Range and Max/Min — The Algebraic Playbook

"Before you differentiate, ask: can I recognize the form?"

Introduction

Finding the range or extreme values of trigonometric expressions is a high-frequency JEE theme. While calculus provides a general method, it's often inefficient. This article equips you with a pattern-recognition toolkit to solve these problems in seconds by identifying standard forms and applying pre-derived results.

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The Core Toolkit: Six Essential Forms

Form 1: The Linear Combination $a\sin\theta + b\cos\theta$

The Fundamental Result: $\boxed{-\sqrt{a^2 + b^2} \ \leq \ a\sin\theta + b\cos\theta \ \leq \ \sqrt{a^2 + b^2}}$ Maximum: $\sqrt{a^2 + b^2}$, Minimum: $-\sqrt{a^2 + b^2}$

Why it works: It can be rewritten as $R\sin(\theta + \phi)$ or $R\cos(\theta - \alpha)$, where $R = \sqrt{a^2 + b^2}$, and the sine/cosine function is bounded by ±1.

Common Variants:

• $a\sin\theta + b\cos\theta + c$: Range is $\left[c - \sqrt{a^2+b^2},\ c + \sqrt{a^2+b^2}\right]$
• $a\sin\theta - b\cos\theta$: Same formula, treat as $a\sin\theta + (-b)\cos\theta$.

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Form 2: The "tan² + cot²" Type (AM-GM Application)

The Result: For $a^2\tan^2\theta + b^2\cot^2\theta$ (where $a, b > 0$, and $\theta$ avoids multiples of $\pi/2$), $\boxed{\text{Minimum value} = 2|ab|}$ Equality occurs when $a^2\tan^2\theta = b^2\cot^2\theta$, i.e., $\tan^4\theta = b^2/a^2$ or $\tan^2\theta = |b/a|$.

Derivation (AM ≥ GM): $\frac{a^2\tan^2\theta + b^2\cot^2\theta}{2} \geq \sqrt{a^2\tan^2\theta \cdot b^2\cot^2\theta} = |ab| \sqrt{\tan^2\theta \cdot \cot^2\theta} = |ab|$ Multiply by 2: $a^2\tan^2\theta + b^2\cot^2\theta \geq 2|ab|$.

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Form 3: The "sec² + cosec²" Type

The Result: For $a\sec^2\theta + b\csc^2\theta$ (with $a, b > 0$, $\theta$ not a multiple of $\pi/2$), $\boxed{\text{Minimum value} = (\sqrt{a} + \sqrt{b})^2}$ Equality occurs when $a\sec^2\theta = b\csc^2\theta$, leading to $\tan^2\theta = \sqrt{b/a}$ (after simplification).

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Form 4: Quadratic in $\sin\theta$ or $\cos\theta$

Standard Strategy: Use the substitution $t = \sin\theta$ (or $\cos\theta$), where $t \in [-1, 1]$. The problem reduces to finding the range of a quadratic $f(t) = At^2 + Bt + C$ over $[-1, 1]$.

Procedure:

1. Find the vertex $t_0 = -B/(2A)$.
2. Evaluate $f(t)$ at $t = -1, 1, \text{and } t_0$ (if $t_0 \in [-1, 1]$).

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Form 5: Rational Functions of $\sin\theta$ or $\cos\theta$

For $f(\theta) = \dfrac{A\sin\theta + B}{C\sin\theta + D}$ or similar forms with $\cos\theta$.

Strategy: Substitute $t = \sin\theta$ ($t \in [-1, 1]$). The function becomes $f(t) = \dfrac{At + B}{Ct + D}$. Its range on $[-1, 1]$ is typically found by checking endpoints and ensuring the denominator doesn't vanish. Alternatively, solve $y = f(t)$ for $t$ and impose $-1 \leq t \leq 1$.

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Form 6: Symmetric Polynomials in $\sin^2\theta$ and $\cos^2\theta$

Key Substitution: Let $u = \sin^2\theta$. Then $\cos^2\theta = 1-u$, and $u \in [0, 1]$. Convert the expression to a polynomial in $u$ and find its range on $[0, 1]$.

Important Special Cases: $\begin{aligned} \sin^4\theta + \cos^4\theta &= 1 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}\sin^2 2\theta \in \left[\frac{1}{2}, 1\right] \\ \sin^6\theta + \cos^6\theta &= 1 - 3\sin^2\theta\cos^2\theta = 1 - \frac{3}{4}\sin^2 2\theta \in \left[\frac{1}{4}, 1\right] \end{aligned}$

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Strategic Problem-Solving Framework

Use this flowchart to select the right tool instantly.

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Applied Mastery: JEE Problems Decoded

Problem Type 1: Direct Linear Combination

Question (JEE Main 2021): Number of integral 'k' for which $3\sin x + 4\cos x = k + 1$ has a solution.

Solution:

1. Form: $a\sin x + b\cos x$ with $a=3, b=4$.
2. Range: $[-\sqrt{3^2+4^2}, \sqrt{3^2+4^2}] = [-5, 5]$.
3. Condition: $k+1 \in [-5, 5] \Rightarrow -6 \leq k \leq 4$.
4. Integral k: $-6, -5, \ldots, 4$ → 11 values.

Key Insight: The problem tests both the range formula and counting integers in an interval.

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Problem Type 2: AM-GM for tan² + cot²

Question: Minimum value of $9\tan^2\theta + 4\cot^2\theta$.

Solution:

1. Form: $a^2\tan^2\theta + b^2\cot^2\theta$ with $a=3, b=2$.
2. Apply Formula: Minimum = $2 \times 3 \times 2 = \boxed{12}$.

Verification: Equality when $9\tan^2\theta = 4\cot^2\theta \Rightarrow \tan^4\theta = 4/9 \Rightarrow \tan^2\theta = 2/3$.

