Sum, Double-Angle and Half-Angle Formulas

What this move does

Everything starts from one formula:

sin (α + β) = sin α cos β + cos α sin β, cos (α + β) = cos α cos β - sin α sin β .

Replace $β$ by $- β$ to get the difference formulas; divide one by the other to get $tan (α \pm β) = \frac{tan α \pm tan β}{1 \mp tan α tan β}$ ; set $β = α$ to get the double-angle formulas

sin 2 α = 2 sin α cos α, cos 2 α = cos^{2} α - sin^{2} α = 2 cos^{2} α - 1 = 1 - 2 sin^{2} α .

Read the last two equalities backwards and you have the power-reduction formulas $cos^{2} α = \frac{1 + cos 2 α}{2}$ and $sin^{2} α = \frac{1 - cos 2 α}{2}$ (lower the power, double the angle). Replace $α$ by $\frac{α}{2}$ and you have the half-angle formulas. To use them, go through three steps:

Look at the angles: can the angle you want be built from the angles you know? For example $β = (α + β) - α$ , $2 α = (α + β) + (α - β)$ , $α = 2 \cdot \frac{α}{2}$ .
Choose the formula: to combine angles, use the sum and difference formulas; to double or halve an angle, use the double- or half-angle formulas; if a power is too high, reduce it.
Fix the sign: every time you take a square root, first pin down the range of the angle, then decide between plus and minus.

Predict first

$α$ and $β$ are both acute. Which is larger, $sin (α + β)$ or $sin α + sin β$ ?

sin (α + β)

is larger

sin α + sin β

is largerit depends on

α, β

Choose an answer first; then you can see the result.

Work through one

A proof by picture: sin(α+β) and cos(α+β)

Step 1 / 7

Build a right triangle
$O P = 1$ makes the angle $α + β$ with the $x$ -axis, so the height of $P$ is $sin (α + β)$ . On the terminal side of $α$ take $Q$ with $∠ O QP = 9 0^{\circ}$ . The hypotenuse is $1$ and the angle at $O$ is $β$ , so $O Q = cos β$ and $QP = sin β$ .

$O Q = cos β, QP = sin β$
The lower triangle
The small top triangle: α appears again
Put them together
Read it across: cosine
Negative β: the difference formula
β = α: the double-angle formula

When it fails: a sine alone does not fix the angle

After “look at the angles” comes “pin down the range”. Getting $cos (α + β)$ from $sin (α + β)$ needs a square root, and its sign depends on whether $α + β$ is acute or obtuse.

Break it

Counterexample lab: remove one condition

Conclusion: $sin (α + β) = \frac{3}{2}$ gives $cos (α + β) = \frac{1}{2}$

$α = 4 0^{\circ}$ , $β = 2 0^{\circ}$ : the reading card shows $sin 6 0^{\circ} = \frac{3}{2}$ . Switch to cosine and you get exactly $\frac{1}{2}$ .

Practice

Exercise 1Basic

Find $cos 8 0^{\circ} cos 2 0^{\circ} + sin 8 0^{\circ} sin 2 0^{\circ}$ .

Exercise 2Basic

Given $tan α = 2$ , find $tan 2 α$ .

Exercise 3Advanced

$α$ and $β$ are acute, $cos α = \frac{4}{5}$ and $cos (α + β) = \frac{5}{13}$ . Find $cos β$ .

Exercise 4Challenge

(A classic.) Find $cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{4 π}{7}$ .

What this move does

Work through one

A proof by picture: sin(α+β) and cos(α+β)

When it fails: a sine alone does not fix the angle

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Distributing sin as if it were multiplication

Taking only the positive root in a half-angle formula