🌓Half-Angle & Product Identities

Monk Tau keeps one page pinned above the door of the observatory on Radian Ridge, and it holds a single line:

cos 2u = 1 - 2 sin² u

Everything on this page comes out of that line, read from right to left instead of left to right.

Solve it for the square. Move the sine term across, divide by 2, and you get

sin² u = (1 - cos 2u)/2

The other version of the double-angle formula, cos 2u = 2 cos² u - 1, does the same job for the cosine:

cos² u = (1 + cos 2u)/2

Two identities, one derivation, and they answer two different questions depending on which way you read them.

Read left to right and they are the power-reducing formulas: a square disappears, and the price is that the angle doubles. That is what makes a squared trigonometric function integrable, averageable, graphable — sin² t is a wave about the line 1/2, and this is the line that says so.

Read right to left with u = t/2 and they are the half-angle formulas. Put u = t/2, so 2u = t, and take a square root:

sin(t/2) = +sqrt((1 - cos t)/2) or -sqrt((1 - cos t)/2)

cos(t/2) = +sqrt((1 + cos t)/2) or -sqrt((1 + cos t)/2)

Now one exact value in the whole unit circle produces another. You know cos 30° exactly, so you know sin 15° exactly. You know cos 45°, so you know cos 22.5°. Angles that no diagram on the wall shows become reachable, and they stay exact — no calculator, no rounding.

The two square roots are the whole difficulty, and the next section is about nothing else.

A square root only knows the size. It cannot know whether the answer is positive or negative, because squaring destroyed that information. Something outside the formula has to put it back, and that something is the quadrant that t/2 lands in.

The trap is that the question hands you a quadrant for t, and it is very easy to use that one. It is the wrong one. Halve the interval first.

180° < t < 270° divide every part by 2 90° < t/2 < 135°

So t sits in quadrant III, where the sine is negative — and t/2 sits in quadrant II, where the sine is positive. Two different quadrants, two different signs, and the answer follows the second.

Halving stretches an interval in an interesting way. The four ordinary quadrants for t all halve into the top half of the circle:

• ◆0° to 90° halves to 0° to 45°, quadrant I
• ◆90° to 180° halves to 45° to 90°, quadrant I
• ◆180° to 270° halves to 90° to 135°, quadrant II
• ◆270° to 360° halves to 135° to 180°, quadrant II

That is why a half-angle sine is never negative for an angle inside one full turn, while a half-angle cosine is negative for the whole second half of the turn. If a question sets t between 360° and 720° — a second lap — then t/2 lands in the bottom half and both can turn negative. The rule never changes: halve the interval, read the quadrant, then choose the sign.

The tangent escapes the whole problem. There are two forms of it, and neither has a square root:

tan(t/2) = sin t/(1 + cos t) = (1 - cos t)/sin t

No plus-or-minus appears, because the signs of sin t and cos t already carry the answer. If you are ever unsure of a sign, this is the safe road: work out the tangent, and let it tell you what the sine and cosine must have been.

Used forwards, the same two identities strip powers out of an expression. Each square becomes a plain cosine of double the angle:

sin² t = (1 - cos 2t)/2 cos² t = (1 + cos 2t)/2

Two details matter more than the formulas themselves.

The angle that doubles is whatever sits inside. Not t, but the argument of the function being squared:

cos² 3t = (1 + cos 6t)/2 sin² (t/2) = (1 - cos t)/2

The second of those is the half-angle formula again, wearing different clothes.

Higher powers need the formula twice. A fourth power is a square of a square, and squaring a reduced expression brings back a cos², which must be reduced in its turn:

sin⁴ t = (sin² t)² = ((1 - cos 2t)/2)² = (1 - 2cos 2t + cos² 2t)/4

The cos² 2t on the end is still squared, so reduce it as well: cos² 2t = (1 + cos 4t)/2. Substituting and tidying gives

sin⁴ t = (3 - 4cos 2t + cos 4t)/8

Notice what happened to the angle: it doubled twice, from t to 2t to 4t. That is the pattern for every power. Each reduction costs one doubling.

Add the sum and difference formulas for the cosine and something cancels:

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

Add those two lines and the sine terms vanish, leaving 2 cos A cos B on the right. Divide by 2 and you have the first product-to-sum formula. Subtract them instead and the cosine terms vanish. Doing the same to the two sine formulas gives the rest:

cos A cos B = (1/2)[cos(A + B) + cos(A - B)]

sin A sin B = (1/2)[cos(A - B) - cos(A + B)]

sin A cos B = (1/2)[sin(A + B) + sin(A - B)]

Every one carries a 1/2, and only two angles ever appear: the sum and the difference. Dropping the 1/2 is the single commonest slip here, and it doubles the answer.

Run the same formulas backwards — put A + B = P and A - B = Q, so that A is the half-sum and B is the half-difference — and you get the sum-to-product family:

sin P + sin Q = 2 sin((P + Q)/2) cos((P - Q)/2)

sin P - sin Q = 2 cos((P + Q)/2) sin((P - Q)/2)

cos P + cos Q = 2 cos((P + Q)/2) cos((P - Q)/2)

cos P - cos Q = -2 sin((P + Q)/2) sin((P - Q)/2)

Here every one carries a 2, and the last one carries a minus as well.

These are worth more than they look. An equation such as sin 3t + sin t = 0 is unsolvable as it stands, because a sum tells you nothing. Turn it into 2 sin 2t cos t = 0 and it falls apart at once: a product is zero exactly when one of its factors is zero, so sin 2t = 0 or cos t = 0, and both of those you can solve on sight.

A verification is not an equation. You are not allowed to move terms from one side to the other, because that assumes the thing you are trying to show. Instead you take the busier side alone and rewrite it until it turns into the other one.

Take (1 - cos 2t)/sin 2t and show it equals tan t. The top and the bottom are both doubled angles, so replace both:

• ◆1 - cos 2t = 1 - (1 - 2 sin² t) = 2 sin² t
• ◆sin 2t = 2 sin t cos t

So the whole fraction becomes (2 sin² t)/(2 sin t cos t). The 2 cancels, one sine cancels, and what is left is sin t/cos t, which is tan t. Done — and nothing crossed the equals sign.

Two habits make these quick. Choose the right form of cos 2t: it has three, and picking the one written in the function you want to keep saves a whole line. Look for the identity hiding in a 1: 1 - cos 2t and 1 + cos 2t are both power-reducing formulas in disguise, and cos⁴ t - sin⁴ t is a difference of two squares whose first bracket collapses to 1.

If you are stuck for which of two answers is right, test a friendly angle. At t = 60° every expression on this page is computable by hand, and two different functions will disagree there.