Trigonometric Identities and Solving Equations in a Given Interval

Solve $\tan\theta = \sqrt{3}$ for $0° \leqslant \theta < 360°$. Which is the complete solution set?

• $\theta = 60°$ and $\theta = 240°$

  Correct. $\arctan(\sqrt3) = 60°$, and tan is positive in quadrants 1 and 3 — its two solutions per revolution are always exactly 180° apart, so the second is $60° + 180° = 240°$.

• B$\theta = 60°$ only

  arctan returns one principal value by definition, but tan has period 180°, so a second genuine solution exists inside a 360° interval and has to be generated, not assumed absent.

• C$\theta = 60°$ and $\theta = 120°$

  $180° - 60° = 120°$ is the second-solution rule for a positive SINE, not tangent — it uses the sin/cos reflection pattern where tan instead needs a 180° shift. Check: $\tan(120°) = -\sqrt3$, negative, so 120° cannot be a solution of $\tan\theta = \sqrt3$ at all.

• D$\theta = 60°$ and $\theta = 300°$

  $360° - 60° = 300°$ is the second-solution rule for a positive COSINE. Check: $\tan(300°) = -\sqrt3$, negative — the sign is wrong, because 300° is in quadrant 4, where tan is negative, not the quadrant tan's positive-value rule actually points to.

Traps tested: Calculator principal value treated as complete answer · Sin second solution rule applied to tan · Cos second solution rule applied to tan

Given that $\cos\theta = -\frac{3}{5}$ and $\theta$ is obtuse ($90° < \theta < 180°$), what is $\sin\theta$?

• $\frac{4}{5}$, since $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$, and $\sin\theta > 0$ in the second quadrant

  Correct. The sign of sinθ is decided by the QUADRANT (positive throughout the second quadrant, 90° to 180°), not by the sign of cosθ — the two are independent facts once the identity has given you the magnitude.

• B$-\frac{4}{5}$, since $\cos\theta$ is negative, $\sin\theta$ must be negative too

  There is no rule linking the sign of sinθ to the sign of cosθ directly — each function's sign depends only on which quadrant θ is in. In the second quadrant cosθ is negative AND sinθ is positive, simultaneously; that combination is exactly what makes it the second quadrant rather than the third.

• C$\frac{2}{5}$, from $1 - \frac{3}{5}$

  This uses the identity as $\sin\theta = 1 - \cos\theta$, a direct subtraction of the two functions — but the real identity relates their SQUARES, sin²θ + cos²θ = 1, not the functions themselves.

• D$\frac{16}{25}$, stopping at $\sin^2\theta$

  $\frac{16}{25}$ is $\sin^2\theta$; the square root — the step that actually answers the question — hasn't been taken yet.

Traps tested: Assumes same sign for sin and cos · Identity misapplied as linear not squared · Square root not taken

Solve $3\sin\theta\cos\theta = \sin\theta$ for $0° \leqslant \theta < 360°$.

• $\theta = 0°, 70.5°, 180°, 289.5°$ — factorise as $\sin\theta(3\cos\theta - 1) = 0$ first, since dividing by $\sin\theta$ would silently discard the $\theta = 0°$ and $\theta = 180°$ solutions

  Correct. $\sin\theta = 0$ gives $\theta = 0°, 180°$; $\cos\theta = \frac{1}{3}$ gives $\theta = 70.5°, 289.5°$ (1 d.p.) — four genuine solutions, all satisfying the original equation.

• B$\theta = 70.5°, 289.5°$ only, from dividing both sides by $\sin\theta$ to get $\cos\theta = \frac{1}{3}$

  Dividing by $\sin\theta$ silently assumes $\sin\theta \neq 0$ — but $\theta = 0°$ and $\theta = 180°$, where $\sin\theta = 0$, are genuine solutions of the ORIGINAL equation (both sides equal zero there). Cancelling a factor that can be zero throws those away.

• C$\theta = 0°, 180°$ only, since $\sin\theta = 0$ is the obvious solution

  Correct as far as it goes, but incomplete: factorising gives TWO brackets, $\sin\theta = 0$ and $3\cos\theta - 1 = 0$, and both have to be solved. Stopping after the first bracket leaves an entire second branch of solutions unfound.

