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A question asks for the maximum of 3 sin x + 4 cos x, and the quick answer "3 + 4 = 7" is wrong, because sin x and cos x cannot both equal 1 at the same x. The right answer is 5, and one rewrite explains why. This post shows that rewrite, works through four question types JEE likes, and lists the slips that cost marks.
Where does this appear in the syllabus?
The JEE (Main) 2026 syllabus lists "Trigonometrical identities and trigonometrical functions" in Unit 14. The JEE (Advanced) 2026 brochure adds "Trigonometric functions, their periodicity and graphs, addition and subtraction formulae". The tool you need is the sum formula sin(x + y) = sin x cos y + cos x sin y, which is in NCERT's Class 11 chapter on trigonometric functions and in the CBSE 2026-27 mathematics curriculum. Check the current documents for your exam year.
Why is the range from −√(a² + b²) to √(a² + b²)?
Let r = √(a² + b²) and pick an angle α with cos α = a/r and sin α = b/r. This works because (a/r)² + (b/r)² = 1. Then
a sin x + b cos x = r(sin x cos α + cos x sin α) = r sin(x + α).
As x runs over all real numbers, sin(x + α) takes every value from −1 to 1, so the expression takes every value from −r to r.
A second way to see it: a sin x + b cos x is the dot product of the vector (a, b) with the unit vector (sin x, cos x). A dot product with a unit vector is largest when the two point the same way, and then it equals the length of (a, b), which is r.
Worked example 1: a constant added
Find the range of f(x) = 3 sin x + 4 cos x + 2.
Here r = √(9 + 16) = 5, so 3 sin x + 4 cos x runs from −5 to 5, and f runs from −3 to 7.
Check the maximum: it happens when x + α = π/2, so sin x = cos α = 3/5 and cos x = sin α = 4/5. Then 3(3/5) + 4(4/5) = 9/5 + 16/5 = 5, and adding 2 gives 7.
Worked example 2: hidden in a sum
Find the range of g(x) = 5 cos x + 3 cos(x + π/3) + 3.
Expand: 3 cos(x + π/3) = (3/2) cos x − (3√3/2) sin x. So g(x) = (13/2) cos x − (3√3/2) sin x + 3.
Now r² = 169/4 + 27/4 = 196/4 = 49, so r = 7. The range is from 3 − 7 to 3 + 7, that is −4 to 10.
Worked example 3: a reciprocal
Find the range of h(x) = 1/(3 sin x + 4 cos x + 7).
The denominator runs from 7 − 5 = 2 to 7 + 5 = 12 and is never zero. A reciprocal of positive numbers reverses the order, so h runs from 1/12 to 1/2. If the constant were 4 instead of 7, the denominator would run from −1 to 9, pass through zero, and h would have no maximum or minimum.
Worked example 4: a restricted interval
Find the range of k(x) = sin x + √3 cos x for 0 ≤ x ≤ π/2.
Here r = 2, and cos α = 1/2, sin α = √3/2 give α = π/3, so k(x) = 2 sin(x + π/3). As x goes from 0 to π/2, the angle x + π/3 goes from π/3 to 5π/6. On that stretch, sine rises from √3/2 to 1 (at π/2) and falls to 1/2 (at 5π/6). Its smallest value is 1/2, so k runs from 1 to 2. The full-period answer, −2 to 2, would be wrong.
Squared terms reduce to the same form. For example, sin²x + 2√3 sin x cos x + 3 cos²x becomes 2 + cos 2x + √3 sin 2x using the double-angle formulas, which runs from 2 − 2 = 0 to 2 + 2 = 4.
Which mistakes cost the most marks?
Adding the maxima: 3 + 4 = 7 assumes sin x and cos x both reach 1 at once.
Ignoring the domain: a restricted interval for x can cut the range, as in example 4.
Flipping a reciprocal without checking for zero in the denominator.
Expanding only part of a sum: in example 2, both cosine terms carry cos x.
A practice routine
Solve each problem two ways. Use the r sin(x + α) rewrite, then the dot-product view, and compare. In a randomised study of 70 seventh-grade students, Rittle-Johnson and Star (2007) found that comparing solution methods side by side led to greater gains in procedural knowledge and flexibility than studying the same methods one at a time.
Mix the four types in sets of six: constant added, hidden sum, reciprocal, restricted interval.
Check one value by substitution for every answer, as in example 1.
Review the identities. Our guide to practising trigonometric identities covers the identities you will use to reduce squared terms.
If you are not sure where a range question goes wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the step that cut or stretched your range.
Tonight, take three range questions from past papers and write the value of r for each before you do anything else.
