Half angle formulas — derivations from double angle: Question bank

Throughout, $A$ is the "full" angle you are given and $\frac{A}{2}$ is the half angle you want. The symbol $\pm$ just means "we must still decide plus or minus"; a square root by itself always outputs a non-negative number.

True or false — justify

The sign in

sin \frac{A}{2} = \pm \frac{1 - c o s A}{2}

is decided by the quadrant of

A

False. The formula outputs

sin \frac{A}{2}

, so the sign is set by where

\frac{A}{2}

lands — halving can move you into a different quadrant than

A

sin \frac{A}{2} = \frac{1}{2} sin A

for all

A

False. Trig functions are nonlinear, so you cannot "distribute" the half onto the angle. Test

A = 18 0^{\circ}

sin 9 0^{\circ} = 1

but

\frac{1}{2} sin 18 0^{\circ} = 0

cos^{2} \frac{A}{2} = \frac{1 + c o s A}{2}

is always true, with no

\pm

ambiguity.

True. Squaring throws away sign, so the squared identity holds exactly for every

A

; the

\pm

only appears when you take the root to get

cos \frac{A}{2}

itself.

The right side

\frac{1 - c o s A}{2}

can be negative, so

sin^{2} \frac{A}{2}

could be negative.

False. Since

- 1 \leq cos A \leq 1

, we have

1 - cos A \geq 0

, so the quotient is always

\geq 0

— consistent with it being a square.

tan \frac{A}{2} = \frac{s i n A}{1 + c o s A}

needs a

\pm

just like the square-root form does.

False. This form is exact and sign-safe:

sin A

already carries the correct sign and

1 + cos A \geq 0

, so the ratio's sign is automatically right.

cos A = 1

then all three half-angle formulas are still well-defined.

Partly false.

sin \frac{A}{2} = 0

and

cos \frac{A}{2} = \pm 1

are fine, but

tan \frac{A}{2} = \frac{1 - c o s A}{s i n A} = \frac{0}{0}

and

\frac{s i n A}{1 + c o s A} = \frac{0}{2} = 0

— the second form survives, so

tan \frac{A}{2} = 0

, matching

A /2

being a multiple of

18 0^{\circ}

The two radical-free tangent forms

\frac{1 - c o s A}{s i n A}

and

\frac{s i n A}{1 + c o s A}

can give different answers.

False. They are algebraically identical (multiply one by a conjugate to get the other); they only look different in which denominator might vanish for a given

A

cos 2 θ = 1 - 2 sin^{2} θ

and

cos 2 θ = 2 cos^{2} θ - 1

are two different theorems.

False. They are the same identity rewritten using

sin^{2} + cos^{2} = 1

; each just "kills" a different squared function. See Double angle formulas.

Spot the error

"Given

cos A = \frac{7}{25}

with

A

in quadrant IV,

A /2

is also in quadrant IV, so

sin \frac{A}{2} < 0

Error:

A

in QIV means

27 0^{\circ} < A < 36 0^{\circ}

, so

13 5^{\circ} < \frac{A}{2} < 18 0^{\circ}

— that is quadrant II, where

sin > 0

. Always compute the range of

\frac{A}{2}

before choosing the sign.

sin^{2} \frac{A}{2} = \frac{1 - c o s A}{2}

, therefore

sin \frac{A}{2} = \frac{1 - c o s A}{2}

Error: they forgot the square root and the

\pm

. Correct is

sin \frac{A}{2} = \pm \frac{1 - c o s A}{2}

; dropping the root changes the whole magnitude.

cos 1 5^{\circ} = \frac{1 + c o s 3 0 ^{\circ}}{2}

Error: that expression equals

cos^{2} 1 5^{\circ}

, not

cos 1 5^{\circ}

. You must square-root:

cos 1 5^{\circ} = \frac{1 + c o s 3 0 ^{\circ}}{2} = \frac{2 + 3}{2}

"To halve

A = 4 5^{\circ}

using

tan \frac{A}{2} = \frac{1 - c o s A}{s i n A}

, plug

A = 22. 5^{\circ}

Error: you substitute the full angle

A = 4 5^{\circ}

into the formula; the formula's own

\frac{A}{2}

then equals

22. 5^{\circ}

. Substituting

22. 5^{\circ}

would compute

tan 11.2 5^{\circ}

"Since

\frac{1 - c o s A}{1 + c o s A}

has a

\pm

, and

\frac{s i n A}{1 + c o s A}

does not, the two disagree."

Error: they are the same number; the radical form merely fails to record the sign, while the

\frac{s i n A}{1 + c o s A}

form encodes it. No disagreement, just a sign the radical loses.

cos A = 1 - 2 sin^{2} \frac{A}{2}

, so

sin^{2} \frac{A}{2} = \frac{c o s A - 1}{2}

Error: sign flipped while solving. From

cos A = 1 - 2 sin^{2} \frac{A}{2}

you get

2 sin^{2} \frac{A}{2} = 1 - cos A

, hence

sin^{2} \frac{A}{2} = \frac{1 - c o s A}{2}

— the

1

comes first.

