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high importance~3 Q in Tier 119 formulas⚡ 11 shortcuts5 subtopics

Maximum and minimum values

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For 0∘≤θ≤90∘0^\circ\le\theta\le90^\circ:

  • 0≤sin⁡θ,cos⁡θ≤10\le\sin\theta,\cos\theta\le1; sec⁡θ,cosecθ≥1\sec\theta,\text{cosec}\theta\ge1; tan⁡θ,cot⁡θ\tan\theta,\cot\theta are unbounded.
  • sin⁡θ+cos⁡θ∈[1,2]\sin\theta+\cos\theta\in[1,\sqrt2]; sin⁡θ−cos⁡θ∈[−1,1]\sin\theta-\cos\theta\in[-1,1]; sin⁡θcos⁡θ≤12\sin\theta\cos\theta\le\frac12.
  • asin⁡2θ+bcos⁡2θa\sin^2\theta+b\cos^2\theta has minimum min⁡(a,b)\min(a,b) and maximum max⁡(a,b)\max(a,b).
  • atan⁡2θ+bcot⁡2θ (and asin⁡2+b cosec2)a\tan^2\theta+b\cot^2\theta\ (\text{and } a\sin^2+b\,\text{cosec}^2) has minimum 2ab2\sqrt{ab} by AM–GM.
  • sec⁡2θ+cosec2θ\sec^2\theta+\text{cosec}^2\theta has minimum 2+2=42+2=4 (at 45∘45^\circ).

Detailed notes

The amplitude formula (the workhorse)

For any angle θ\theta: −a2+b2 ≤ asin⁡θ+bcos⁡θ ≤ +a2+b2-\sqrt{a^2 + b^2} \ \le\ a\sin\theta + b\cos\theta \ \le\ +\sqrt{a^2 + b^2} The maximum of asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta is a2+b2\sqrt{a^2+b^2} and the minimum is its negative (unrestricted θ\theta); restricted to acute angles the range sits between the endpoint values — usually stated so the interior extremum is the answer. Instant values: 4sin⁡θ+3cos⁡θ→54\sin\theta + 3\cos\theta \to 5, 5sin⁡θ+12cos⁡θ→135\sin\theta + 12\cos\theta \to 13, 8sin⁡θ−15cos⁡θ→178\sin\theta - 15\cos\theta \to 17 — Pythagorean pairs every time. On 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ: sin⁡θ+cos⁡θ\sin\theta + \cos\theta ranges from 11 to 2\sqrt2 — max 2\sqrt2 at 45∘45^\circ, min 11 at the endpoints.

The product bound

sin⁡θcos⁡θ≤12,2sin⁡θcos⁡θ=sin⁡2θ≤1\sin\theta\cos\theta \le \frac12, \qquad 2\sin\theta\cos\theta = \sin 2\theta \le 1 Both maxima occur at θ=45∘\theta = 45^\circ. Any multiple ksin⁡θcos⁡θk\sin\theta\cos\theta peaks at k2\frac{k}{2}.

Self-reciprocal sums (AM-GM)

tan⁡2θ+cot⁡2θ≥2,sec⁡2θ+cosec⁡2θ≥4,tan⁡θ+cot⁡θ≥2\tan^2\theta + \cot^2\theta \ge 2,\qquad \sec^2\theta + \cosec^2\theta \ge 4,\qquad \tan\theta + \cot\theta \ge 2 All attained at θ=45∘\theta = 45^\circ. The sec⁡2+cosec⁡2\sec^2 + \cosec^2 version: 1sin⁡2cos⁡2≥4\frac{1}{\sin^2\cos^2} \ge 4 since sin⁡2cos⁡2≤14\sin^2\cos^2 \le \frac14. Any x+1x≥2x + \frac1x \ge 2 shape with x>0x > 0 follows the same rule, equality at x=1x = 1.

Ordering and comparison facts

On (0∘,90∘)(0^\circ, 90^\circ): sin⁡\sin rises 0→10 \to 1, cos⁡\cos falls 1→01 \to 0, tan⁡\tan rises 0→∞0 \to \infty. So sin⁡60∘>sin⁡45∘>sin⁡30∘\sin 60^\circ > \sin 45^\circ > \sin 30^\circ; sin⁡θ>cos⁡θ  ⟺  θ>45∘\sin\theta > \cos\theta \iff \theta > 45^\circ; tan⁡θ=cot⁡θ\tan\theta = \cot\theta only at 45∘45^\circ. Also 11+sin⁡2θ\frac{1}{1+\sin^2\theta} is smallest when sin⁡2θ\sin^2\theta is largest (=1= 1), giving minimum 12\frac12.

