Question bank — Double angle formulas — sin 2A, cos 2A (three forms), tan 2A

False. It only holds when $\cos A=1$, i.e. $A=0^{\circ },360^{\circ },\dots$, because the true formula is $\sin 2A=2\sin A\cos A$ and the two agree only when the extra factor $\cos A$ equals $1$.

True. They are equal for every $A$ because you get the second from the first by the swap ${\sin }^{2}A=1-{\cos }^{2}A$; they differ only in appearance, not in value.

False. Since $\sin 2A=2\sin A\cos A=\sin A\cdot (2\cos A)$, it exceeds $\sin A$ only when $2\cos A>1$, i.e. $\cos A>\frac{1}{2}$; for $A$ near $90^{\circ }$, $\cos A$ is tiny, so $\sin 2A$ is much smaller than $\sin A$ (e.g. at $A=80^{\circ }$, $\sin 2A=\sin 160^{\circ }\approx 0.34<\sin 80^{\circ }\approx 0.98$).

False. The $2$ is not a free multiplier: $\sin A$ and $\cos A$ can't both be $1$ at once. The product peaks at $A=45^{\circ }$ giving $2\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=1$, so the maximum is $1$.

Approximately, not exactly. For small $A$, ${\tan }^{2}A\approx 0$, so $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}\approx 2\tan A$ — but it is never exactly equal unless $\tan A=0$.

False (incomplete). $\cos 2A=1$ means $2A=0^{\circ },\pm 360^{\circ },\pm 720^{\circ },\dots$, so $A=0^{\circ },\pm 180^{\circ },\pm 360^{\circ },\dots$ — any integer multiple of $180^{\circ }$, not only $A=0^{\circ }$.

False. That would need ${\sin }^{2}2A+{\cos }^{2}2A=0$, contradicting the Pythagorean identity which forces the sum to be $1$; at least one is always non-zero.

True. Replacing $A$ by $2A$ compresses the graph horizontally by a factor of $2$, so over any interval $\sin 2A$ oscillates twice as fast as $\sin A$ — exactly what the second figure shows (orange wave packs two humps into the blue wave's one).

==Reasoning is fine, but the WHY is being skipped.== Here is the missing step: ${\sin }^{2}A\le 1$, so $2{\sin }^{2}A\le 2$, so $1-2{\sin }^{2}A\ge 1-2=-1$; and ${\sin }^{2}A\ge 0$ gives $1-2{\sin }^{2}A\le 1$. Hence $1-2{\sin }^{2}A\in [-1,1]$ — the bound comes precisely from $0\le {\sin }^{2}A\le 1$, so that inequality is the check.

==Error: the sign of $\cos A$ was assumed.== $\cos A=\pm \frac{4}{5}$ depending on the quadrant. Luckily $\cos 2A=1-2{\sin }^{2}A$ doesn't need $\cos A$, so it is $\frac{7}{25}$ regardless — but the stated reasoning (via a fixed $\cos A$) is unsafe.

==Error: $\tan 90^{\circ }$ is undefined, so you cannot plug it in.== Instead go back to $2A=180^{\circ }$ and evaluate $\tan 180^{\circ }=0$ directly; the formula's ingredients broke before the formula did.

Error: no such symmetry. Using $1-\cos 2A=2{\sin }^{2}A$ gives $\frac{2\sin A\cos A}{2{\sin }^{2}A}=\frac{\cos A}{\sin A}=\cot A$, not $\tan A$ (valid where $\sin A\ne 0$). The $+$ and $-$ versions produce reciprocal results.

Error: it's a nonlinear transform, not a shift. You square $\cos A$, scale by $2$, then drop by $1$ — this changes the shape and frequency, producing a wave of double the frequency, not a vertical translation. The first figure contrasts the two curves so you can see they are not parallel copies.

Half-right, the "nothing else" is wrong. $\sin 2A=0$ also when $\cos A=0$ (e.g. $A=90^{\circ }$, where $\sin 2A=\sin 180^{\circ }=0$). Both factors offer routes to zero.

