Double Angle Formulae

sin 2A, cos 2A (three forms), tan 2A — Derivation and Application
9709 Paper 3 · §3.3 Trigonometry · Lesson 5
💡 Lead In: Set B = A

The Scenario

You already own the compound-angle formulae:

$$\sin(A+B)=\sin A\cos B+\cos A\sin B, \qquad \cos(A+B)=\cos A\cos B-\sin A\sin B.$$

Now ask a natural question: what happens when the two angles are the same? Replace $B$ with $A$, so that $A+B$ becomes $2A$. Old machinery, new output.

Review · Activating Prior Knowledge

Quick check — pick the correct statement in each question before we double the angles.

Review 1 — Which compound-angle formula is correct?
$\sin(A+B)=\sin A+\sin B$
$\cos(A-B)=\cos A\cos B-\sin A\sin B$
$\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\tan(A+B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

$\sin(A+B)=\sin A\cos B+\cos A\sin B$. Option A forgets the cross terms, B has the wrong sign for $\cos(A-B)$ (it is $\cos A\cos B+\sin A\sin B$), and D uses the minus version — correct is $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Review 2 — Which is the master identity that lets us rewrite $\cos 2A$ in three forms?
$\sin A+\cos A=1$
$\sin^2 A-\cos^2 A=1$
$\tan^2 A+1=\cosec^2 A$
$\sin^2 A+\cos^2 A=1$

The master identity is $\sin^2 A+\cos^2 A=1$. Solving for $\sin^2 A$ or $\cos^2 A$ and substituting into $\cos 2A=\cos^2 A-\sin^2 A$ produces the two extra forms $2\cos^2 A-1$ and $1-2\sin^2 A$.

The Question

Set $B=A$ in $\sin(A+B)$, $\cos(A+B)$ and $\tan(A+B)$. What three double-angle formulae do you obtain? Using $\sin^2 A+\cos^2 A=1$, can you write $\cos 2A$ in two more different ways?

💡 The Bridge (click to reveal)

$$\sin 2A=2\sin A\cos A.$$

$$\cos 2A=\cos^2 A-\sin^2 A=2\cos^2 A-1=1-2\sin^2 A.$$

$$\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}.$$

These double-angle formulae — especially the three forms of $\cos 2A$ — are today's focus.

📖 Key Concepts: Double Angle Formulae

The double angle formulae are special cases of the compound angle formulae where $A = B$. Substitute $B = A$ into $\sin(A+B)$, $\cos(A+B)$, and $\tan(A+B)$ to obtain the three double angle identities.

Double angle formula for sine: $$\sin 2A = 2 \sin A \cos A$$

Derived from $\sin(A+A) = \sin A \cos A + \cos A \sin A$.

Double angle formula for cosine (three equivalent forms): $$\cos 2A = \cos^2 A - \sin^2 A$$ $$\cos 2A = 2\cos^2 A - 1$$ $$\cos 2A = 1 - 2\sin^2 A$$

Form 1 comes directly from $\cos(A+A)$. Forms 2 and 3 use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to eliminate one variable. Choose the form that matches your equation.

Double angle formula for tangent: $$\tan 2A = \dfrac{2 \tan A}{1 - \tan^2 A}$$

Derived from $\tan(A+A) = \dfrac{\tan A + \tan A}{1 - \tan A \tan A}$. Note: $\tan 2A$ is undefined when $\tan^2 A = 1$, i.e. $A = 45^{\circ}, 135^{\circ}, \ldots$

When solving equations, choose the form of $\cos 2A$ that produces only one trig function. If the equation has $\sin\theta$, use $\cos 2\theta = 1 - 2\sin^2\theta$. If it has $\cos\theta$, use $\cos 2\theta = 2\cos^2\theta - 1$.

Activity 1 · Pair B2 Pattern Detective: Discovering Double Angle Formulae

Pair activity (8 min): Follow the detective trail — five clues that lead from the compound angle formulae to the double angle formulae. At each clue, discuss with your partner and identify the correct expression from the suspects.

Known formulae: $\sin(A+B) = \sin A \cos B + \cos A \sin B$, $\quad \cos(A+B) = \cos A \cos B - \sin A \sin B$, $\quad \tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$

Clue 1 / 5
Activity 2 · Individual A1 Interactive Manipulative: Predict, Then Verify

Individual activity (5 min): First predict the value of $2\sin A \cos A$ for a fixed angle. Then use the slider to verify that the double angle formula $\sin 2A = 2\sin A \cos A$ holds for every value of $A$.

Predict: Before You Slide

The angle is fixed at $A = 30°$. Using the double angle formula $\sin 2A = 2\sin A \cos A$, predict the value of $2\sin A \cos A$.

0.866
0.500
1.000
0.707
Group · Guided Worked Example 1: Solving Double Angle Equations Exercise 3C Q1(i)(ii)(iii)

Solve the following equations for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$.

