Compound Angle and Double Angle Formulae: From sin(A±B) to sin2A, cos2A and tan2A

Which expression is equal to $\cos 2A$ for every value of $A$?

• $\cos^2 A - \sin^2 A$

  Correct — exactly what substituting $B=A$ into $\cos(A+B) \equiv \cos A\cos B - \sin A\sin B$ produces, before any further rewriting with $\cos^2 A + \sin^2 A \equiv 1$. (Also equal to $2\cos^2 A - 1$ and $1-2\sin^2 A$ — all three are the same quantity.)

• B$\cos^2 A - 1$

  A real, recorded exam error — examiners note candidates writing exactly this in place of one of the genuine forms (Oct 2020, Q1). It's missing a factor of 2: the real form built from $\cos^2 A + \sin^2 A \equiv 1$ is $2\cos^2 A - 1$, not $\cos^2 A - 1$.

• C$1 - \cos A$

  This treats doubling the angle as if it directly changed the OUTPUT by a fixed amount, the way $f(x)+1$ shifts a graph up — but $\cos 2A$ is a genuinely different function of $A$, not $\cos A$ plus or minus a constant, and no such simple shift exists for cosine.

• D$2\cos A - 1$

  Close to a real form but missing a square: the correct identity is $2\cos^2 A - 1$, with $\cos A$ squared. Dropping the square is an easy slip once the '2' and the '−1' are remembered correctly but the exponent isn't.

Traps tested: Wrong double angle cosine form · Double angle treated as additive shift · Cos squared exponent dropped

$\sin A = \frac{3}{5}$, where $A$ is acute ($0° < A < 90°$). What is $\cos 2A$?

• $\frac{7}{25}$

  Correct. Since $A$ is acute, $\cos A > 0$, so $\cos A = \sqrt{1-\left(\frac{3}{5}\right)^2} = \frac{4}{5}$ (the 3-4-5 triangle). Then $\cos 2A = 1-2\sin^2 A = 1-2\left(\frac{9}{25}\right) = \frac{7}{25}$ — or, checking with the other form, $\cos^2 A - \sin^2 A = \frac{16}{25}-\frac{9}{25} = \frac{7}{25}$. Both agree, because they're the same quantity.

• B$\frac{24}{25}$

  This is $2\sin A\cos A = 2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)$ — the formula for $\sin 2A$, applied to a $\cos 2A$ question. Sharing the same two ingredients, $\sin A$ and $\cos A$, doesn't make the two formulae interchangeable.

• C$\frac{1}{25}$

  This is $(\cos A - \sin A)^2 = \left(\frac{4}{5}-\frac{3}{5}\right)^2$, not $\cos^2 A - \sin^2 A$. A difference of two squares, $x^2-y^2$, is not the same expression as the square of a difference, $(x-y)^2$ — expanding the second gives $x^2 - 2xy + y^2$, an extra cross term the first doesn't have.

• D$\frac{43}{25}$

  This is $1+2\sin^2 A$ — the sign inside the identity flipped. It's also a value cosine can never take: $\cos 2A$ is bounded between $-1$ and $1$ for every real angle, and $\frac{43}{25} > 1$ is an immediate sign something went wrong before further checking is even needed.

Traps tested: Sin2A formula used for cos2A · Difference of squares confused with square of difference · Sign error in double angle cosine form

$\tan A = 2$. What is $\tan 2A$?

• $-\frac{4}{3}$

  Correct. $\tan 2A = \dfrac{2\tan A}{1-\tan^2 A} = \dfrac{2(2)}{1-2^2} = \dfrac{4}{1-4} = \dfrac{4}{-3} = -\dfrac{4}{3}$.

• B$-4$

  The denominator is missing its square: this computes $\frac{2(2)}{1-2}$, using $\tan A$ alone where the formula needs $\tan^2 A$. The denominator of $\tan 2A$ is $1-\tan^2 A$, not $1-\tan A$.

• C$\frac{4}{5}$

  The sign in the denominator is flipped: this uses $1+\tan^2 A$ instead of $1-\tan^2 A$. The '+' belongs to $\sec^2 A \equiv 1+\tan^2 A$ (a different identity, spec 2.2) — the double-angle tangent formula's denominator is a minus.

• D$8$

  This computes $2\tan^2 A = 2(2)^2$ — squaring the wrong instance of $\tan A$ and dropping the fraction entirely. $\tan 2A$ is a ratio, $\dfrac{2\tan A}{1-\tan^2 A}$, never a product.