Why questions

Why do we use the cosine double-angle identity to build half-angle formulas, not the sine one?

Because

cos 2 θ

can be written using a single squared function (

sin^{2}

cos^{2}

) alone, which we can isolate and solve;

sin 2 θ = 2 sin θ cos θ

mixes two functions and cannot be solved for a lone half-angle.

Why does the

\pm

attach to

sin \frac{A}{2}

and

cos \frac{A}{2}

but never to

sin^{2} \frac{A}{2}

Squaring destroys sign information, so the squared identity is unambiguous; taking a root re-introduces the two possible signs, which only the quadrant of

\frac{A}{2}

can resolve.

Why does

\frac{s i n A}{1 + c o s A}

escape the

\pm

problem entirely?

Because it was built without ever taking a square root of a squared quantity — the sign lives naturally inside

sin A

, so nothing was thrown away to recover.

Why is

\frac{A}{2}

's quadrant, not

A

's, the deciding factor?

The formulas produce a value of

\frac{A}{2}

, and a trig function's sign depends on the quadrant of its own argument — that argument is

\frac{A}{2}

Why does the Weierstrass substitution

t = tan \frac{x}{2}

prefer the

\frac{s i n A}{1 + c o s A}

form?

Because it is a single rational expression with correct sign and no case-splitting, which is exactly what you need to turn every trig function of

x

into a rational function of

t

Why can't we just memorize

cos 1 5^{\circ}

from a table instead of deriving it?

1 5^{\circ}

is not a standard-table angle, but

3 0^{\circ}

is and

15 = 30/2

, so the half-angle route is the natural way to reach Exact trig values like

1 5^{\circ}, 22. 5^{\circ}, 7 5^{\circ}

Why does the Pythagorean identity appear in this derivation at all?

It is the tool that converts

cos^{2} θ - sin^{2} θ

into a one-function form (

2 cos^{2} θ - 1

1 - 2 sin^{2} θ

), which is the whole reason a lone half-angle can be isolated.

Edge cases

What is

tan \frac{A}{2}

when

A = 18 0^{\circ}

, and which formula must you use?

Here

cos A = - 1

, so

\frac{s i n A}{1 + c o s A} = \frac{0}{0}

is undefined, but

\frac{1 - c o s A}{s i n A} = \frac{2}{0}

diverges — consistent with

tan 9 0^{\circ}

being undefined. The value genuinely does not exist; both forms warn you correctly.

What is

tan \frac{A}{2}

when

A = 0^{\circ}

\frac{1 - c o s A}{s i n A} = \frac{0}{0}

is indeterminate, but

\frac{s i n A}{1 + c o s A} = \frac{0}{2} = 0

gives the answer cleanly:

tan 0^{\circ} = 0

. This is exactly why we keep both forms.

A = 36 0^{\circ}

, does

sin \frac{A}{2} = sin 18 0^{\circ} = 0

come out of the formula?

Yes:

cos 36 0^{\circ} = 1

, so

sin^{2} \frac{A}{2} = \frac{1 - 1}{2} = 0

, giving

sin 18 0^{\circ} = 0

— the degenerate case checks out.

For

A

in quadrant I, must

\frac{A}{2}

also be in quadrant I?

Yes here, since

0^{\circ} < A < 9 0^{\circ}

forces

0^{\circ} < \frac{A}{2} < 4 5^{\circ}

. But this "same-quadrant" comfort is precisely the habit that fails for QII–QIV, so never generalize it.

Can

cos \frac{A}{2}

be negative even when

cos A

is positive?

Yes. Take

A = 30 0^{\circ}

(QIV,

cos A = \frac{1}{2} > 0

); then

\frac{A}{2} = 15 0^{\circ}

is in QII where

cos < 0

. The sign of

cos A

tells you nothing about the sign of

cos \frac{A}{2}

A = 9 0^{\circ}

, evaluate all three half-angle expressions and confirm consistency.

cos A = 0

, so

sin^{2} 4 5^{\circ} = \frac{1}{2}

cos^{2} 4 5^{\circ} = \frac{1}{2}

, and

tan 4 5^{\circ} = \frac{1 - 0}{1} = 1

— all match the known values of

4 5^{\circ}

, a clean sanity check.

Recall One-line summary of every trap

The three recurring sins: (1) reading the quadrant of $A$ instead of $\frac{A}{2}$ , (2) forgetting the root or the $\pm$ , (3) forgetting that one tangent form may be $\frac{0}{0}$ while the other stays finite. Keep both tangent forms and always locate $\frac{A}{2}$ first.

Connections

Double angle formulas — the seed identities every trap here bends.
Pythagorean identity — the tool that makes one-function forms possible.
Weierstrass substitution — why the sign-safe tangent form matters.
Exact trig values — where half-angle results get used.
Product-to-sum formulas — sibling manipulations of the same seed.