The 45° equality case and restricted domains

Every bound in this block is attained at θ=45∘\theta = 45^\circ: the amplitude expressions reach ±a2+b2\pm\sqrt{a^2+b^2} only for unrestricted θ\theta (the peak of asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta sits at tan⁡θ=ab\tan\theta = \frac{a}{b}, which is acute for positive a,ba, b — so on 0∘0^\circ to 90∘90^\circ the max a2+b2\sqrt{a^2+b^2} IS reachable, while the negative minimum is cut off at the endpoint −b-b). Always read the domain: "for all θ\theta" allows the negative amplitude; "0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ" floors the minimum at the endpoint value. Options often include both — the domain decides.

Quick revision

  • asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta: max a2+b2\sqrt{a^2+b^2}, min −a2+b2-\sqrt{a^2+b^2}.
  • sin⁡θcos⁡θ≤12\sin\theta\cos\theta \le \frac12; 2sin⁡θcos⁡θ≤12\sin\theta\cos\theta \le 1, both at 45∘45^\circ.
  • tan⁡2+cot⁡2≥2\tan^2 + \cot^2 \ge 2; sec⁡2+cosec⁡2≥4\sec^2 + \cosec^2 \ge 4; x+1x≥2x + \frac1x \ge 2 with equality at x=1x=1.
  • sin⁡\sin up, cos⁡\cos down on (0∘,90∘)(0^\circ, 90^\circ); crossover at 45∘45^\circ.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Max/min of a sinθ ± b cosθvery common4 practice Q
How to spot it:

'Find the maximum of 4sinθ + 3cosθ' — a linear combination with Pythagorean coefficients.

Max $= +\sqrt{a^2+b^2}$, min $= -\sqrt{a^2+b^2}$ for $a\sin\theta + b\cos\theta$.
  1. Read off (a,b)(a, b); compute a2+b2\sqrt{a^2 + b^2} — expect a triplet.
  2. Maximum positive, minimum negative (unless θ\theta is restricted; then check endpoints).
  3. Quote both if 'range' is asked.

Why: the expression is a single wave with amplitude a2+b2\sqrt{a^2+b^2} — the formula IS the answer.

Example: The maximum value of 4sin⁡θ+3cos⁡θ4\sin\theta + 3\cos\theta is:

42+32=5\sqrt{4^2 + 3^2} = 5. Minimum: −5-5 (unrestricted θ\theta).

Type 2: Bounds of sinθ cosθcommon2 practice Q
How to spot it:

Max/min of sinθcosθ or 2sinθcosθ (sometimes with a coefficient or a plus-one).

$\sin\theta\cos\theta \le \frac12$; $2\sin\theta\cos\theta = \sin2\theta \le 1$; both peak at $45^\circ$.
  1. Recognise the product form; halve the coefficient for the max of ksin⁡θcos⁡θk\sin\theta\cos\theta.
  2. Or convert to sin⁡2θ\sin 2\theta and use the wave range.
  3. Minimum on acute angles is 00 (endpoint); quote only if asked.

Why: the product is the double-angle formula in disguise — one conversion bounds it.

Example: The maximum value of 2sin⁡θcos⁡θ2\sin\theta\cos\theta (θ\theta acute) is:

2sin⁡θcos⁡θ=sin⁡2θ≤12\sin\theta\cos\theta = \sin 2\theta \le 1, attained at θ=45∘\theta = 45^\circ.

Type 3: AM-GM self-reciprocal sumscommon3 practice Q
How to spot it:

Minimum of tan²+cot², sec²+cosec², tan+cot — or an equation whose sum equals the bound.

$x + \frac1x \ge 2$; $\sec^2\theta + \cosec^2\theta = \frac{1}{\sin^2\theta\cos^2\theta} \ge 4$.
  1. Identify the self-reciprocal pair; quote the AM-GM bound.
  2. sec⁡2+cosec⁡2\sec^2 + \cosec^2: combine to 1sin⁡2cos⁡2\frac{1}{\sin^2\cos^2}, then use sin⁡cos⁡≤12\sin\cos \le \frac12.
  3. Equality case: θ=45∘\theta = 45^\circ in every one of these.