Because $\cos 2A={\cos }^{2}A-{\sin }^{2}A$ contains both squares, so the Pythagorean Identity lets you eliminate either one to get two alternatives; $\sin 2A=2\sin A\cos A$ has no square to trade.

Because $1+(2{\cos }^{2}A-1)=2{\cos }^{2}A$: the constant $+1$ exactly cancels the form's $-1$, collapsing everything to a single clean term.

Rearranging $\cos 2A=1-2{\sin }^{2}A$ gives ${\sin }^{2}A=\frac{1-\cos 2A}{2}$, which lowers a squared trig function to a first-power one — exactly what Integration of sin²x and cos²x needs. See Power-Reduction / Half-Angle Formulas.

Because the denominator $1-{\tan }^{2}A$ hits $0$ when ${\tan }^{2}A=1$; geometrically $2A=90^{\circ }$, where $\tan$ itself is undefined, so the formula faithfully reports that singularity.

We can, but the addition formulas are the shortest path: setting $X=Y=A$ in Compound (Addition) Angle Formulas gives all of them in one substitution, so re-proving from scratch just re-does that work.

$\cos 2A=1-2{\sin }^{2}A$ depends only on ${\sin }^{2}A$, immune to sign; but $\sin 2A=2\sin A\cos A$ is linear in $\sin A$, so its sign follows.

It is exactly $\frac{\sin 2A}{\cos 2A}$; dividing $\frac{2\sin A\cos A}{{\cos }^{2}A-{\sin }^{2}A}$ top and bottom by ${\cos }^{2}A$ produces $\frac{2\tan A}{1-{\tan }^{2}A}$, so the "$\tan$-only" form is that same ratio in disguise.

$\sin 0^{\circ }=2\sin 0^{\circ }\cos 0^{\circ }=2\cdot 0\cdot 1=0$, and all three ways of viewing it agree — the degenerate angle collapses cleanly.

All give $0$: Form 1 is $\frac{1}{2}-\frac{1}{2}$, Form 2 is $2\cdot \frac{1}{2}-1$, Form 3 is $1-2\cdot \frac{1}{2}$ — a good sanity check that the forms are genuinely equal.

From below (${\tan }^{2}A<1$) the denominator is a small positive number, so $\tan 2A\to +\infty$; from above (${\tan }^{2}A>1$) it is small negative, so $\tan 2A\to -\infty$ — a genuine sign flip across the asymptote.

For all $A$ except where the denominator vanishes: $1+\cos 2A=0\Rightarrow \cos 2A=-1\Rightarrow 2A=180^{\circ },540^{\circ },\cdots \Rightarrow A=90^{\circ },270^{\circ },\dots$ (odd multiples of $90^{\circ }$). At exactly those angles $\tan A$ is undefined too, so both sides fail together — the identity holds everywhere it is defined.

Its denominator $1-\cos 2A=0\Rightarrow \cos 2A=1\Rightarrow 2A=0^{\circ },360^{\circ },\cdots \Rightarrow A=0^{\circ },180^{\circ },\dots$ (integer multiples of $180^{\circ }$). There $\cot A$ is undefined as well, so this identity holds only away from multiples of $180^{\circ }$.

Yes — their product is positive, matching $\sin 2A$ for $2A$ landing in the first/second quadrant; the formula is an identity, valid in every quadrant without sign fixes.

$\cos 2A=1-2(1{)}^2=-1$, correctly reporting $\cos 180^{\circ }=-1$; the extreme input still lands on the right answer.

No — $\tan 90^{\circ }$ is undefined, so the formula has no valid input. You must fall back to $\tan 2A=\tan 180^{\circ }=0$ computed directly.

Yes: numerator $\sin 90^{\circ }=1$, denominator $1+\cos 90^{\circ }=1$, ratio $1=\tan 45^{\circ }$. Since $45^{\circ }$ is not an odd multiple of $90^{\circ }$, the denominator is safe.