Harmonic Form: Writing a cos t + b sin t as R cos(t ± a)

Express $5\cos\theta - 12\sin\theta$ in the form $R\cos(\theta + \alpha)$, where $R > 0$ and $0 < \alpha < 90°$. What is $\alpha$ to 1 decimal place?

• $67.4°$

  Correct. $R\cos(\theta+\alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$, so $R\cos\alpha = 5$ and $R\sin\alpha = 12$; $\tan\alpha = \frac{12}{5}$, giving $\alpha = 67.4°$. ($R = \sqrt{5^2+12^2} = 13$.)

• B$22.6°$

  This comes from $\tan\alpha = \frac{5}{12}$ — the ratio inverted. It's $\frac{R\sin\alpha}{R\cos\alpha} = \frac{b}{a}$, always the sinθ-coefficient over the cosθ-coefficient for a cos-led form, never the other way round.

• C$-67.4°$

  The magnitude is right and the sign convention isn't followed: the question fixes $0 < \alpha < 90°$, and both $R\cos\alpha = 5$ and $R\sin\alpha = 12$ came out positive, which places α in the first quadrant, not as a negative angle.

• D$112.6°$

  This is $180° - 67.4°$, the supplementary angle — a habit worth having when solving a trig EQUATION, but wrong here: this question is building an IDENTITY, and $0 < \alpha < 90°$ was fixed by the question. A supplementary angle would make Rcosα negative, contradicting Rcosα = 5.

Traps tested: Tan ratio inverted · Negative alpha not converted to range · Supplementary angle used instead of reference angle

What is the maximum value of $7\cos\theta + 24\sin\theta$, and at what value of $\theta$ in $[0°, 360°)$ does it occur (to 1 d.p.)?

• $25$, at $\theta = 73.7°$

  Correct. $7\cos\theta + 24\sin\theta = 25\cos(\theta - 73.7°)$ (R = √(7²+24²) = 25, tanα = 24/7). The maximum of a cos-led form R cos(θ−α) is R, at θ = α.

• B$25$, at $\theta = 0°$

  R is right, but the phase shift has been ignored — the maximum of plain cosθ is at θ=0, but R cos(θ−α) is plain cosine SHIFTED by α, so its maximum moves to θ=α along with it.

• C$31$, at $\theta = 73.7°$

  R has been computed as 7 + 24 = 31, not $\sqrt{7^2+24^2}$. The two coefficients combine by Pythagoras, since they are R cosα and R sinα — legs of a right triangle with hypotenuse R — not two lengths simply added.

• D$25$, at $\theta = 163.7°$

  R is right; the location has picked up an extra 90°, as if this were the sin-led form's rule ('maximum where the argument is 90°') rather than the cos-led one ('maximum where the argument is 0°'). For R cos(θ−α), the maximum is at θ = α exactly, with nothing added.

Traps tested: Phase shift ignored maximum assumed at zero · R computed as sum not pythagorean · Cos form location confused with sin form location

The equation $2\cos\theta + 5\sin\theta = 4$ is to be solved for $\theta$ in $0° \leq \theta < 360°$, after writing the left side as $R\cos(\theta - \alpha)$. Which statement about the solutions is correct?

• There are exactly two solutions, because $4 < R$ and cosine is two-to-one over a full period

  Correct. $R = \sqrt{2^2+5^2} = \sqrt{29} \approx 5.39$, and $4 < 5.39$, so $\cos(\theta-\alpha) = \frac{4}{\sqrt{29}} \approx 0.743$ is a genuinely reachable value, strictly between −1 and 1 — which is exactly the condition for two solutions in one period.

• BThere are no solutions, because 4 is not achievable — cos(θ−α) would need to exceed 1

  $\frac{4}{R} = \frac{4}{\sqrt{29}} \approx 0.743$, which is comfortably below 1, not above it. The equation is achievable; the threshold to check is whether the constant on the right exceeds R, and 4 does not exceed √29 ≈ 5.39.

• CThere is exactly one solution, because inverse cosine only returns one value

  That is a fact about the cos⁻¹ button, not about the equation. Cosine's two-to-one behaviour over a full period means a second solution always exists alongside the calculator's principal value whenever the target lies strictly between −R and R, as it does here.

• DThere are infinitely many solutions, because the equation is periodic

  The underlying wave is periodic, but θ is restricted to one 360° interval here, which catches exactly one period's worth of solutions — two of them, not an unbounded number.

Traps tested: Solvability threshold misjudged · Only principal solution found · Periodicity treated as infinite within a bounded interval

For $a\cos\theta + b\sin\theta$ with $a > 0$ and $b < 0$, which form keeps $R > 0$ and $0 < \alpha < 90°$?

• $R\cos(\theta + \alpha)$

  Correct. This expands to $R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$, giving $R\cos\alpha = a > 0$ and $R\sin\alpha = -b > 0$ (since $b < 0$) — both positive, so a first-quadrant α exists.

• B$R\cos(\theta - \alpha)$

  This form gives $R\cos\alpha = a$ and $R\sin\alpha = b$. With $b < 0$, that second value is negative, which no angle in $(0°, 90°)$ can produce — the default minus-form only works when both a and b are positive.

• C$R\sin(\theta - \alpha)$

  A sin-led form is the right family when it's the sinθ term that needs isolating for a later step, but switching to sin doesn't fix the sign problem on its own — the matching equations for THIS form would need checking exactly the same way, and simply picking 'sin' without checking is a choice made without the actual sign check.

