Double Angle Formulae (AA SL)
The double angle formulas turn \(\sin2\theta\) and \(\cos2\theta\) into expressions built entirely from \(\sin\theta\) and \(\cos\theta\), which is what makes them so useful once you already know one ratio. Cosine has three equivalent forms and picking the right one saves a step, so this page focuses on when to reach for each. It's part of the broader Identities & Exact Values topic.
21 questions on this sub-topic.
Double angle for sine and cosine
Covered under IB syllabus reference SL3.6: the double angle identities for sine and cosine, alongside the Pythagorean identity they're derived from.
Double angle for sine
\[\sin2\theta=2\sin\theta\cos\theta\]
Only one form - always the product of sine and cosine, doubled.
Double angle for cosine
\[\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta\]
Pick whichever form matches the information you're given - both are equivalent via the Pythagorean identity.
Need the Pythagorean identity these come from? See Pythagorean and Reciprocal Identities, or the full syllabus table at Identities & Exact Values.
Worked examples
Given \(\sin\theta=\tfrac35\) and \(\cos\theta=\tfrac45,\) find \(\sin 2\theta.\)
Worked solution
Use \(\sin 2\theta = 2\sin\theta\cos\theta.\) M1
\(\sin2\theta = 2\cdot\tfrac35\cdot\tfrac45 = \tfrac{24}{25}.\) A1
Given \(\cos\theta=\tfrac{1}{3}\), find \(\cos2\theta\).
Worked solution
Given \(\cos\theta\), the fastest form is \(\cos2\theta=2\cos^2\theta-1.\) M1
\(\cos2\theta=2\cdot\tfrac19-1=\tfrac29-1=-\tfrac79.\) A1
Given \(\cos\theta=\tfrac{5}{13}\) with \(\theta\) acute, find \(\sin2\theta\).
Worked solution
\(\sin\theta=\sqrt{1-\tfrac{25}{169}}\) M1 \(=\tfrac{12}{13}\) (positive, acute). A1
\(\sin 2\theta=2\sin\theta\cos\theta=2\cdot\tfrac{12}{13}\cdot\tfrac{5}{13}\): M1 \(=\tfrac{120}{169}.\) A1
Given \(\sin\theta=\tfrac13\) and \(\cos\theta=\tfrac{2\sqrt2}{3}\), find \(\sin2\theta.\)
Worked solution
\(\sin 2\theta = 2\sin\theta\cos\theta.\) M1
\(= 2\cdot\tfrac13\cdot\tfrac{2\sqrt2}{3}.\) A1
Evaluate: M1
\(= \tfrac{4\sqrt2}{9}.\) A1
Given \(\cos\theta=\tfrac34\), find \(\cos2\theta.\)
Worked solution
\(\cos 2\theta = 2\cos^2\theta - 1.\) M1
\(= 2\cdot\tfrac{9}{16} - 1.\) A1
\(\tfrac98 - 1\) M1
\(= \tfrac18.\) A1
Show that \(\sin^2 x=\dfrac{1-\cos2x}{2}.\)
Worked solution
\(\cos 2x = 1 - 2\sin^2 x.\) M1 A1
\(2\sin^2 x = 1 - \cos 2x\) M1 \(\Rightarrow \sin^2 x = \dfrac{1 - \cos 2x}{2}.\) A1 AG ∎
Common mistakes
- Writing \(\cos2\theta=2\cos\theta\). Doubling the angle is not the same as doubling the ratio - the correct identity is \(\cos2\theta=2\cos^2\theta-1\) (or an equivalent form).
- Mixing up the three forms of \(\cos2\theta\). All three - \(\cos^2\theta-\sin^2\theta\), \(2\cos^2\theta-1\), \(1-2\sin^2\theta\) - give the same value, but only the one matching the ratio you're given avoids an extra step to find the other ratio first.
- Forgetting to square before doubling. \(\sin2\theta=2\sin\theta\cos\theta\) has no squared terms, but \(\cos2\theta\) always does - dropping a square is an easy slip when switching between the two formulas mid-question.
Ready to practise properly?
23 double-angle questions, marked instantly like the real exam.
Quick answers
What is the double angle formula for sine?
\(\sin2\theta=2\sin\theta\cos\theta\). There is only one form, so it's usually the easier of the two double-angle rules to apply.
Why does \(\cos2\theta\) have three different forms?
\(\cos2\theta=\cos^2\theta-\sin^2\theta\), and applying the Pythagorean identity to swap one ratio for the other gives the equivalent forms \(2\cos^2\theta-1\) and \(1-2\sin^2\theta\). Pick whichever form matches the ratio you already know.