What happens when both angles in a compound angle formula are the same? You get the double angle formulas, a powerful special case that shows up everywhere from integration to projectile motion to signal processing. There are three different forms for $\cos(2A)$ alone, and knowing which one to reach for is half the skill. In this lesson you'll derive them all, understand why they exist, and learn to apply each form strategically.
How could you find $\sin(2\theta)$ if you know $\sin(\theta + \theta)$? Think about this without using any formula, what do you expect will happen when you apply the compound angle formula for $\sin(A+B)$ with $A = B = \theta$? Write your prediction before reading on.
Every double angle problem has exactly two moves, and the key is recognising which form of $\cos(2A)$ makes the algebra cleanest.
Move 1, Substitute and simplify. The double angle formulas are derived by setting $B = A$ in the compound angle formulas. For $\sin(2A)$: $\sin(A+A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$. For $\cos(2A)$: substitute $B=A$ into $\cos(A+B)$ to get $\cos^2 A - \sin^2 A$. For $\tan(2A)$: set $B=A$ in the tangent formula.
Move 2, Choose the right form of $\cos(2A)$. There are three equivalent forms. Pick the one that matches what information you have: use $\cos^2 A - \sin^2 A$ when you know both; use $2\cos^2 A - 1$ when you know $\cos A$; use $1 - 2\sin^2 A$ when you know $\sin A$. Choosing the wrong form forces extra substitution steps and risks errors.
Setting $B = A$ in the compound angle formulas gives all three double angle formulas directly.
Derivation of $\sin(2A)$:
$\sin(2A) = \sin(A+A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$
Derivation of $\cos(2A)$:
$\cos(2A) = \cos(A+A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A$
Using $\sin^2 A = 1 - \cos^2 A$: $\cos(2A) = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1$
Using $\cos^2 A = 1 - \sin^2 A$: $\cos(2A) = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A$
Derivation of $\tan(2A)$:
$\tan(2A) = \tan(A+A) = \dfrac{\tan A + \tan A}{1 - \tan A \cdot \tan A} = \dfrac{2\tan A}{1 - \tan^2 A}$
(2A) = 2 A A, always a product of both and; (2A) = ^2 A - ^2 A = 2^2 A - 1 = 1 - 2^2 A
Pause, copy all three forms of $\cos 2A$ into your book: $\cos^2A-\sin^2A$, $2\cos^2A-1$, $1-2\sin^2A$; and the single form $\sin 2A = 2\sin A\cos A$.
Quick check: Which of the following is a correct alternative form of $\cos(2A)$?
We just saw that $\sin 2A = 2\sin A\cos A$ and $\cos 2A$ has three equivalent forms from setting $B=A$. That raises a question: when a quadrant condition like $\frac{\pi}{2}<\theta<\pi$ is given, how does that restrict the sign of a missing ratio before we can substitute? This card answers it → identify the quadrant, then $\cos\theta = -\sqrt{1-\sin^2\theta}$ (negative in Q2) before substituting into the double angle formula.
When a quadrant condition is given (e.g., $\frac{\pi}{2} < \theta < \pi$), you must use it to determine the correct signs of the missing trig ratios before substituting.
Quadrant sign summary:
Procedure:
Always identify the quadrant before finding missing ratios, it determines the sign; In Q2: is negative; in Q3: both and are negative
Pause, copy the sign-determination rule into your book: in Q2 $\cos\theta < 0$, so use $\cos\theta = -\sqrt{1-\sin^2\theta}$; identify the quadrant from the inequality constraint before substituting into the double angle formula.
Did you get this? True or false: if $\sin\theta = \frac{3}{5}$ and $\theta$ is in Q2, then $\cos\theta = \frac{4}{5}$.
Worked examples · 3 in a row, reveal as you go
If $\sin\theta = \dfrac{3}{5}$ and $\dfrac{\pi}{2} < \theta < \pi$, find $\sin(2\theta)$.
Express $\cos^2\theta$ in terms of $\cos(2\theta)$. Hence find $\cos^2 30°$ exactly.
If $\tan\theta = 2$, find $\tan(2\theta)$.
Fill the gap: Starting from $\cos(2\theta) = 1 - 2\sin^2\theta$, rearranging gives $\sin^2\theta = \dfrac{1 - \square}{2}$, where $\square =$ . This is called the formula.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $\tan\theta = 3$, then $\tan(2\theta) = \dfrac{6}{1 - 9} = -\dfrac{3}{4}$.
