots to get the half-angle formulas:

\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}, \qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}},

where the sign is decided by the quadrant of $\dfrac{\theta}{2}$ , not of $\theta$ . For example, $\dfrac{\pi}{8}$ is half of $\dfrac{\pi}{4}$ and lies in Quadrant I, so

\cos\frac{\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt2}{2}}{2}} = \frac{\sqrt{2 + \sqrt2}}{2}.

Tip

When a problem mixes $\sin 2x$ with $\sin x$ or $\cos x$ , rewrite everything in terms of the single angle $x$ first. Mixed angles are the main thing that makes identities and equations look harder than they are.