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Problem Type 3: Quadratic Transformation

Question (JEE Main 2022): Let $f(x) = 3\sin^4 x + 10\sin^2 x \cos^2 x + 7\cos^4 x$. Find max-min difference.

Solution:

1. Substitute: Let $t = \sin^2 x \Rightarrow \cos^2 x = 1-t, \ t \in [0,1]$. $f = 3t^2 + 10t(1-t) + 7(1-t)^2$
2. Simplify: $f = 3t^2 + 10t - 10t^2 + 7 - 14t + 7t^2 = 7 - 4t$.
3. Range on [0,1]: Linear decreasing function.
   • Max at $t=0$: $f=7$
   • Min at $t=1$: $f=3$
4. Difference: $7 - 3 = \boxed{4}$.

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Problem Type 4: Rational Function (Bounded Variable)

Question: Range of $f(\theta) = \dfrac{1}{3 - 2\cos\theta}$.

Solution:

1. Bounded Variable: $\cos\theta \in [-1, 1]$.
2. Denominator Range: $3 - 2\cos\theta \in [3-2(1), 3-2(-1)] = [1, 5]$.
3. Reciprocal Range: $f(\theta) \in \left[\frac{1}{5}, 1\right]$.

Check Continuity: Denominator is positive, so $f$ is continuous, and the range is the closed interval $[1/5, 1]$.

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Problem Type 5: Constrained Optimization (Advanced)

Question (JEE Advanced 2019): Minimum of $\sec^4\alpha + \csc^4\alpha$, $\alpha \in (0, \pi/2)$.

Solution (Elegant Approach):

1. Use known identity: $\sec^2\alpha + \csc^2\alpha = \sec^2\alpha\csc^2\alpha = \dfrac{4}{\sin^2 2\alpha}$. Minimum occurs when $\sin 2\alpha = 1$ (i.e., $\alpha = \pi/4$), giving $\sec^2\alpha + \csc^2\alpha = 4$.
2. Let $x = \sec^2\alpha, y = \csc^2\alpha$. We have $x + y = 4$ and need to minimize $x^2 + y^2$.
3. For fixed sum, sum of squares is minimized when $x = y$: $x = y = 2$.
4. Minimum value: $x^2 + y^2 = 2^2 + 2^2 = \boxed{8}$.

Alternative (Direct substitution): At $\alpha = \pi/4$, $\sec\alpha = \csc\alpha = \sqrt{2}$, so $\sec^4\alpha + \csc^4\alpha = 4 + 4 = 8$.

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Quick Reference: Special Ranges & Minima

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Common Pitfalls & Safeguards

Pitfall 1: Ignoring the Domain

Example: Finding the minimum of $\tan^2\theta + \cot^2\theta$ over $\theta \in (0, \pi/2)$ is valid. Over $\theta \in \mathbb{R}$, the function is undefined at multiples of $\pi/2$, and the "minimum" of 2 is never actually attained for a continuous domain.

Safeguard: Always state the domain explicitly in your answer. If the domain excludes points where the expression is defined, the range might be an open interval.

Pitfall 2: Misapplying AM-GM

AM-GM requires non-negative terms. For $\tan^2\theta + \cot^2\theta$, this holds. For $\sin\theta + \csc\theta$, it does not always hold because $\sin\theta$ can be negative. AM-GM applies only for $\theta$ in intervals where $\sin\theta > 0$.

Safeguard: Check term positivity before applying AM-GM. If in doubt, use calculus or other methods.

Pitfall 3: Forgetting the ± in the Linear Formula

The range is $\pm\sqrt{a^2+b^2}$. A common error is to write only the positive maximum.

Safeguard: Remember the sine/cosine function oscillates between -1 and +1.

Pitfall 4: Overlooking Simple Substitution

Before attempting complex manipulations, check if the expression simplifies dramatically with a substitution like $u = \sin^2\theta$ or $v = \sin\theta + \cos\theta$.

Safeguard: Always spend 10 seconds looking for a simplifying substitution.

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Practice Problems (Time Target: 2 minutes each)

1. Find the range of $f(\theta) = 5\cos\theta - 12\sin\theta + 7$.

2. Determine the minimum value of $25\sec^2\theta + 16\csc^2\theta$ for $\theta \in (0, \pi/2)$.

3. Find the maximum and minimum of $\sin^2\theta - 3\sin\theta + 5$. (Hint: Quadratic in $\sin\theta$).

4. If $y = \dfrac{2\sin\theta + 3}{4\sin\theta + 5}$, find its range.

5. Prove that $8\cos^4\theta - 8\cos^2\theta + 1 \geq -1$ and find its maximum. (Hint: Substitute $u = \cos^2\theta$).

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Final Strategic Summary

1. Recognition is Key: The first 15 seconds should be spent matching the problem to one of the six forms.
2. Linear Combination is #1: The $\sqrt{a^2+b^2}$ rule is the single most important result. Know it cold.
3. AM-GM for Specific Forms: Use it confidently for $\tan^2+\cot^2$ and $\sec^2+\csc^2$ types, but verify term positivity.
4. Substitution Simplifies: Reducing a trig problem to an algebraic problem in $t$ (where $t$ is bounded) is a powerful simplification.
5. Calculus as a Last Resort: Only use differentiation if no standard form applies and substitution doesn't simplify sufficiently.
6. Domain Dictates Range: The answer can change if the domain is restricted (e.g., $\theta \in [0, \pi/4]$ vs. $\mathbb{R}$). Always note the domain.

By internalizing these forms and strategies, you transform range problems from tedious explorations into quick, confident identifications. This not only saves time but also drastically reduces computational errors.

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"Efficiency in problem-solving doesn't mean cutting corners; it means knowing the straightest path to the answer."