Sine, Cosine and Tangent Graphs — Period vs Domain Confusion

What is the period of $y = \tan x$?

• $\pi$

  Correct. $\tan(x + \pi) = \tan x$ because the half-turn identities $\sin(x+\pi) = -\sin x$ and $\cos(x+\pi) = -\cos x$ flip both the numerator and denominator's sign together, and those two minus signs cancel in the ratio — a shift smaller than π has no equivalent cancellation.

• B$2\pi$, the same as $\sin x$ and $\cos x$

  Sin and cos individually need the full 2π to return to themselves, but their ratio needs only half of that — a shift of π already makes both the numerator and the denominator flip sign together, which cancels in the division.

• C$-\frac{\pi}{2} < x < \frac{\pi}{2}$

  This is a real, documented wrong answer to this exact question: "Quite frequently the answer was given as an interval, including $-\frac{\pi}{2} < x < \frac{\pi}{2}$" (Jan 2023, Q9(a)). It correctly describes the domain of one branch of the graph — a SET of x-values — but a period is a single number, not an interval.

• D$\frac{\pi}{2}$

  This is the gap from one zero of tan x to the nearest asymptote (e.g. x = 0 to x = π/2) — a quarter of the way through the pattern, not a full repeat. The graph does not look the same at x = π/2 as it does at x = 0; the shift that actually reproduces the whole shape is twice this.

Traps tested: Period of sin cos assumed for tan · Period confused with domain or interval · Asymptote spacing halved in error

Which best describes the symmetry of $y = \cos x$?

• It is an even function — line symmetry about the y-axis

  Correct. $\cos(-x) = \cos x$ for every x, which is exactly the algebraic statement that reflecting the graph in the y-axis maps it onto itself.

• BIt is an odd function — point symmetry about the origin

  That describes $\sin x$ and $\tan x$, not $\cos x$. Check a single value: $\cos(-\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = 0.5$, the same sign and size — not the negative, which is what an odd function would require.

• CIt is periodic, so it has no reflective or rotational symmetry

  Being periodic (repeating along the x-axis) and having a line or point of symmetry are two independent properties — a curve can, and cos x does, have both at once. Periodicity doesn't rule symmetry out.

• DIt is symmetric about the line $y = x$

  Symmetry about $y = x$ is the property that makes a function its own inverse when reflected — an unrelated kind of symmetry from either of the two that actually apply to trig graphs (line symmetry about a vertical axis, or point symmetry about a point).

Traps tested: Sin symmetry attributed to cos · Periodicity mistaken for absence of symmetry · Symmetry axis misidentified

The graph of $y = \sin\left(x + \frac{\pi}{6}\right)$ is obtained from $y = \sin x$ by...

• a translation LEFT by $\frac{\pi}{6}$

  Correct. Check against the peak: $\sin x$ reaches its maximum at $x = \frac{\pi}{2}$; $\sin\left(x+\frac{\pi}{6}\right)$ reaches the same maximum when $x + \frac{\pi}{6} = \frac{\pi}{2}$, i.e. at $x = \frac{\pi}{3}$ — a smaller x-value, so the peak arrives sooner and the graph has moved left.

• Ba translation RIGHT by $\frac{\pi}{6}$

  The sign is reversed. $y = f(x + a)$ moves the graph by $-a$, i.e. LEFT when $a$ is positive — the same sign-flip that costs marks on completing-the-square turning points, applied here to a horizontal shift instead of a vertex.

• Ca translation UP by $\frac{\pi}{6}$

  $y = f(x) + a$ (added OUTSIDE the function) is a vertical shift; $y = f(x + a)$ (added INSIDE, to x itself) is a horizontal one. The $\frac{\pi}{6}$ here is inside the bracket, next to x — it shifts the graph sideways, not up.

• Da horizontal stretch, scale factor $\frac{\pi}{6}$

  A number ADDED to x inside the function is a translation; a stretch comes from a number that MULTIPLIES x, as in sin 2x. Nothing here multiplies x, so there is no stretch to describe.

Traps tested: Translation direction sign not flipped · F(x+a) confused with f(x)+a · Translation confused with stretch

What is the period of $y = \sin 2x$?

• $\pi$

  Correct. $y = \sin 2x$ is a horizontal stretch of $y = \sin x$, scale factor $\frac{1}{2}$ — the graph is compressed to half its width, so its whole repeat-length, $2\pi$, is halved too: $2\pi \times \frac{1}{2} = \pi$.

• B$2\pi$, unchanged from $\sin x$

  This treats the ×2 as if it only changed the graph's height, not its width — the same misconception a real report records for the closely related cos(x) → cos(2x): mistaking a stretch that acts on the input (horizontal) for one that acts on the output (vertical). The coefficient is attached to x, so the effect is horizontal, and it does change the period.

• C$4\pi$

  This applies the scale factor as if it were $2$ rather than its reciprocal, $\frac{1}{2}$ — doubling the period instead of halving it. For $y = f(ax)$ the horizontal scale factor is $\frac{1}{a}$, not $a$ itself.

• D$\frac{\pi}{2}$

  This is what you get by applying the halving twice — once correctly, from $2\pi$ to $\pi$, and then a second time by mistake. The scale factor $\frac{1}{2}$ is applied to the period exactly once.

Traps tested: Stretch direction inverted y for x · Stretch scale factor inverted · Period and scale factor conflated

What is the range of $y = 3\sin x$?

• $[-3, 3]$

  Correct. The $3$ multiplies the OUTPUT of $\sin x$ after it has been calculated, so every y-value $\sin x$ could produce, from $-1$ to $1$, is scaled by 3 as well — including the negative half.

• B$[-1, 1]$, unchanged

  This is the range of $\sin x$ itself, before the transformation is applied at all. $y = af(x)$ genuinely changes the range whenever $a \neq 1$ — it is not decorative.

• C$[0, 3]$

  Only the positive half of the output has been scaled here. Multiplying by 3 acts on every value the function produces, including the negative ones — $\sin x = -1$ becomes $3\sin x = -3$, not $0$.

• D$[-3, 1]$

  The two ends of the range have been scaled inconsistently — the negative end multiplied by 3 but the positive end left alone. Whatever operation is applied to the output has to apply to the whole range, both ends alike.

Traps tested: Amplitude scaling ignored · Only positive part scaled · Asymmetric scaling error

For $0 \leqslant x < 2\pi$, at how many values of $x$ is $\tan x$ undefined, and what are they?

• $2$, at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

  Correct. $\tan x = \frac{\sin x}{\cos x}$ is undefined exactly where $\cos x = 0$, and in one full revolution that happens twice — at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, which is also exactly the gap of π the period predicts between one asymptote and the next.

• B$1$, at $x = \frac{\pi}{2}$ only

  This is what you'd get by assuming tan x only breaks once per 2π, the way you might expect from a period of 2π. Since the real period is π, there is a second, equally genuine break exactly π further along, at $\frac{3\pi}{2}$.

• C$4$, at $x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$

  $x = \pi$ and $x = 2\pi$ are where $\sin x = 0$, which makes $\tan x = \frac{0}{\cos x} = 0$ — a perfectly ordinary, well-defined value, not a break. The undefined points come from a zero DENOMINATOR, not a zero numerator.

• D$2$, at $x = 0$ and $x = \pi$

  These are where $\sin x = 0$ (the numerator), not where $\cos x = 0$ (the denominator) — the opposite condition from the one that actually makes a fraction undefined. tan x is perfectly defined at both of these points; it equals 0.

Traps tested: Period of sin cos assumed for tan · Zeros confused with asymptotes · Sin zeros substituted for cos zeros