CDS Mathematics · Trigonometric Ratios and Identities

Maximum, Minimum & Impossible Values

The greatest and least values of trigonometric expressions, and the equations that can never hold — all settled by the range of a ratio, by t + 1/t ≥ 2, or by a sin θ + b cos θ ≤ √(a² + b²).

All Trigonometric Ratios and Identities notes CDS Mathematics strategy

Why this matters

Thirty-seven PYQs — the largest page in the chapter and one of its two hardest, with thirteen HARD. Four tools cover all of them. The hard ones are hard only because the tool is disguised: an expression that is secretly t + 1/t, or an equation that is secretly at its maximum.

Concept 1 of 5: Expressions linear in sin²θ or sin θ

9\sin^2\theta + 16\cos^2\theta

looks like two moving parts, but

\cos^2\theta = 1 - \sin^2\theta

turns it into

16 - 7\sin^2\theta

: one moving part,

\sin^2\theta

, which runs from

0

1

. The extremes are then the two ends.

Definition

$a\sin^2\theta + b\cos^2\theta = b + (a - b)\sin^2\theta$ , so it runs between $a$ and $b$ .
$p + q\sin\theta$ runs from $p - |q|$ to $p + |q|$ — or over a smaller interval if $\theta$ is restricted.
Restricted range: on $0 \le \theta \le \dfrac{\pi}{2}$ , $\sin\theta$ runs over $[0, 1]$ , not $[-1, 1]$ . Recompute the ends.

The two ends

a\sin^2\theta + b\cos^2\theta \in [\min(a,b),\, \max(a,b)]

Worked example

Find the greatest and least values of

5\cos^2\theta + 2\sin^2\theta

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q56Moderate

Example 1 · Trigonometric Ratios and Identities · Maximum, Minimum and Impossible Values

What is the least value of

9 \sin^2 \theta + 16 \cos^2 \theta

A restricted range moves the ends

0 \le \theta \le \dfrac{\pi}{2}

\sin\theta

never goes below

0

. The minimum of

6 + 4\sin\theta

there is

6

, not

2

. Always take the ends of the ratio over the stated range.

Drill 8 more on expressions linear in sin²θ or sin θ

Concept 2 of 5: t + 1/t ≥ 2, and weighted forms by AM–GM

A positive number plus its reciprocal is never below

2

, because

t + \dfrac1t - 2 = \dfrac{(t - 1)^2}{t}

. Trig ratios come in reciprocal pairs —

\tan

and

\cot

\sin

and

\operatorname{cosec}

— so this one inequality settles a large share of the page.

Definition

For $t > 0$ : $t + \dfrac1t \ge 2$ , with equality only at $t = 1$ .
AM–GM: for $x, y > 0$ , $x + y \ge 2\sqrt{xy}$ . So $a\tan^2\theta + b\cot^2\theta \ge 2\sqrt{ab}$ .
$a\sec^2\theta + b\operatorname{cosec}^2\theta = (a + b) + a\tan^2\theta + b\cot^2\theta \ge a + b + 2\sqrt{ab} = (\sqrt a + \sqrt b)^2$ .
Equality needs the two terms equal. If the range forbids that (open interval ending at $t = 1$ ), the bound is not reached, and the answer is 'greater than', not 'at least'.

The bounds

t + \frac1t \ge 2, \qquad a\sec^2\theta + b\operatorname{cosec}^2\theta \ge (\sqrt a + \sqrt b)^2

Worked example

Find the minimum of

16\sec^2\theta + 9\operatorname{cosec}^2\theta

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q60Moderate

Example 2 · Trigonometric Ratios and Identities · Maximum, Minimum and Impossible Values

What is the minimum value of

9 \tan^2 \theta + 4 \cot^2 \theta

On an open interval the bound may not be reached

For

0 < x < \dfrac{\pi}{2}

\sin x + \operatorname{cosec} x

would equal

2

only at

\sin x = 1

, which the open interval excludes. So the correct statement is '

> 2

', and the option '

\ge 2

' is the trap.

Drill 11 more on t + 1/t ≥ 2, and weighted forms by am–gm

Concept 3 of 5: Quadratics in sin θ or cos²θ

\sin^2\theta + \cos^4\theta

is not linear in anything — but with

c = \cos^2\theta

it is

c^2 - c + 1

, a parabola in

c

. Complete the square to find its lowest point, then remember that

c

only runs over

[0, 1]

, so the highest value sits at an end.

Definition

Substitute one variable: $s = \sin\theta$ in $[-1, 1]$ , or $c = \cos^2\theta$ in $[0, 1]$ .
Complete the square: $as^2 + bs + k = a\left(s + \dfrac{b}{2a}\right)^2 + \ldots$ .
The extreme is at the vertex if the vertex lies in the variable's range; otherwise at the nearer end. Always check the ends too.

Vertex, then ends

c^2 - c + 1 = \left(c - \tfrac12\right)^2 + \tfrac34, \quad c \in [0, 1]

Worked example

Find the greatest value of

6\sin\theta - 3\sin^2\theta

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2016 · CDS (II) 2016 — Elementary Mathematics · Q78Hard

Example 3 · Trigonometric Ratios and Identities · Maximum, Minimum and Impossible Values

A = \sin^2\theta + \cos^4\theta

where

0 \le \theta < \frac{\pi}{2}

, then which one of the following is correct ?