• D$\theta = 0°, 90°, 180°, 270°$ — treating $\cos\theta = \frac{1}{3}$ as if it landed on a multiple of $90°$

  $\cos\theta = \frac{1}{3}$ is not one of the standard exact angles — its reference angle, $\arccos(\frac{1}{3}) \approx 70.5°$, actually has to be computed, not guessed from familiar round numbers.

Traps tested: Divides out a common trig factor · Second factor branch not pursued · Reference angle not actually computed

The spec's own worked example for this topic asks you to solve $\sin(x + \frac{\pi}{2}) = \frac{3}{4}$ for $0 < x < 2\pi$. A student's calculator gives $\arcsin(0.75) \approx 48.6°$, which they convert to $\approx 0.848$ radians and give as their only answer.

Which option correctly identifies every problem with this response?

• It stops at the calculator's one value for the bracket $(x + \frac{\pi}{2})$, missing the second solution of the same equation, AND $0.848$ is a value of $(x + \frac{\pi}{2})$, not of $x$ — the shift back to $x$ was never applied

  Correct — two separate, compounding errors. The equation is in $(x + \frac{\pi}{2})$, which needs its own two solutions found first (the compound angle isolated and treated as a single unknown, exactly as in the compound-angle trap above); only THEN does $\frac{\pi}{2}$ get subtracted from each to recover the values of $x$ the question actually asked for.

• BNothing — arcsin always returns the only solution in a given interval

  arcsin returns exactly one number because it must, by definition of what a function is — but that number is one representative of a many-to-one relationship, not proof that only one solution exists in a wider interval.

• COnly that the answer should have stayed in degrees, since the interval was given using $\pi$

  This is backwards, and also not the real problem: an interval written with $\pi$ in it is a radian interval, so an answer genuinely does need to be in radians here — but fixating on units misses the two substantive errors (one solution missing, and the wrong variable being reported) entirely.

• DOnly that a second solution is missing; $0.848$ radians is otherwise a correct value of $x$

  This catches half the problem. $0.848$ is a value of the BRACKET $(x + \frac{\pi}{2})$, not of $x$ itself — even taken as a lone solution, it still needs $\frac{\pi}{2}$ subtracted from it before it answers the question that was actually asked.

Traps tested: Calculator principal value treated as complete answer · Surface formatting mistaken for the real error · Compound angle shift not applied

Why is $\tan\theta$ undefined at $\theta = 90°$ and $\theta = 270°$, but not at $\theta = 0°$ or $\theta = 180°$?

• Because $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and $\cos\theta = 0$ exactly at those two angles — the two points on the unit circle where the radius OP is vertical, so its gradient is undefined

  Correct. tan θ is literally the gradient of the line from the origin to the point (cos θ, sin θ); a vertical line has no defined gradient, and OP is vertical exactly when its x-coordinate, cos θ, is zero — which happens at 90° and 270°, nowhere else in a revolution.

• BBecause $\sin\theta = 0$ at those two angles, and dividing by a zero numerator is undefined

  This has numerator and denominator swapped, and also picks the wrong angles: sinθ = 0 at 0° and 180°, where tan is actually perfectly defined (it equals 0, since a zero numerator over a nonzero denominator is just zero). The undefined case is a zero DENOMINATOR.

• CBecause $\tan\theta$ repeats every $90°$, and those are the boundary points of each repeat

  tan's period is 180°, not 90° — check $\tan(45°) = 1$ and $\tan(225°) = 1$, exactly 180° apart, not 90°. The real mechanism is the vertical-line/zero-denominator argument, not a boundary of a (mis-stated) period.

• DBecause those two angles aren't defined on the unit circle

  Every real angle defines a point on the unit circle — 90° and 270° are perfectly ordinary points, at the top and bottom of the circle. The issue is specific to the RATIO sinθ/cosθ at those points, not to whether the angle or the point exists.

Traps tested: Numerator and denominator swapped · Period of tan misstated · Unit circle domain misunderstood