Why: reciprocal pairs cannot be small simultaneously — the product x⋅1x=1x \cdot \frac1x = 1 forces the sum up.

Example: The minimum value of sec⁡2θ+cosec⁡2θ\sec^2\theta + \cosec^2\theta is:

=1sin⁡2θcos⁡2θ≥1(1/2)2=4= \frac{1}{\sin^2\theta\cos^2\theta} \ge \frac{1}{(1/2)^2} = 4, at θ=45∘\theta = 45^\circ.

Type 4: Ordering and crossover comparisonscommon3 practice Q
How to spot it:

'Which is largest: sin 30, sin 45, sin 60?'; 'sinθ > cosθ for which θ?'; minimum of 1/(1+sin²θ).

On $(0^\circ, 90^\circ)$: sin increases, cos decreases; crossover at $45^\circ$; denominators largest → fraction smallest.
  1. Monotonicity orders same-function comparisons instantly.
  2. sin vs cos comparisons pivot at 45∘45^\circ.
  3. For 11+sin⁡2θ\frac{1}{1+\sin^2\theta}: minimise by maximising the denominator (sin⁡2θ=1\sin^2\theta = 1).

Why: the monotone staircase of the three functions on the acute quadrant decides everything without computation.

Example: For 0∘<θ<90∘0^\circ < \theta < 90^\circ, sin⁡θ>cos⁡θ\sin\theta > \cos\theta holds when:

θ>45∘\theta > 45^\circ — the crossover of the two graphs is at 45∘45^\circ.

Formulas

Ranges
0≤sin⁡θ≤1,sec⁡θ≥1,sin⁡θ+cos⁡θ∈[1,2]0\le\sin\theta\le1,\quad \sec\theta\ge1,\quad \sin\theta+\cos\theta\in[1,\sqrt2]
Weighted square form
asin⁡2θ+bcos⁡2θ∈[min⁡(a,b),max⁡(a,b)]a\sin^2\theta+b\cos^2\theta\in[\min(a,b),\max(a,b)]
AM-GM forms
atan⁡2θ+bcot⁡2θ≥2ab,asin⁡2θ+b cosec2θ≥2aba\tan^2\theta+b\cot^2\theta\ge2\sqrt{ab},\quad a\sin^2\theta+b\,\text{cosec}^2\theta\ge2\sqrt{ab}
Combined sec-cosec
sec⁡2θ+cosec2θ=2+tan⁡2θ+cot⁡2θ≥4\sec^2\theta+\text{cosec}^2\theta=2+\tan^2\theta+\cot^2\theta\ge4
Linear combination
asin⁡θ+bcos⁡θ has maximum a2+b2a\sin\theta+b\cos\theta\text{ has maximum }\sqrt{a^2+b^2}

Shortcut tricks

⚡ Match the question to a range

Bounded pairs (sin⁡+cos⁡\sin+\cos etc.) → memorised range. Weighted square sums → min of coefficients or 2ab2\sqrt{ab}.

Example: Find the minimum value of 4tan⁡2θ+9cot⁡2θ4\tan^2\theta+9\cot^2\theta for 0∘<θ<90∘0^\circ<\theta<90^\circ.

24×9=122\sqrt{4\times9}=12 (at tan⁡2θ=32\tan^2\theta=\frac32).

⚡ Weighted sin-cos squares are bounded by coefficients

2sin⁡2θ+3cos⁡2θ=2+cos⁡2θ2\sin^2\theta+3\cos^2\theta=2+\cos^2\theta, so it ranges from 2 to 3 — min 2, max 3, no calculus.

Example: Find the minimum value of 5sin⁡2θ+12cos⁡2θ5\sin^2\theta+12\cos^2\theta.

=5+7cos⁡2θ≥5=5+7\cos^2\theta\ge5; minimum 55 (at θ=0\theta=0).

Where students lose marks

  • AM-GM applied to asin⁡θ+bcosecθa\sin\theta+b\text{cosec}\theta where the equality point may be unreachable — for the square versions it is safe, check before using.

  • Quoting 2\sqrt2 as the minimum of sin⁡+cos⁡\sin+\cos — 2\sqrt2 is the maximum.

  • sec⁡2+cosec2\sec^2+\text{cosec}^2 minimum quoted as 2 instead of 4.

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.