The vertex may lie outside the variable's range

For

s^2 + 4s

with

s = \sin\theta

, the vertex

s = -2

is not a possible sine. The minimum is then at the nearer end,

s = -1

, giving

-3

— not the vertex value

-4

Drill 5 more on quadratics in sin θ or cos²θ

Concept 4 of 5: a sin θ + b cos θ is at most √(a² + b²)

a\sin\theta + b\cos\theta

is a single sine wave of height

\sqrt{a^2 + b^2}

. So when a question says

11\sin\theta + 60\cos\theta = 61

and

61 = \sqrt{11^2 + 60^2}

, the equation is sitting exactly at its maximum — which pins

\sin\theta

and

\cos\theta

completely.

Definition

$-\sqrt{a^2 + b^2} \le a\sin\theta + b\cos\theta \le \sqrt{a^2 + b^2}$ .
If $a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2}$ , then $\sin\theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ and $\cos\theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ — nothing else fits.
If the given value exceeds $\sqrt{a^2 + b^2}$ , the equation has no solution.
In particular $\sin\theta + \cos\theta \le \sqrt2$ .

Amplitude bound

|a\sin\theta + b\cos\theta| \le \sqrt{a^2 + b^2}

Worked example

8\sin\theta + 15\cos\theta = 17

, find

\tan\theta + \cot\theta

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (II) 2026 — Elementary Mathematics · Q46Moderate

Example 4 · Trigonometric Ratios and Identities · Maximum, Minimum and Impossible Values

5\sin\theta + 12\cos\theta = 13

, where

0 < \theta < \frac{\pi}{2}

, then what is

\tan\theta + \cot\theta

equal to ?

Check whether the given value IS the maximum

Before squaring and solving a quadratic, compute

\sqrt{a^2 + b^2}

. If it equals the right-hand side, the sine and cosine are read off immediately; solving the long way wastes minutes and invites a spurious root.

Drill 2 more on a sin θ + b cos θ is at most √(a² + b²)

Concept 5 of 5: Equations that can never hold

Many statements ask whether something like

\sin\theta = x + \dfrac1x

is possible. The right side is at least

2

in size, and a sine is at most

1

, so it never is. Put a bound on each side, and see whether the two ranges overlap.

Definition

$\left|x + \dfrac1x\right| \ge 2$ for every real $x \ne 0$ , so it can never equal a sine or cosine.
$\dfrac{a + b}{2\sqrt{ab}} \ge 1$ for $a, b > 0$ , with equality only at $a = b$ ; so it can equal $\sin\theta$ only if $a = b$ .
$\dfrac{(x + y)^2}{4xy} > 1$ for positive unequal $x, y$ , so it cannot be $\sin^2\theta$ .
A product like $(\sin\alpha + 2)(\sin\alpha - 2)$ is never zero: $\pm 2$ is out of reach.
A quadratic $x^2 + y^2 - 2xy\sin^2\theta = 0$ has real solutions only when its discriminant is not negative, which forces $x = y$ .

The key bound

\left|x + \frac1x\right| \ge 2, \qquad |\sin\theta|, |\cos\theta| \le 1

Worked example

For how many values of

\theta

(\cos\theta - 3)(\cos\theta + 1.5) = 0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q49Moderate

Example 5 · Trigonometric Ratios and Identities · Maximum, Minimum and Impossible Values

Consider the following statements : 1.

\sin\theta = x + \dfrac{1}{x}

is possible for some real value of x. 2.

\cos\theta = x + \dfrac{1}{x}

is possible for some real value of x. Which of the above statements is/are correct ?

A secant CAN exceed 1 — the bound runs the other way

\dfrac{a^2 + b^2}{2ab} \ge 1

can never be a sine or cosine, but it is a perfectly good secant or cosecant. Match the bound to the ratio: sine and cosine live inside

[-1, 1]

, secant and cosecant outside it.

Drill 6 more on equations that can never hold

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Expressions linear in sin²θ or sin θ
The two ends
$a\sin^2\theta + b\cos^2\theta \in [\min(a,b),\, \max(a,b)]$
t + 1/t ≥ 2, and weighted forms by AM–GM
The bounds
$t + \frac1t \ge 2, \qquad a\sec^2\theta + b\operatorname{cosec}^2\theta \ge (\sqrt a + \sqrt b)^2$
Quadratics in sin θ or cos²θ
Vertex, then ends
$c^2 - c + 1 = \left(c - \tfrac12\right)^2 + \tfrac34, \quad c \in [0, 1]$
a sin θ + b cos θ is at most √(a² + b²)
Amplitude bound
$|a\sin\theta + b\cos\theta| \le \sqrt{a^2 + b^2}$
Equations that can never hold
The key bound
$\left|x + \frac1x\right| \ge 2, \qquad |\sin\theta|, |\cos\theta| \le 1$

Watch out for (5)

A restricted range moves the ends
→ Expressions linear in sin²θ or sin θ
On an open interval the bound may not be reached
→ t + 1/t ≥ 2, and weighted forms by AM–GM
The vertex may lie outside the variable's range
→ Quadratics in sin θ or cos²θ
Check whether the given value IS the maximum
→ a sin θ + b cos θ is at most √(a² + b²)
A secant CAN exceed 1 — the bound runs the other way
→ Equations that can never hold

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