Equations with Double/Compound Angle (AA HL)
Once an equation involves \(\sin2\theta\), \(\cos2\theta\), or an expression like \(\cos75^\circ\) that isn't a standard angle, you first have to rewrite it using a compound or double angle identity before any of your usual solving techniques apply. This page focuses on that rewriting step and the equations it unlocks. It's part of the broader Identities & Equations topic.
27 questions on this sub-topic.
The key identities
Covered under IB syllabus reference AHL3.10: compound angle identities, and the double angle identities derived from them. All four are in the formula booklet.
Compound angle
\(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\)
\(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\)
Use to expand an angle that's a sum or difference of two known angles - e.g. \(75^\circ = 45^\circ+30^\circ\).
Double angle
\(\sin2\theta=2\sin\theta\cos\theta\)
\(\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta\)
Set \(A=B=\theta\) in the compound formulas. Pick whichever \(\cos2\theta\) form leaves your equation in a single trig function.
Need the full syllabus wording and formula-booklet reference table? See Identities & Equations.
Worked examples
Find the exact value of \(\cos75^\circ.\)
Worked solution
\(\cos 75^\circ = \cos(45^\circ + 30^\circ)\) M1 \(= \cos45\cos30 - \sin45\sin30.\) A1
\(\tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} - \tfrac{\sqrt2}{2}\cdot\tfrac12\) M1 \(= \dfrac{\sqrt6 - \sqrt2}{4}.\) A1
Solve \(\cos2x + \sin x = 0\) for \(0\le x<2\pi.\)
(a)(i) Give the solution with \(x<2.618\).
(a)(ii) Give the solution with \(2.618
Worked solution
(a)(i) \(\sin x = 1 \Rightarrow x = \tfrac\pi2\) A1
(a)(ii) \(\sin x = -\tfrac12 \Rightarrow x = \tfrac{7\pi}{6}\) A1
(a)(iii) or \(x=\tfrac{11\pi}{6}.\) A1
Prove that \(\dfrac{1-\cos2\theta}{\sin2\theta}=\tan\theta.\)
Worked solution
\(1 - \cos2\theta = 2\sin^2\theta\) M1 A1 and \(\sin2\theta = 2\sin\theta\cos\theta.\) M1
\(\dfrac{2\sin^2\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta}\) A1 \(= \tan\theta.\) A1 AG ∎
Given \(\sin\theta=\tfrac35\) with \(\theta\) acute, find \(\sin2\theta.\)
Worked solution
\(\cos\theta = \tfrac45\) (acute). M1
\(\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\tfrac35\cdot\tfrac45\) A1
\(= \tfrac{24}{25}.\) A1
Common mistakes
- Splitting \(\sin(A+B)\) into \(\sin A+\sin B\). Sine and cosine don't distribute over addition - the compound angle formula has four terms, not two.
- Picking the wrong \(\cos2\theta\) form. If the rest of the equation is in \(\sin\theta\) only, use \(\cos2\theta=1-2\sin^2\theta\) so everything becomes a quadratic in one variable - substituting the \(\cos^2\theta-\sin^2\theta\) form instead leaves two different functions in play.
- Dropping solutions when solving in an interval. After forming a quadratic in \(\sin\theta\) or \(\cos\theta\), both roots usually give valid angles - check every root against the interval before writing a final answer list.
Ready to practise properly?
27 compound/double-angle equation questions, marked instantly like the real exam.
Quick answers
What is the double angle formula for sine?
\(\sin2\theta=2\sin\theta\cos\theta\). It follows directly from the compound angle formula \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) by setting \(A=B=\theta\).
Why does \(\cos2\theta\) have three different forms?
\(\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta\). The last two come from substituting the Pythagorean identity, and each is useful for a different kind of equation - the sin-only or cos-only forms are what let you form a quadratic in one variable. For GDC-friendly graphical checks, see the parent topic's GDC guidance.