(i) $2\sin 2\theta = \cos \theta$   (ii) $\tan 2\theta = 4 \tan \theta$   (iii) $\cos 2\theta + \sin \theta = 0$

👁 Reveal Solution
Part (i): $2\sin 2\theta = \cos\theta$

Replace $\sin 2\theta = 2\sin\theta\cos\theta$:

$2(2\sin\theta\cos\theta) = \cos\theta$

$4\sin\theta\cos\theta = \cos\theta$

$4\sin\theta\cos\theta - \cos\theta = 0$

$\cos\theta(4\sin\theta - 1) = 0$

$\cos\theta = 0$: $\theta = 90^{\circ},\ 270^{\circ}$

$4\sin\theta = 1 \Rightarrow \sin\theta = \tfrac{1}{4}$: $\theta = 14.5^{\circ},\ 165.5^{\circ}$

$\boxed{\theta = 14.5^{\circ},\ 90^{\circ},\ 165.5^{\circ},\ 270^{\circ}}$

Part (ii): $\tan 2\theta = 4\tan\theta$

Replace $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$:

$\dfrac{2\tan\theta}{1-\tan^2\theta} = 4\tan\theta$

$2\tan\theta = 4\tan\theta(1 - \tan^2\theta)$   (multiply by $1 - \tan^2\theta$)

$2\tan\theta = 4\tan\theta - 4\tan^3\theta$

$0 = 2\tan\theta - 4\tan^3\theta = 2\tan\theta(1 - 2\tan^2\theta)$

$\tan\theta = 0$: $\theta = 0^{\circ},\ 180^{\circ},\ 360^{\circ}$

$\tan^2\theta = \tfrac{1}{2} \Rightarrow \tan\theta = \pm\tfrac{1}{\sqrt{2}}$:

$\tan\theta = \tfrac{1}{\sqrt{2}}$: $\theta = 35.3^{\circ},\ 215.3^{\circ}$

$\tan\theta = -\tfrac{1}{\sqrt{2}}$: $\theta = 144.7^{\circ},\ 324.7^{\circ}$

Check: $1 - \tan^2\theta \neq 0$ for all solutions (since $\tan^2\theta \in \{0, \tfrac{1}{2}\}$), so no extraneous solutions.

$\boxed{\theta = 0^{\circ},\ 35.3^{\circ},\ 144.7^{\circ},\ 180^{\circ},\ 215.3^{\circ},\ 324.7^{\circ},\ 360^{\circ}}$

Part (iii): $\cos 2\theta + \sin\theta = 0$

Use $\cos 2\theta = 1 - 2\sin^2\theta$ (chooses the form in $\sin\theta$):

$1 - 2\sin^2\theta + \sin\theta = 0$

$2\sin^2\theta - \sin\theta - 1 = 0$

$(2\sin\theta + 1)(\sin\theta - 1) = 0$

$2\sin\theta + 1 = 0 \Rightarrow \sin\theta = -\tfrac{1}{2}$: $\theta = 210^{\circ},\ 330^{\circ}$

$\sin\theta = 1$: $\theta = 90^{\circ}$

$\boxed{\theta = 90^{\circ},\ 210^{\circ},\ 330^{\circ}}$

Individual · Guided Worked Example 2: Triple Angle via Double Angle Exercise 3C Q3

By first writing $\sin 3\theta$ as $\sin(2\theta + \theta)$, express $\sin 3\theta$ in terms of $\sin\theta$. Hence solve the equation $\sin 3\theta = \sin\theta$ for $0 \leqslant \theta \leqslant 2\pi$.

👁 Reveal Solution
Step 1: Expand $\sin 3\theta = \sin(2\theta + \theta)$

$\sin 3\theta = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta$

Substitute $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$:

$= 2\sin\theta\cos\theta \cdot \cos\theta + (1 - 2\sin^2\theta)\sin\theta$

$= 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta$

Replace $\cos^2\theta = 1 - \sin^2\theta$:

$= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta$

$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta$

$\boxed{\sin 3\theta = 3\sin\theta - 4\sin^3\theta}$

Step 2: Solve $\sin 3\theta = \sin\theta$

Substitute the expansion:

$3\sin\theta - 4\sin^3\theta = \sin\theta$

$2\sin\theta - 4\sin^3\theta = 0$

$2\sin\theta(1 - 2\sin^2\theta) = 0$

Step 3: Solve each factor

$\sin\theta = 0$: $\theta = 0,\ \pi,\ 2\pi$

$1 - 2\sin^2\theta = 0 \Rightarrow \sin^2\theta = \tfrac{1}{2} \Rightarrow \sin\theta = \pm\tfrac{1}{\sqrt{2}}$:

