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Module 3 · L7 of 15 ~30 min ⚡ +90 XP available

Half Angle Formulas

From $\cos(2A) = 2\cos^2 A - 1$, a simple substitution unlocks a powerful set of identities — half angle formulas that let you find exact trig values for angles like $22.5°$ and $15°$ that no standard table covers. In this lesson you'll derive all three and learn when the $\pm$ sign trips students up most.

Today's hook — What is the exact value of $\sin(22.5°)$? A calculator gives a decimal, but there is a closed-form surd answer. By the end of this lesson you'll be able to derive it in four lines using nothing but double angle formulas you already know.

0/5QUESTS

Recall — your gut answer first

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If $\cos(2A) = 2\cos^2 A - 1$, rearrange this to find $\cos A$ in terms of $\cos(2A)$. Then substitute $A = \frac{\theta}{2}$ — what does this give you for $\cos\frac{\theta}{2}$? Write your prediction before reading on.

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The key idea

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Every half angle formula comes from one source: the double angle identities with a clever substitution $A = \frac{\theta}{2}$. There is only one trap — the sign ambiguity — and you must resolve it from the quadrant every time.

Start with $\cos(2A) = 1 - 2\sin^2 A$ and $\cos(2A) = 2\cos^2 A - 1$. Replace $A$ with $\frac{\theta}{2}$ (so $2A$ becomes $\theta$). Isolate the squared trig function, then take a square root. The $\pm$ records the fact that both roots are algebraically valid — context picks the sign.

$\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{2}}$

Sin half-angle

From $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$, so $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$.

Cos half-angle

From $\cos\theta + 1 = 2\cos^2\frac{\theta}{2}$, so $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$.

Tan half-angle

Divide: $\tan\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$.

What you'll master

Know

Key facts

$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$
$\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$
$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$

Understand

Concepts

How half angle formulas follow directly from double angle identities
Why the $\pm$ sign exists and how the quadrant of $\frac{\theta}{2}$ resolves it
Why $\sin\frac{\theta}{2} \neq \frac{1}{2}\sin\theta$

Can do

Skills

Derive half angle formulas from double angle identities
Find exact trig values at non-standard angles (e.g. $22.5°$, $15°$)
Determine the correct sign by identifying the quadrant of $\frac{\theta}{2}$

Key terms

Half angleAn angle of the form $\frac{\theta}{2}$, half the value of $\theta$.

Sign ambiguityThe $\pm$ in half angle formulas; must be resolved by determining which quadrant $\frac{\theta}{2}$ lies in.

Double angle identityIdentities of the form $\sin(2A)$, $\cos(2A)$, $\tan(2A)$ — the source of all half angle formulas.

Exact valueA surd or rational expression (not a decimal) representing the precise value of a trig function.

The half angle formulas — derivation

core concept

Start from two forms of the double angle formula for cosine:

$$\cos(2A) = 1 - 2\sin^2 A \qquad \Rightarrow \qquad \sin^2 A = \frac{1 - \cos(2A)}{2}$$

$$\cos(2A) = 2\cos^2 A - 1 \qquad \Rightarrow \qquad \cos^2 A = \frac{1 + \cos(2A)}{2}$$

Now substitute $A = \frac{\theta}{2}$ (so $2A = \theta$) and take square roots:

$$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$$

$$\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$

$$\tan\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}$$

The $\pm$ arises because taking a square root introduces both roots. The quadrant of $\frac{\theta}{2}$ determines which sign applies. Note that the two alternative forms for $\tan\frac{\theta}{2}$ are always unambiguous — they carry no $\pm$.

Sign rule summary. If $\frac{\theta}{2}$ is in Q1 or Q2, $\sin\frac{\theta}{2} > 0$. If in Q1 or Q4, $\cos\frac{\theta}{2} > 0$. For example, if $0 < \theta < \pi$, then $0 < \frac{\theta}{2} < \frac{\pi}{2}$ (Q1), so both sin and cos of $\frac{\theta}{2}$ are positive — take the positive square root.

{2} = {1-{2}}; {2} = {1+{2}}; {2} = 1-{} = {1+} (no sign ambiguity)

Pause — copy all three half-angle formulas into your book: $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$, $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$, and $\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta}$ (no sign ambiguity in the $\tan$ form).

Quick check: Which formula gives $\cos\frac{\theta}{2}$ directly?