Traps tested: Tan squared not formed in denominator · Sign of denominator flipped · Tan2A formula invented

To rewrite $4\sin^2\theta - \cos 2\theta = 1$ as an equation entirely in $\sin\theta$, which form of $\cos 2\theta$ should replace it?

• $1-2\sin^2\theta$

  Correct. The equation already has a $\sin^2\theta$ term; substituting the form of $\cos 2\theta$ that's ALSO written in $\sin^2\theta$ means nothing new is introduced. (It gives $4\sin^2\theta - (1-2\sin^2\theta) = 1$, i.e. $6\sin^2\theta = 2$ — one function, one equation.)

• B$\cos^2\theta - \sin^2\theta$

  True, but unhelpful here: it introduces a fresh $\cos^2\theta$ term into an equation that otherwise has none, which then has to be eliminated with $\sin^2\theta+\cos^2\theta\equiv 1$ anyway — extra work a better choice avoids entirely.

• C$2\cos^2\theta - 1$

  The same problem as the other cosine-only form: it adds a $\cos^2\theta$ term the equation didn't have, instead of matching the $\sin^2\theta$ that's already there.

• D$2\cos\theta$

  This isn't one of the three real forms at all — it treats doubling the angle as doubling the function's value directly, the way multiplying $x$ by 2 doubles $2x$. Cosine doesn't scale that way; $\cos 2\theta$ is a different function of $\theta$, not twice $\cos\theta$.

Traps tested: Unhelpful cos2a form chosen · Double angle treated as linear scaling

Which response to 'Prove that $(\sin\theta+\cos\theta)^2 \equiv 1+\sin 2\theta$' is certain to earn full marks?

• Expands the bracket to three terms, applies $\sin^2\theta+\cos^2\theta\equiv 1$ to two of them, then identifies the remaining term as $\sin 2\theta$ — every line shown

  Correct — the complete method, with the identity that does the actual work (spotting $2\sin\theta\cos\theta \equiv \sin 2\theta$) shown explicitly, not just asserted.

• BWrites '$(\sin\theta+\cos\theta)^2 = 1+\sin 2\theta$, as required' directly, with no working shown

  This is the exact failure documented on real 'prove that' questions on this topic — jumping straight to the given answer. A 'show that'/'prove' instruction is marked on the presence of the intermediate lines, not just on whether the final line matches; with no working, no method mark is available to award.

• CSubstitutes $\theta=45°$ into both sides, confirms they're equal, and concludes the identity is proved

  One numeric value matching is consistent with the identity being true — it isn't proof that it's true for every $\theta$. An identity claims equality for ALL values of the variable; checking one value can only ever disprove an identity (by finding a mismatch), never establish one.

• DExpands the bracket correctly, but writes 'which equals 1' after $\sin^2\theta+\cos^2\theta$ with no reference to the identity used to get there

  The value is right and the justification is missing — $\sin^2\theta+\cos^2\theta\equiv 1$ is itself a memorised identity being invoked, and 'show that' marking expects that invocation to be visible, not just the arithmetic it enables.

Traps tested: Incomplete working on a prove that · Proof by single example

Using $\cos 2A \equiv 1-2\sin^2 A$ with $A=\frac{\theta}{2}$, which expression equals $1-\cos\theta$?

• $2\sin^2\left(\frac{\theta}{2}\right)$

  Correct. Setting $A=\frac{\theta}{2}$ in $\cos 2A \equiv 1-2\sin^2 A$ gives $\cos\theta \equiv 1-2\sin^2\left(\frac{\theta}{2}\right)$; rearranging for $1-\cos\theta$ gives exactly $2\sin^2\left(\frac{\theta}{2}\right)$.

• B$2\cos^2\left(\frac{\theta}{2}\right)$

  This is the OTHER half-angle identity — it equals $1+\cos\theta$, not $1-\cos\theta$, and comes from the $2\cos^2 A - 1$ form of $\cos 2A$ instead of the $1-2\sin^2 A$ form. The two half-angle results are mirror images, and it's easy to reach for the wrong one.

• C$\sin\left(\frac{\theta}{2}\right)$

  Missing both the square and the factor of 2. The identity relates $1-\cos\theta$ to $\sin^2\left(\frac{\theta}{2}\right)$, doubled — not to $\sin\left(\frac{\theta}{2}\right)$ on its own.

• D$2\sin\theta\cos\theta$

  This is the double-angle formula for $\sin 2\theta$, built from the FULL angle $\theta$ — the wrong identity entirely for a question about the HALF angle $\frac{\theta}{2}$.

Traps tested: Half angle form confused with 1 plus cos · Half angle square and factor dropped · Sin2A formula misapplied to half angle question