$\sin\theta = \tfrac{1}{\sqrt{2}}$: $\theta = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$

$\sin\theta = -\tfrac{1}{\sqrt{2}}$: $\theta = \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}$

$\boxed{\theta = 0,\ \dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \pi,\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4},\ 2\pi}$

✏️ Concept Quiz
Score: 0 / 4
1. Which of the following is the correct expansion of $\sin 2A$?
$2 \sin A \cos A$
$2 \sin A$
$2 \cos A$
$2 \sin A + 2 \cos A$
$\sin 2A = \sin(A+A) = \sin A \cos A + \cos A \sin A = 2 \sin A \cos A$. The factor of 2 multiplies the product $\sin A \cos A$, not $\sin A$ alone.
2. Which of the following is NOT a valid form of $\cos 2A$?
$2 \sin^2 A - 1$
$1 - 2 \sin^2 A$
$2 \cos^2 A - 1$
$\cos^2 A - \sin^2 A$
The three valid forms are $\cos^2 A - \sin^2 A$, $2\cos^2 A - 1$, and $1 - 2\sin^2 A$. The expression $2\sin^2 A - 1$ is the negative of the third form (i.e. $-\cos 2A$), so it is NOT valid.
3. Which of the following is the correct expansion of $\tan 2A$?
$2 \tan A \over 1 - \tan^2 A$
$2 \tan A \over 1 + \tan^2 A$
$2 \tan A$
$2 \tan A \over \tan^2 A - 1$
$\tan 2A = \tan(A+A) = \dfrac{\tan A + \tan A}{1 - \tan A \tan A} = \dfrac{2\tan A}{1 - \tan^2 A}$. The denominator is $1 - \tan^2 A$ (minus sign), and $\tan 2A$ is undefined when $\tan^2 A = 1$.
4. Simplify $\dfrac{1 + \cos 2A}{\sin 2A}$.
$2 \cos^2 A$
$2 \sin A \cos A$
$\tan A$
$1$
Using $\cos 2A = 2\cos^2 A - 1$: $1 + \cos 2A = 1 + 2\cos^2 A - 1 = 2\cos^2 A$. So $\dfrac{1+\cos 2A}{\sin 2A} = \dfrac{2\cos^2 A}{2\sin A \cos A} = \dfrac{\cos A}{\sin A} = \cot A$. (Note: the question asks for the simplified numerator $1 + \cos 2A = 2\cos^2 A$; the full fraction simplifies to $\cot A$.)
🎉
You scored 0 / 4
📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 3C Q1(iv)(v) Standard

Solve the following equations for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$.

(iv) $\tan \theta \tan 2\theta = 1$   (v) $2\cos 2\theta = 1 + \cos\theta$

💡 Hint

(iv) Replace $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$, then solve the resulting equation in $\tan\theta$.

(v) Replace $\cos 2\theta = 2\cos^2\theta - 1$ (chooses the form in $\cos\theta$), then solve the quadratic in $\cos\theta$.

👁 Reveal Solution
Part (iv): $\tan\theta \cdot \tan 2\theta = 1$

Substitute $\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}$:

$\tan\theta \cdot \dfrac{2\tan\theta}{1-\tan^2\theta} = 1$

$\dfrac{2\tan^2\theta}{1-\tan^2\theta} = 1$

$2\tan^2\theta = 1 - \tan^2\theta$

$3\tan^2\theta = 1 \;\Rightarrow\; \tan^2\theta = \dfrac{1}{3} \;\Rightarrow\; \tan\theta = \pm\dfrac{1}{\sqrt{3}}$

$\tan\theta = \dfrac{1}{\sqrt{3}}$: $\theta = 30^{\circ},\ 210^{\circ}$

$\tan\theta = -\dfrac{1}{\sqrt{3}}$: $\theta = 150^{\circ},\ 330^{\circ}$

Check: $1 - \tan^2\theta = 1 - \tfrac{1}{3} = \tfrac{2}{3} \neq 0$. All solutions valid.

$\boxed{\theta = 30^{\circ},\ 150^{\circ},\ 210^{\circ},\ 330^{\circ}}$

Part (v): $2\cos 2\theta = 1 + \cos\theta$

Use $\cos 2\theta = 2\cos^2\theta - 1$:

$2(2\cos^2\theta - 1) = 1 + \cos\theta$

$4\cos^2\theta - 2 = 1 + \cos\theta$

$4\cos^2\theta - \cos\theta - 3 = 0$

$(4\cos\theta + 3)(\cos\theta - 1) = 0$

$\cos\theta = 1$: $\theta = 0^{\circ},\ 360^{\circ}$

$4\cos\theta + 3 = 0 \Rightarrow \cos\theta = -\tfrac{3}{4}$: $\theta = 138.6^{\circ},\ 221.4^{\circ}$

$\boxed{\theta = 0^{\circ},\ 138.6^{\circ},\ 221.4^{\circ},\ 360^{\circ}}$

Exercise 3C Q2(i)(ii)(iii) Standard

Solve the following equations for $-\pi \leqslant \theta \leqslant \pi$.