Resolving the sign — quadrant method

core concept

We just saw that $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$ and $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$. That raises a question: how do we determine which sign to choose, since both $+$ and $-$ satisfy the equation algebraically? This card answers it → by halving the interval for $\theta$ to find which quadrant $\frac{\theta}{2}$ lies in, then applying ASTC to pick the correct sign.

The most common error with half angle formulas is ignoring sign ambiguity. Here is the systematic method:

Identify the range of $\theta$.
Halve it to find the range of $\frac{\theta}{2}$.
Determine the quadrant of $\frac{\theta}{2}$.
Apply the ASTC rule to determine the sign of $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$.

Apply to the half-angle $\frac{\theta}{2}$, not to $\theta$ itself.

Example: If $0 < \theta < \frac{\pi}{2}$, then $0 < \frac{\theta}{2} < \frac{\pi}{4}$ — still in Q1, so both $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$ are positive.

Example: If $\pi < \theta < 2\pi$, then $\frac{\pi}{2} < \frac{\theta}{2} < \pi$ — Q2, so $\sin\frac{\theta}{2} > 0$ but $\cos\frac{\theta}{2} < 0$.

Step 1: find the range of {2} by halving the range of; Step 2: use ASTC on the half angle to determine signs

Pause — copy the two-step sign method into your book: (1) halve the range of $\theta$ to find the range of $\frac{\theta}{2}$; (2) apply ASTC to $\frac{\theta}{2}$ to select $+$ or $-$ for each half-angle formula.

Did you get this? True or false: if $\pi < \theta < 2\pi$, then $\sin\frac{\theta}{2}$ is always negative.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · EXACT VALUE

Find the exact value of $\sin(22.5°)$.

$22.5° = \dfrac{45°}{2}$, so use the half angle formula with $\theta = 45°$

Identify which double angle to use: $22.5° = \frac{1}{2}(45°)$, and $\cos 45° = \frac{\sqrt{2}}{2}$ is known.

PROBLEM 2 · GIVEN COS THETA

If $\cos\theta = \dfrac{1}{3}$ and $0 < \theta < \pi$, find $\sin\dfrac{\theta}{2}$.

Since $0 < \theta < \pi$, we have $0 < \dfrac{\theta}{2} < \dfrac{\pi}{2}$ (Q1), so $\sin\dfrac{\theta}{2} > 0$

Halving the range of $\theta$ gives the quadrant of $\frac{\theta}{2}$.

PROBLEM 3 · PROOF

Show that $\tan\dfrac{\theta}{2} = \dfrac{1-\cos\theta}{\sin\theta}$.

$\tan\dfrac{\theta}{2} = \dfrac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \dfrac{\pm\sqrt{\frac{1-\cos\theta}{2}}}{\pm\sqrt{\frac{1+\cos\theta}{2}}}$

Write $\tan = \frac{\sin}{\cos}$ and substitute the half angle formulas.

Fill the gap: If $\cos\theta = \frac{3}{5}$ and $0 < \theta < \frac{\pi}{2}$, then $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \frac{3}{5}}{2}} = $ .

Misconceptions to fix · the traps that cost marks

Trap 01

$\sin\frac{\theta}{2} \neq \frac{1}{2}\sin\theta$

This is the single most common error. For example, $\sin 45° = \frac{\sqrt{2}}{2} \approx 0.707$, but $\frac{1}{2}\sin 90° = \frac{1}{2} \neq 0.707$. Trig functions are nonlinear — you cannot factor out constants.

Trap 02

Always taking the positive root

The $\pm$ in half angle formulas is real. If $\theta = \frac{3\pi}{2}$, then $\frac{\theta}{2} = \frac{3\pi}{4}$ (Q2), where $\cos < 0$. Always check the quadrant before choosing the sign.

Trap 03

Applying the quadrant rule to $\theta$ instead of $\frac{\theta}{2}$

The quadrant that matters is the quadrant of $\frac{\theta}{2}$. If $\theta$ is in Q2, that only tells you $\sin\theta > 0$ — it says nothing directly about $\sin\frac{\theta}{2}$. Halve the range first.

Did you get this? True or false: $\sin\frac{\theta}{2}$ is always positive when $0 < \theta < 2\pi$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the exact value of $\cos(22.5°)$ using the half angle formula.

If $\cos\theta = -\frac{1}{2}$ and $\pi < \theta < 2\pi$, find $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$.