• DIt cannot be written in this form at all when $b$ is negative

  Every a cosθ + b sinθ has a harmonic form for any real, nonzero a and b — it's a question of choosing the RIGHT one of the four sign combinations, not of harmonic form being unavailable.

Traps tested: Sign of b ignored when choosing the form · Sin led form chosen without checking signs · Negative coefficient assumed to block harmonic form

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Express $3\cos\theta + 4\sin\theta$ in the form $R\cos(\theta - \alpha)$ and hence solve $3\cos\theta + 4\sin\theta = 2$ for $0° \leq \theta < 360°$.

Which response is certain to earn full marks under the printed instruction above?

• R = 5, α = 53.1° shown via $R\cos\alpha = 3$, $R\sin\alpha = 4$, then $\cos(\theta-53.1°) = 0.4$ solved to give θ = 119.6° and θ = 346.7°, with every step shown

  Correct — the full method is visible: the matching equations, R and α derived from them, the substitution back into the equation, and both solutions extracted, exactly what the printed instruction is checking for.

• Bθ = 119.6° and θ = 346.7°, produced directly from the calculator's equation solver, with no R, α, or intermediate working shown

  Correct final answers with no visible method. The printed instruction removes the usual benefit of the doubt entirely — the general marking guidance is explicit that answers without working may not gain full credit, and this question was flagged precisely to exclude this response.

• CR = 5, α = 53.1° with working shown, then only θ = 119.6° given as 'the' solution

  Part (a)'s work is fine; the equation is only half-solved. cos(θ−α) = 0.4 has a second solution in the interval, and the mark for finding it is separate from the mark for finding the first one.

• Dθ = 119.6° only, with the note 'calculator confirms no other solutions in range'

  This is the exact response the printed instruction was written to exclude — naming the calculator as the method for confirming a solution count makes explicit what the instruction disallows, and it is also mathematically wrong: a second solution does exist.

Traps tested: Answer without working · Missing second solution in range · Calculator cited as the method

In part (a) you meant to write α = 53.1° but transposed the digits and wrote 51.3°. Part (b) says 'hence' state the maximum value of $g(\theta)$ and where it occurs. What is the best thing to do?

• Use your own 51.3° consistently in part (b): maximum 5 at θ = 51.3°, and carry on

  Correct. B marks marked 'ft' are awarded for correctly applying your own earlier value, right or wrong — a correct method applied consistently to a wrong number is still paid for, and abandoning the question forfeits marks the scheme was going to award regardless.

• BLeave part (b) blank, since part (a) is wrong

  The most expensive response available. It converts a one-digit transposition into the loss of an entire part, when the follow-through marks in (b) were available without redoing anything.

• CWrite θ = 53.1° in part (b) without correcting part (a), since that's the value you now suspect is right

  This leaves the script internally inconsistent — part (b) no longer follows from part (a), so there's no working for a follow-through mark to attach to. If (a) is suspected wrong, correct (a) itself and let (b) follow from the corrected value.

• DCross out the whole question and restart it from part (a), whatever it costs in time

  Worth it only if spotted immediately with time to spare. As a default it's the wrong trade — the ft marks in (b) were free, and a 90-minute, 75-mark paper leaves little room to re-run a question already partly paid for.

Traps tested: Abandons the question after an error · Answer inconsistent with own working · Restarts instead of continuing

A real WMA13 question of this exact type (Oct 2020, Q7(a)) reads: 'Express cos x + 4 sin x in the form R cos(x − α) where R > 0 and 0 < α < π/2. Give the exact value of R and give the value of α, in radians, to 3 decimal places.' Its real mark scheme awards B1 for R = √17 with the note 'do not allow decimals for this mark', and states for α = awrt 1.326 that 'the degree equivalent α = awrt 75.96° is A0' — the same angle, in the wrong unit.

Applying that same real mark scheme's rules: $2\cos\theta + 3\sin\theta$ is written as $R\cos(\theta-\alpha)$, with $R>0$ and $0 < \alpha < \frac{\pi}{2}$, and the question asks for the EXACT value of $R$ and the value of $\alpha$ in RADIANS to 3 decimal places. Which pair earns full marks?

• $R = \sqrt{13}$, $\alpha = 0.983$

  Correct. $R = \sqrt{2^2+3^2} = \sqrt{13}$ — left exact, as asked — and $\tan\alpha = \frac{3}{2}$ gives $\alpha = 0.983$ (3 d.p.) in radians, as asked. Both the form and the unit match what the question specified.

• B$R = 3.606$, $\alpha = 0.983$

  R has been rounded to a decimal, and the question asked for the EXACT value. On the real question this trap is verified against, the mark scheme states plainly for R = √17: 'do not allow decimals for this mark' — a correct decimal, however precise, is not an acceptable substitute for the exact surd when the exact value was what was asked for.

• C$R = \sqrt{13}$, $\alpha = 56.3°$

  α has been given in degrees, and the question asked for radians. On the real question this trap is verified against, the mark scheme states for exactly this situation: 'the degree equivalent α = awrt 75.96° is A0' — the identical angle, correctly computed, earns nothing because it isn't in the unit the question specified.

• D$R = 13$, $\alpha = 0.983$

  This is $R^2$, with the square root left off — $R^2 = 2^2+3^2 = 13$, so $R = \sqrt{13}$, not 13 itself. The same slip as forgetting the square root anywhere else this identity is built.

Traps tested: R rounded to decimal when exact value required · Alpha given in degrees when radians specified · Forgot square root of a squared plus b squared