(i) $\sin 2\theta = 2\sin\theta$   (ii) $\tan 2\theta = 2\tan\theta$   (iii) $\cos 2\theta - \cos\theta = 0$

👁 Reveal Solution
Part (i): $\sin 2\theta = 2\sin\theta$

$2\sin\theta\cos\theta = 2\sin\theta$

$2\sin\theta(\cos\theta - 1) = 0$

$\sin\theta = 0$: $\theta = -\pi,\ 0,\ \pi$

$\cos\theta = 1$: $\theta = 0$ (already listed)

$\boxed{\theta = -\pi,\ 0,\ \pi}$

Part (ii): $\tan 2\theta = 2\tan\theta$

$\dfrac{2\tan\theta}{1-\tan^2\theta} = 2\tan\theta$

$2\tan\theta = 2\tan\theta(1 - \tan^2\theta)$

$2\tan\theta = 2\tan\theta - 2\tan^3\theta$

$2\tan^3\theta = 0 \;\Rightarrow\; \tan\theta = 0$

$\theta = -\pi,\ 0,\ \pi$

Check: $1 - 0 = 1 \neq 0$. All valid.

$\boxed{\theta = -\pi,\ 0,\ \pi}$

Part (iii): $\cos 2\theta - \cos\theta = 0$

Use $\cos 2\theta = 2\cos^2\theta - 1$:

$2\cos^2\theta - 1 - \cos\theta = 0$

$2\cos^2\theta - \cos\theta - 1 = 0$

$(2\cos\theta + 1)(\cos\theta - 1) = 0$

$\cos\theta = -\dfrac{1}{2}$: $\theta = -\dfrac{2\pi}{3},\ \dfrac{2\pi}{3}$

$\cos\theta = 1$: $\theta = 0$

$\boxed{\theta = -\dfrac{2\pi}{3},\ 0,\ \dfrac{2\pi}{3}}$

Exercise 3C Q5 Medium

Simplify $\dfrac{1 + \cos 2\theta}{\sin 2\theta}$.

💡 Hint

Use the form $\cos 2\theta = 2\cos^2\theta - 1$ to simplify the numerator, and $\sin 2\theta = 2\sin\theta\cos\theta$ for the denominator. Then cancel common factors.

👁 Reveal Solution
Step 1: Simplify the numerator

Using $\cos 2\theta = 2\cos^2\theta - 1$:

$1 + \cos 2\theta = 1 + (2\cos^2\theta - 1) = 2\cos^2\theta$

Step 2: Simplify the denominator

$\sin 2\theta = 2\sin\theta\cos\theta$

Step 3: Combine and cancel

$\dfrac{1 + \cos 2\theta}{\sin 2\theta} = \dfrac{2\cos^2\theta}{2\sin\theta\cos\theta} = \dfrac{\cos\theta}{\sin\theta} = \boxed{\cot\theta}$

Exercise 3C Q4 Medium

Solve $\cos 3\theta = 1 - 3\cos\theta$ for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$.

💡 Hint

Write $\cos 3\theta = \cos(2\theta + \theta)$ and expand using the compound angle formula. Use $\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$, then replace $\sin^2\theta = 1 - \cos^2\theta$ to get an expression in $\cos\theta$ only.

👁 Reveal Solution
Step 1: Expand $\cos 3\theta = \cos(2\theta + \theta)$

$\cos 3\theta = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$

Substitute $\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$:

$= (2\cos^2\theta - 1)\cos\theta - 2\sin\theta\cos\theta \cdot \sin\theta$

$= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta$

Replace $\sin^2\theta = 1 - \cos^2\theta$:

$= 2\cos^3\theta - \cos\theta - 2(1-\cos^2\theta)\cos\theta$

$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$

$= 4\cos^3\theta - 3\cos\theta$

Step 2: Set up and solve the equation

$4\cos^3\theta - 3\cos\theta = 1 - 3\cos\theta$

$4\cos^3\theta = 1$

$\cos^3\theta = \dfrac{1}{4} \;\Rightarrow\; \cos\theta = \sqrt[3]{\dfrac{1}{4}} = \dfrac{1}{\sqrt[3]{4}} \approx 0.6300$

Step 3: Solve in the given range

Since $\cos\theta > 0$, $\theta$ is in quadrants I and IV.

$\theta = \arccos(0.6300) \approx 51.0^{\circ}$

$\theta = 360^{\circ} - 51.0^{\circ} = 309.0^{\circ}$

$\boxed{\theta \approx 51.0^{\circ},\ 309.0^{\circ}}$

Exit Ticket

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