Setting B equal to A in the addition formulae collapses two angles into one — and hands you three new identities almost for free.
9 min read4 worked examples
Deriving the double angle formulae
Substituting $B=A$ into the addition formulae from the previous lesson turns a compound angle $A+B$ into simply $2A$ — producing the double angle formulae.
Derivationsin 2θ, from sin(A+B) with B = A
1
Start from $\sin(A+B) = \sin A\cos B + \cos A\sin B$.
The sine addition formula.
2
Set $B=A$ (and rename $A$ to $\theta$): $\sin(\theta+\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta$.
Substituting equal angles collapses the sum to 2θ.
3
$\sin2\theta = 2\sin\theta\cos\theta$.
Combine the two identical terms on the right.
▪ QED
Derivationcos 2θ, from cos(A+B) with B = A, in three equivalent forms
1
Start from $\cos(A+B) = \cos A\cos B - \sin A\sin B$, and set $B=A$: $\cos2\theta = \cos^2\theta - \sin^2\theta$.
Having three equivalent forms of $\cos2\theta$ is deliberate: pick whichever matches the rest of an equation. If an equation already involves $\cos\theta$ only, use $2\cos^2\theta-1$; if it involves $\sin\theta$ only, use $1-2\sin^2\theta$; if both appear, the original $\cos^2\theta-\sin^2\theta$ may be cleanest.
Applications: half-angle problems and equations
Because $\cos2\theta$ links directly to $\cos\theta$ or $\sin\theta$ via a quadratic, it's often used to move between an angle $\theta$ and its double $2\theta$ — for instance, finding $\cos\theta$ given $\cos2\theta$, or solving equations that mix $\sin2\theta$/$\cos2\theta$ with $\sin\theta$/$\cos\theta$ by rewriting everything in terms of a single angle first.
Worked example 1
Finding cos θ from cos 2θ
Given that $\cos2\theta = \dfrac{7}{9}$ and $\theta$ is acute, find the exact value of $\cos\theta$.
MethodUse the $2\cos^2\theta-1$ form and solve the resulting quadratic
Rearrange to zero and factorise out sin θ — never divide by sin θ, as that loses solutions.
3
Either $\sin\theta=0 \Rightarrow \theta = 0°, 180°$, or $\cos\theta=\dfrac12 \Rightarrow \theta = 60°, 300°$.
Solve each factor equal to zero separately across the full interval.
4
$\theta = 0°, 60°, 180°, 300°$.
Combine all solutions from both factors.
▪ solved
The key insight
Dividing both sides by $\sin\theta$ early would silently throw away the solutions where $\sin\theta=0$. Factorising instead of dividing is the safe way to solve any equation where a common factor could be zero.
Substitute sin 2θ = 2 sin θ cos θ, and use the 2cos²θ − 1 form of cos 2θ since a "1 +" is already present to cancel with the "−1".
2
$= \dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}$.
The 1 and −1 cancel in the denominator.
3
$= \dfrac{\sin\theta}{\cos\theta}$.
Cancel the common factor of 2cos θ from numerator and denominator.
4
$= \tan\theta$, which is the right-hand side.
sin θ / cos θ = tan θ by definition.
▪ proven
Look deeper — choosing the right form
The choice of $2\cos^2\theta-1$ over the other two equivalent forms of $\cos2\theta$ wasn't arbitrary — it was the one that made the "$1+$" in the denominator collapse cleanly. Scanning ahead for which form will cancel nicely is a key proof-writing skill with these identities.
Common mistake
Writing $\cos2\theta = 2\cos\theta - 1$ or $\sin2\theta = 2\sin\theta$ — dropping the squares, or forgetting to multiply, entirely changes the identity. Always keep $\sin2\theta=2\sin\theta\cos\theta$ and the three equivalent squared forms of $\cos2\theta$ precise; a small slip here cascades through an entire solution.
Check yourself
Solve $\cos2\theta = \cos\theta$ for $0 \le \theta < 360°$. (Hint: use the 2cos²θ − 1 form of cos 2θ to get a quadratic in cos θ.)