The sine and cosine of a sum

A tempting mistake

Is $\cos(x + y)$ the same as $\cos x + \cos y$? Test it with $x = y = \dfrac{\pi}{3}$.

• The left side is $\cos \dfrac{2\pi}{3} = -\dfrac{1}{2}$.
• The right side is $\cos \dfrac{\pi}{3} + \cos \dfrac{\pi}{3} = \dfrac{1}{2} + \dfrac{1}{2} = 1$.

The two are far apart. A trigonometric function does not split over a sum, as a bracket does. We need the true formula, and the unit circle gives it.

The proof on the unit circle

Take the unit circle, with centre $O$ and radius $1$. A point at the angle $\theta$ is $(\cos \theta, \sin \theta)$. Mark four points:

• $P_1 = (\cos x, \sin x)$, at the angle $x$;
• $P_2 = \big(\cos(x + y), \sin(x + y)\big)$, at the angle $x + y$;
• $P_3 = \big(\cos(-y), \sin(-y)\big) = (\cos y, -\sin y)$, at the angle $-y$;
• $P_4 = (1, 0)$, at the angle $0$.

The angle from $P_3$ round to $P_1$ is $x - (-y) = x + y$. The angle from $P_4$ round to $P_2$ is also $x + y$. Equal angles at the centre cut off equal chords, so $P_1P_3 = P_2P_4$. Now square both lengths with the distance formula.

• $P_1P_3^2 = (\cos x - \cos y)^2 + (\sin x + \sin y)^2 = 2 - 2(\cos x \cos y - \sin x \sin y)$.
• $P_2P_4^2 = \big(\cos(x + y) - 1\big)^2 + \sin^2(x + y) = 2 - 2\cos(x + y)$.

Each time, $\sin^2 \theta + \cos^2 \theta = 1$ turns a pair of squares into $1$. Set the two equal, and the key formula appears:

Notice the minus sign. It is the most common slip in this chapter.

Move both sliders. The two chords keep the same length, and the last two readouts always agree.

Five more formulae from one

Change $y$ to $-y$. Since $\cos(-y) = \cos y$ and $\sin(-y) = -\sin y$, the sign in the middle flips:

Swap sine and cosine. Put $x = \dfrac{\pi}{2}$ in that formula. As $\cos \dfrac{\pi}{2} = 0$ and $\sin \dfrac{\pi}{2} = 1$, we get $\cos\left(\dfrac{\pi}{2} - y\right) = \sin y$. Writing $\dfrac{\pi}{2} - y$ for $y$ gives $\sin\left(\dfrac{\pi}{2} - y\right) = \cos y$.

The sine of a sum. Now $\sin(x + y) = \cos\left(\left(\dfrac{\pi}{2} - x\right) - y\right)$. Expand with the formula for $\cos(x - y)$ and use the two swaps:

• $\sin(x + y) = \sin x \cos y + \cos x \sin y$
• $\sin(x - y) = \sin x \cos y - \cos x \sin y$

The tangent. Divide $\sin(x + y)$ by $\cos(x + y)$, and then divide the top and the bottom by $\cos x \cos y$:

• $\tan(x + y) = \dfrac{\tan x + \tan y}{1 - \tan x \tan y}$
• $\tan(x - y) = \dfrac{\tan x - \tan y}{1 + \tan x \tan y}$

These hold when none of the tangents is undefined and the bottom is not $0$.

Allied angles

Put a special angle into the formulae, and the allied angles follow. For example, $\cos\left(\dfrac{\pi}{2} + x\right) = \cos \dfrac{\pi}{2} \cos x - \sin \dfrac{\pi}{2} \sin x = -\sin x$.

• $\sin\left(\dfrac{\pi}{2} + x\right) = \cos x$ and $\cos\left(\dfrac{\pi}{2} + x\right) = -\sin x$
• $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$
• $\sin(\pi + x) = -\sin x$ and $\cos(\pi + x) = -\cos x$
• $\sin\left(\dfrac{3\pi}{2} - x\right) = -\cos x$ and $\cos\left(\dfrac{3\pi}{2} - x\right) = -\sin x$
• $\sin\left(\dfrac{3\pi}{2} + x\right) = -\cos x$ and $\cos\left(\dfrac{3\pi}{2} + x\right) = \sin x$

A rule to remember them. With $\pi$ or $2\pi$, the function stays the same. With $\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$, sine and cosine swap. For the sign, imagine $x$ small and acute, see which quadrant the angle lands in, and use the sign of the first function there.

Exact values

The formulae give exact values for angles made of $30^\circ$, $45^\circ$ and $60^\circ$:

As $75^\circ = 90^\circ - 15^\circ$, the allied angles give $\cos 75^\circ = \sin 15^\circ$, the same number.

A worked example

$\cos A = \dfrac{12}{13}$ with $A$ in the fourth quadrant, and $\sin B = \dfrac{3}{5}$ with $B$ in the second quadrant. Find $\cos(A - B)$.

1. The missing ratios. $\sin A = -\dfrac{5}{13}$, negative in the fourth quadrant. $\cos B = -\dfrac{4}{5}$, negative in the second quadrant.
2. The formula. $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
3. The numbers. $\dfrac{12}{13} \cdot \left(-\dfrac{4}{5}\right) + \left(-\dfrac{5}{13}\right) \cdot \dfrac{3}{5} = -\dfrac{48}{65} - \dfrac{15}{65} = -\dfrac{63}{65}$.

The signs come from the quadrants, never from the formula alone.

Practise with new numbers

Try it in Python

A formula that is true must hold at every angle. The program tests the formula for $\cos(x + y)$ at a thousand random pairs of angles, and also the wrong guess $\cos x + \cos y$.

The error of the formula is about $10^{-15}$: only the rounding of the computer. The guess misses by almost $3$.

Common mistakes

• Splitting over a sum. $\cos(x + y) \neq \cos x + \cos y$, and $\sin 2x \neq 2 \sin x$.
• The sign of the cosine formula. $\cos(x + y)$ has a minus sign, and $\cos(x - y)$ has a plus sign. The sine formulae keep the sign of the angle.
• Forgetting the quadrant. $\sin^2 A + \cos^2 A = 1$ gives the size of the missing ratio. The quadrant gives its sign.
• Using the tangent formula at $\dfrac{\pi}{2}$. $\tan \dfrac{\pi}{2}$ is not defined, so $\tan\left(\dfrac{\pi}{2} - x\right)$ needs sine and cosine. It equals $\cot x$.

Remember these

• $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$
• $\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$
• $\tan(x \pm y) = \dfrac{\tan x \pm \tan y}{1 \mp \tan x \tan y}$
• With $\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$, sine and cosine swap; with $\pi$ they stay. The quadrant gives the sign.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

2. Show that $\sin(x + y)\sin(x - y) = \sin^2 x - \sin^2 y$.

3. Write $\tan(\pi - x)$ in terms of $\tan x$.

4. Ravi says that $\cos(x + y) = \cos x + \cos y$. Test his rule at $x = y = 0$.

5. Without a calculator, find $\sin 75^\circ \cos 15^\circ + \cos 75^\circ \sin 15^\circ$.