Product-to-sum, sum-to-product and 3D trigonometry

From products to sums

You already know the formulae for $\sin(A \pm B)$ and $\cos(A \pm B)$ (formulae 4 and 5 in the Appendix to the syllabus). Write the two sine formulae side by side:

Add them and the $\cos A \sin B$ terms cancel. Subtract them and the $\sin A \cos B$ terms cancel. Do the same with the two cosine formulae. This gives four results:

Each one turns a product into a sum or difference. Why is that useful? Sums are easier to work with than products when you want exact values or exact solutions.

Try the widget below. The two curves are the product and its equivalent sum, so they lie on top of each other whatever values you choose.

From sums to products

Now read the same four results backwards. In the first formula, let $P = A + B$ and $Q = A - B$. Then $A = \frac{P+Q}{2}$ and $B = \frac{P-Q}{2}$. Substituting gives a formula that turns a sum into a product:

All of these are in the Appendix of the syllabus, and they are in your Formulae and Tables booklet. You will still need to recognise which one to use. The pattern is:

• the sum of the angles over 2 goes in the first factor, the difference over 2 in the second;

• $\sin + \sin$ gives $\sin\cdot\cos$, $\cos + \cos$ gives $\cos\cdot\cos$;

• $\cos P - \cos Q$ is the only one with a minus sign out the front.

When $a$ and $b$ are close, the product form shows the slowly changing factor $\cos\frac{a-b}{2}x$. This slow factor is the envelope of the wave.

Back to the tuning forks. The sound is $\sin(2\pi\cdot 440\,t) + \sin(2\pi\cdot 444\,t)$. With $P = 888\pi t$ and $Q = 880\pi t$ this becomes

You hear a note of 442 Hz whose loudness follows $|\cos(4\pi t)|$. That peaks 4 times a second, which is exactly the "wah-wah" beat you hear.

Using the formulae to solve equations

An equation like $\cos 5x + \cos x = \cos 3x$ mixes three different angles. You cannot solve it directly. Convert the sum to a product, and the equation will factorise.

Trigonometry in three dimensions

A 3D problem looks scary, but the method is always the same: find right-angled or general triangles that lie flat in one plane, and use the usual rules on them.

For these problems you need three tools:

• Right-angled triangles: $\sin$, $\cos$, $\tan$ and Pythagoras' theorem.

• The sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$.

• The cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$.

The mast $TB$ is perpendicular to every line on the ground through $B$. So $TBA$ and $TBC$ are right-angled at $B$. The ground triangle $ABC$ is a separate flat triangle. Solve one triangle at a time and pass lengths from one to the next.

Angle between a line and a plane

The angle between a line and a plane is the angle between the line and its projection (its shadow, straight down) on the plane. For a sloping edge $TA$ with $T$ directly above $B$, the angle is $\angle TAB$.

Angle between two planes

Find the line where the planes meet. Then draw, in each plane, a line perpendicular to this line of intersection meeting it at the same point. The angle between those two lines is the angle between the planes.

Summary

• Product to sum: $2\sin A\cos B = \sin(A+B)+\sin(A-B)$, $2\cos A\cos B = \cos(A+B)+\cos(A-B)$, $2\sin A\sin B = \cos(A-B)-\cos(A+B)$, $2\cos A\sin B = \sin(A+B)-\sin(A-B)$. They come from adding or subtracting the $\sin(A\pm B)$ and $\cos(A\pm B)$ formulae.

• Sum to product: use the half-sum and the half-difference of the angles, for example $\sin P+\sin Q = 2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$.

• To solve an equation, turn a sum into a product, move everything to one side, factorise (never divide by a trig factor), and list all solutions in the given interval.

• In 3D, break the solid into plane triangles, look for right angles (a vertical line is perpendicular to the ground), and use trig ratios, the sine rule and the cosine rule.

• The angle between a line and a plane is measured against the line's projection. The angle between two planes is measured with two lines perpendicular to the line of intersection.