Using $\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta}$, find $\tan(22.5°)$ exactly.

Explain why $\cos\frac{\theta}{2} = \frac{1}{2}\cos\theta$ is incorrect. Give a counterexample.

Find the exact value of $\sin(15°)$ using a half angle formula with $\theta = 30°$.

Revisit your thinking

Earlier you were asked to rearrange $\cos(2A) = 2\cos^2 A - 1$ and substitute $A = \frac{\theta}{2}$. The answer is $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$.

Why can't we just write $\sin\frac{\theta}{2} = \frac{1}{2}\sin\theta$? Because trig functions do not distribute over scalar multiples in the argument. The formula $\sin(2A) = 2\sin A\cos A$ involves a product of trig functions, not a simple doubling of the output. The half angle formula for sine depends on $\cos\theta$, not $\sin\theta$ — a surprising asymmetry that reflects the geometry of the unit circle.

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Odd one out: Three of these are valid forms for $\tan\frac{\theta}{2}$. Which one is NOT?

Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. Find the exact value of $\cos(15°)$ using a half angle formula. (2 marks)

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ApplyBand 42 marks

Q2. If $\cos\theta = \dfrac{1}{3}$ and $0 < \theta < \pi$, find $\sin\dfrac{\theta}{2}$. (2 marks)

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AnalyseBand 52 marks

Q3. Show that $\tan\dfrac{\theta}{2} = \dfrac{1-\cos\theta}{\sin\theta}$. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\cos(22.5°) = \sqrt{\frac{1+\cos 45°}{2}} = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2+\sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2}$

2. $\theta \in (\pi, 2\pi)$ so $\frac{\theta}{2} \in (\frac{\pi}{2}, \pi)$ — Q2. Thus $\sin\frac{\theta}{2} > 0$, $\cos\frac{\theta}{2} < 0$. $\cos\theta = -\frac{1}{2}$: $\sin\frac{\theta}{2} = \sqrt{\frac{1-(-\frac{1}{2})}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$; $\cos\frac{\theta}{2} = -\sqrt{\frac{1+(-\frac{1}{2})}{2}} = -\sqrt{\frac{1}{4}} = -\frac{1}{2}$

3. $\tan(22.5°) = \frac{1-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2-\sqrt{2}}{\sqrt{2}} = \sqrt{2}-1$

4. If $\theta = 90°$: $\cos\frac{90°}{2} = \cos 45° = \frac{\sqrt{2}}{2} \approx 0.707$, but $\frac{1}{2}\cos 90° = 0$. Clearly different.

5. $\sin(15°) = \sqrt{\frac{1-\cos 30°}{2}} = \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2-\sqrt{3}}{4}} = \frac{\sqrt{2-\sqrt{3}}}{2}$

Q1 (2 marks): $\cos(15°) = +\sqrt{\frac{1+\cos 30°}{2}} = \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2+\sqrt{3}}{4}} = \frac{\sqrt{2+\sqrt{3}}}{2}$ [2]. Positive since $15°$ is in Q1.

Q2 (2 marks): $0 < \theta < \pi \Rightarrow 0 < \frac{\theta}{2} < \frac{\pi}{2}$ (Q1) so positive root [1]. $\sin\frac{\theta}{2} = \sqrt{\frac{1-\frac{1}{3}}{2}} = \sqrt{\frac{1}{3}} = \frac{\sqrt{3}}{3}$ [1].

Q3 (2 marks): $\tan\frac{\theta}{2} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \frac{\sqrt{\frac{1-\cos\theta}{2}}}{\sqrt{\frac{1+\cos\theta}{2}}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ [1]. Rationalise: multiply numerator and denominator by $\sqrt{1-\cos\theta}$: $\frac{\sqrt{(1-\cos\theta)^2}}{\sqrt{1-\cos^2\theta}} = \frac{1-\cos\theta}{\sin\theta}$ [1]. $\square$

Boss battle · The Angle Halver

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Five timed questions on half angle formulas. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Key takeaways

$\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}$ and $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}$
$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ — these forms are sign-free
Always determine the sign by finding the quadrant of $\frac{\theta}{2}$, not $\theta$
$\sin\frac{\theta}{2} \neq \frac{1}{2}\sin\theta$ — never simplify this way
Derived from double angle formulas by substituting $A = \frac{\theta}{2}$

Next lesson: The t-Formulas — a substitution that converts trig equations to polynomial equations.

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