Sums into products

From products to sums

Write the two sine formulae one under the other:

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

Add them, and the second terms cancel: $2 \sin x \cos y = \sin(x + y) + \sin(x - y)$. Subtract them, and the first terms cancel. Doing the same with the two cosine formulae gives four rules:

• $2\sin x \cos y = \sin(x + y) + \sin(x - y)$
• $2\cos x \sin y = \sin(x + y) - \sin(x - y)$
• $2\cos x \cos y = \cos(x + y) + \cos(x - y)$
• $2\sin x \sin y = \cos(x - y) - \cos(x + y)$

Look at the last one. The difference comes in the order $\cos(x - y) - \cos(x + y)$.

From sums to products

Now read the rules the other way. Put $x + y = C$ and $x - y = D$. Then $x = \dfrac{C + D}{2}$ and $y = \dfrac{C - D}{2}$:

• $\sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2}$
• $\sin C - \sin D = 2\cos\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$
• $\cos C + \cos D = 2\cos\dfrac{C + D}{2}\cos\dfrac{C - D}{2}$
• $\cos C - \cos D = -2\sin\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$

The pattern is easy to keep: half the sum and half the difference. Only the last formula has a minus sign in front.

Two waves, one product

When two notes of nearly the same pitch play together, you hear the sound swell and fade. These are beats. The sum of the two waves is a product: a fast wave inside a slow one.

Set $p = 12$ and $q = 11$. The red curves change slowly, because $\dfrac{p - q}{2}$ is small. The blue wave swells and fades between them.

A worked example: an exact value

Find $\sin 75^\circ + \sin 15^\circ$ without a table.

1. Half the sum and half the difference. $\dfrac{75^\circ + 15^\circ}{2} = 45^\circ$ and $\dfrac{75^\circ - 15^\circ}{2} = 30^\circ$.
2. The product. $\sin 75^\circ + \sin 15^\circ = 2\sin 45^\circ \cos 30^\circ = 2 \cdot \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2}$.
3. Simplify. $\dfrac{\sqrt{3}}{\sqrt{2}} = \dfrac{\sqrt{6}}{2}$.

A method for proving identities

1. Start from the side that looks more complicated.
2. Group two terms whose angles add or subtract to give simple halves.
3. Turn each pair into a product.
4. Look for a common factor, and cancel it.

Example. Prove that $\dfrac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$.

1. The top. $\sin 5x + \sin 3x = 2\sin 4x \cos x$.
2. The bottom. $\cos 5x + \cos 3x = 2\cos 4x \cos x$.
3. Cancel. The factor $2\cos x$ cancels, and $\dfrac{\sin 4x}{\cos 4x} = \tan 4x$.

Practise with new numbers

Try it in Python

The program checks the formula for $\sin P + \sin Q$ for three pairs of angles in degrees.

Common mistakes

• Forgetting the halves. $\sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2}$, not $2\sin(C + D)\cos(C - D)$.
• The sign of $\cos C - \cos D$. It is $-2\sin\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$. Check it with $C = \dfrac{\pi}{2}$ and $D = 0$: the left side is $-1$.
• The order in $2\sin x \sin y$. It is $\cos(x - y) - \cos(x + y)$, and the result can be checked at $x = y$.
• Grouping the wrong pairs. Pair terms whose halves give angles you can use, often a common factor.

Remember these

• $2\sin x \cos y = \sin(x + y) + \sin(x - y)$, and three more like it.
• $\sin C \pm \sin D$ and $\cos C \pm \cos D$ become products of the half sum and the half difference.
• $\cos C - \cos D$ is the only one with a minus sign in front.
• To prove an identity, turn pairs of sums into products, and cancel the common factor.

Check yourself

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

1. Write $\cos 6x + \cos 2x$ as a product.

2. Write $2\sin 3x \sin x$ as a difference of two cosines.

3. Find the exact value of $\sin 75^\circ - \sin 15^\circ$.

4. Prove that $\dfrac{\cos x - \cos 3x}{\sin 3x - \sin x} = \tan 2x$.