Odd one out: Three of these are valid expressions for $\cos(2A)$. Which one is incorrect?
Activities · practice with the ideas
If $\cos\theta = \dfrac{12}{13}$ and $0 < \theta < \dfrac{\pi}{2}$, find $\sin(2\theta)$.
Express $\sin^2\theta$ in terms of $\cos(2\theta)$. Use this to find the exact value of $\sin^2 45°$.
If $\tan\theta = 2$, find $\tan(2\theta)$ and explain whether the result is positive or negative.
If $\sin\theta = \dfrac{1}{3}$ and $\theta$ is obtuse (Q2), find the exact value of $\cos(2\theta)$. Choose the most efficient form of $\cos(2A)$.
Explain in your own words why there are three different forms for $\cos(2A)$ and describe a situation where you would choose each one.
Earlier you were asked: How could you find $\sin(2\theta)$ from $\sin(\theta + \theta)$?
Applying $\sin(A+B) = \sin A \cos B + \cos A \sin B$ with $A = B = \theta$: $\sin(\theta+\theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta$. So $\sin(2\theta) = 2\sin\theta\cos\theta$, not just $2\sin\theta$. The factor of $\cos\theta$ is essential and arises naturally from the cross-product terms of the compound angle formula.
Why are there three different forms for $\cos(2A)$? All three are algebraically equivalent, but each suits a different context. $\cos^2 A - \sin^2 A$ shows the symmetry; $2\cos^2 A - 1$ and $1 - 2\sin^2 A$ are power reduction tools that eliminate one trig function entirely, which is exactly what you need when given only $\cos A$ or only $\sin A$.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. If $\cos\theta = \dfrac{12}{13}$ and $0 < \theta < \dfrac{\pi}{2}$, find $\sin(2\theta)$. (2 marks)
Q2. Express $\sin^2\theta$ in terms of $\cos(2\theta)$. (2 marks)
Q3. If $\tan\theta = 2$, find $\tan(2\theta)$ and explain why the three forms of $\cos(2A)$ are all equivalent to each other. (2 marks)
Activity 1: 1. $\sin\theta = \frac{5}{13}$ (Q1, positive); $\sin(2\theta) = 2\cdot\frac{5}{13}\cdot\frac{12}{13} = \frac{120}{169}$ · 2. $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$; $\sin^2 45° = \frac{1-\cos90°}{2} = \frac{1-0}{2} = \frac{1}{2}$ · 3. $\tan(2\theta) = \frac{4}{1-4} = -\frac{4}{3}$; negative because $1 - \tan^2\theta = 1-4 < 0$ makes the denominator negative while the numerator is positive · 4. Use $1-2\sin^2\theta = 1-2\cdot\frac{1}{9} = 1-\frac{2}{9} = \frac{7}{9}$ · 5. All three forms arise from $\cos^2 A - \sin^2 A$ by substituting either $\sin^2 A = 1 - \cos^2 A$ to get $2\cos^2 A - 1$, or $\cos^2 A = 1 - \sin^2 A$ to get $1 - 2\sin^2 A$. Choose $2\cos^2 A - 1$ when you know $\cos A$; choose $1 - 2\sin^2 A$ when you know $\sin A$; use $\cos^2 A - \sin^2 A$ when you know both.
Q1 (2 marks): $\sin\theta = \sqrt{1-\frac{144}{169}} = \frac{5}{13}$ (positive, Q1) [1]; $\sin(2\theta) = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}$ [1].
Q2 (2 marks): $\cos(2\theta) = 1 - 2\sin^2\theta$ [0.5]; rearrange: $2\sin^2\theta = 1 - \cos(2\theta)$ [0.5]; $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$ [1].
Q3 (2 marks): $\tan(2\theta) = \frac{2\times2}{1-4} = \frac{4}{-3} = -\frac{4}{3}$ [1]. The three forms are equivalent because substituting $\sin^2 A = 1-\cos^2 A$ into $\cos^2 A - \sin^2 A$ gives $2\cos^2 A - 1$; substituting $\cos^2 A = 1 - \sin^2 A$ gives $1 - 2\sin^2 A$, so each form is derived from the others via the Pythagorean identity [1].
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