Math Core

You know the exact values of sine and cosine at $30^\circ$ , $45^\circ$ and $60^\circ$ . But what about $75^\circ$ or $15^\circ$ ? Since $75^\circ = 45^\circ + 30^\circ$ , it would be convenient if you could build $\sin 75^\circ$ out of values you already know. The sum and difference formulas do exactly that.

A tempting mistake

It is natural to guess that $\sin(A + B) = \sin A + \sin B$ . Test it with $A = B = 90^\circ$ :

\sin(90^\circ + 90^\circ) = \sin 180^\circ = 0, \qquad \text{but} \qquad \sin 90^\circ + \sin 90^\circ = 2.

The guess is false. Trig functions don't distribute over addition. The true formulas mix sines and cosines together.

Where the cosine formula comes from

Place two angles $A$ and $B$ in standard position. Their terminal sides meet the unit circle at $P = (\cos A, \sin A)$ and $Q = (\cos B, \sin B)$ . The angle between the two rays is $A - B$ .

Compute the squared distance $PQ^2$ in two ways.

Distance formula: $(\cos A - \cos B)^2 + (\sin A - \sin B)^2$ . Expanding and using $\sin^2 + \cos^2 = 1$ twice gives $2 - 2(\cos A\cos B + \sin A \sin B)$ .
Rotate the picture so that $Q$ sits at $(1, 0)$ . Then $P$ moves to $(\cos(A - B), \sin(A - B))$ , and the same computation gives $2 - 2\cos(A - B)$ .

Distances don't change under rotation, so the two results are equal, and

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

Every other formula follows from this one. Replacing $B$ with $-B$ (cosine is even, sine is odd) gives the sum formula for cosine. The cofunction identity $\sin\theta = \cos(90^\circ - \theta)$ turns the cosine formulas into sine formulas, and dividing sine by cosine produces the tangent formulas.

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}

In the sine formula the sign in the middle matches the sign in the angle. In the cosine formula it is opposite.

Common mistake

The cosine formula flips the sign: $\cos(A + B)$ has a minus in the middle, and $\cos(A - B)$ has a plus. Mixing this up is the most common error. If you're unsure, test $A = B = 45^\circ$ : $\cos 90^\circ = 0$ , and $\cos 45^\circ\cos 45^\circ - \sin 45^\circ \sin 45^\circ = \tfrac{1}{2} - \tfrac{1}{2} = 0$ . The minus sign is right.

Exact values

Write the angle as a sum or difference of special angles ( $30^\circ$ , $45^\circ$ , $60^\circ$ and their relatives), then apply a formula.

Worked example: Sine of 75°

Find the exact value of $\sin 75^\circ$ .

Write $75^\circ = 45^\circ + 30^\circ$ :

\begin{aligned} \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} + \sqrt{2}}{4} \end{aligned}

As a check, $\dfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966$ , which matches a calculator's $\sin 75^\circ$ .

Worked example: Tangent of 105°

Find the exact value of $\tan 105^\circ$ .

Write $105^\circ = 60^\circ + 45^\circ$ , with $\tan 60^\circ = \sqrt{3}$ and $\tan 45^\circ = 1$ :

\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}\cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}.

Rationalize by multiplying the top and bottom by $1 + \sqrt{3}$ :

\frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{4 + 2\sqrt{3}}{1 - 3} = -2 - \sqrt{3}.

The answer is negative, as it should be: $105^\circ$ is in Quadrant II, where tangent is negative.

Using the formulas with any angles

The formulas work for all angles, not just special ones. When you know sines and cosines from other information, you can still combine them.

Worked example: Given values in two quadrants

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

First find the missing values with the Pythagorean identity. $\cos A = \dfrac{4}{5}$ (positive in QI) and $\sin B = \dfrac{12}{13}$ (positive in QII).

\begin{aligned} \sin(A + B) &= \sin A \cos B + \cos A \sin B = \frac{3}{5}\cdot\left(-\frac{5}{13}\right) + \frac{4}{5}\cdot\frac{12}{13} = \frac{-15 + 48}{65} = \frac{33}{65} \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B = \frac{4}{5}\cdot\left(-\frac{5}{13}\right) + \frac{3}{5}\cdot\frac{12}{13} = \frac{-20 + 36}{65} = \frac{16}{65} \end{aligned}

Reading a formula backward

Sometimes an expression already has the shape of a formula. Recognize it and collapse it into a single trig value.

Worked example: Collapse into one term

Find the exact value of $\cos 70^\circ \cos 10^\circ + \sin 70^\circ \sin 10^\circ$ .

This is $\cos A\cos B + \sin A \sin B$ with $A = 70^\circ$ and $B = 10^\circ$ , which is $\cos(A - B)$ :

\cos(70^\circ - 10^\circ) = \cos 60^\circ = \frac{1}{2}.

Tip

The formulas also prove shift identities quickly. For instance, $\sin(x + 90^\circ) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$ . That is why the graph of sine shifted left by $90^\circ$ is the graph of cosine.