Previous year questions: Trigonometric Functions

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Six items below are real JEE Main 2026 questions from the official National Testing Agency papers, and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the paper and the marks. The stem and the options of every JEE item were checked against the official PDF page, and the final value against the official key. Try the question before reading the solution; the methods come straight from the formula sheet and the three topic pages.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Given that sin of pi by 18, sin of 5 pi by 18 and sin of 7 pi by 18 multiply to K, find sin of (10 K pi by 3). Choose from (root 3 plus 1) by (2 root 2), (root 3 minus 1) by root 2, root 3 by 2, and 1 by 2.

Answer: The three angles are 10, 50 and 70 degrees. Use the product to sum formula twice. Multiplying sin 10 by sin 50 gives half of (cos 40 minus cos 60), that is half of (cos 40 minus 1 by 2). Multiplying by sin 70 and using the product to sum formula again reduces the expression to 1 by 8. So K equals 1 by 8. Then 10 K pi by 3 equals 10 pi by 24, that is 5 pi by 12, and the sine of 75 degrees is (root 3 plus 1) by (2 root 2). The correct option is the first one, matching the official key.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

For x between 0 and pi, find the maximum value of 16 sin of x by 2 multiplied by the cube of cos of x by 2.

Answer: Write the expression in a form that uses a single angle. Since 2 sin(x/2) cos(x/2) is sin x, the given expression becomes 8 sin x times cos squared of x by 2. Replace cos squared of x by 2 with (1 plus cos x) by 2, so the expression is 4 sin x plus 4 sin x cos x, that is 4 sin x plus 2 sin 2x.

Differentiate with respect to x: 4 cos x plus 4 cos 2x. Setting this to zero gives cos x plus cos 2x equal to 0. Use cos 2x equal to 2 cos squared x minus 1 to get 2 cos squared x plus cos x minus 1 equal to 0, whose roots are cos x equal to 1 by 2 or cos x equal to minus 1. In the interval, x equal to pi by 3 or x equal to pi. At x equal to pi by 3 the expression is 4 times root 3 by 2 times 3 by 2, which is 3 root 3, while at pi it is 0. The maximum is 3 root 3, matching the official key.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Let tan A and tan B, with A and B lying between minus pi by 2 and pi by 2, be the roots of x squared minus 2x minus 5 equal to 0. Find 20 sin squared of (A plus B) by 2.

Answer: From the sum and product of roots, tan A plus tan B equals 2 and tan A tan B equals minus 5. Apply the tangent addition formula: tan(A plus B) equals 2 divided by 1 plus 5, that is 1 by 3. The sum A plus B lies between minus pi and pi, and its tangent is positive and small, so A plus B is the first quadrant angle whose tangent is 1 by 3. Then cos of A plus B equals 3 by root 10, because the cosine of an angle with tangent 1 by 3 is 3 by the square root of 10. Now use the half angle form: sin squared of half the angle equals (1 minus cos) by 2, which is (1 minus 3 by root 10) by 2. Multiplying by 20 gives 10 minus 3 root 10, which is the third option and the official key.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Find the sum of all integral values of p for which 3 sin x plus 12 cos x minus 3 equals p has at least one real solution x.

Answer: The equation has a solution exactly when p plus 3 lies in the range of the expression 3 sin x plus 12 cos x. The range of a sin x plus b cos x is from minus the square root of (a squared plus b squared) to the same positive value. Here a is 3 and b is 12, so the range is from minus root 153 to root 153. Root 153 is about 12.37. Therefore p plus 3 lies between minus 12.37 and 12.37, and p lies between about minus 15.37 and 9.37. The integers in that interval run from minus 15 to 9, and their sum is found by pairing: the average of the first and last is minus 3, and there are 25 integers, so the sum is 25 times minus 3, that is minus 75. That is the third option and the official key.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

For alpha greater than beta greater than zero, define f of theta as alpha tan squared theta plus beta cot squared theta and g of theta as alpha sin squared theta plus beta cos squared theta. Given that the minimum of f on the open interval from 0 to pi by 2 equals the maximum of g on the open interval from 0 to pi, the first term of a geometric progression is alpha by 2 beta and its common ratio is 2 beta by alpha. The sum of the first ten terms is m by n in lowest terms. Find m plus n.

Answer: For the minimum of f, apply the inequality that a positive number plus its reciprocal is at least 2. Write alpha tan squared plus beta cot squared as a sum whose product is alpha beta, so the minimum is 2 root of alpha beta, reached where alpha tan squared equals beta cot squared. For g, rewrite it as beta plus (alpha minus beta) sin squared theta, which increases with sin squared and has maximum alpha. Equating the two gives 2 root alpha beta equal to alpha, so alpha equals 4 beta.

Now the first term is 2 and the common ratio is 1 by 2. The sum of the first ten terms of a geometric progression is 2 times (1 minus 1 by 1024) divided by (1 minus 1 by 2), which is 4 times 1023 by 1024, that is 1023 by 256. So m is 1023 and n is 256, and m plus n is 1279, matching the official answer.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

The number alpha is the sum of 3, 4, 8, 9, 13, 14 and so on up to 40 terms. If (tan beta) to the power alpha by 1020 is a root of the equation x squared plus x minus 2 equal to 0, with beta between 0 and pi by 2, find sin squared beta plus 3 cos squared beta.

Answer: The series groups as pairs: 3 plus 4, then 8 plus 9, then 13 plus 14, and so on. The pair sums are 7, 17, 27, and so on, an arithmetic progression with first term 7 and common difference 10. There are 20 pairs in 40 terms, so the sum is 20 by 2 times (2 times 7 plus 19 times 10), which is 10 times (14 plus 190), that is 2040. So alpha by 1020 equals 2, and the expression is tan squared beta.

The quadratic factors as (x minus 1)(x plus 2), so its roots are 1 and minus 2. Since beta lies in the first quadrant, tan beta is positive, so tan squared beta can only be the positive root 1. That gives beta equal to pi by 4. Then sin squared beta is 1 by 2 and cos squared beta is 1 by 2, so the required value is 1 by 2 plus 3 by 2, that is 2. The official key is 2.

Q7 (NCERT Exemplar, 3 marks)

If cos of (alpha plus beta) equals 4 by 5 and sin of (alpha minus beta) equals 5 by 13, with the angles in the first quadrant and alpha greater than beta, find tan 2 alpha.

Answer: Since alpha plus beta is acute, sin of (alpha plus beta) is 3 by 5, so tan of (alpha plus beta) is 3 by 4. Since alpha minus beta is acute with sine 5 by 13, its cosine is 12 by 13 and its tangent is 5 by 12. Now 2 alpha equals (alpha plus beta) plus (alpha minus beta), so apply the tangent addition formula. The numerator is 3 by 4 plus 5 by 12, that is 14 by 12, and the denominator is 1 minus the product 15 by 48, that is 33 by 48. The quotient is 56 by 33. Source: NCERT Exemplar Class 11 Mathematics, Chapter 3, Exercise 3.3, Short Answer Type, Q4.

Q8 (NCERT Exemplar, 3 marks)

If a cos theta plus b sin theta equals m and a sin theta minus b cos theta equals n, show that a squared plus b squared equals m squared plus n squared.

Answer: Square both equations and add. The square of the first is a squared cos squared theta plus b squared sin squared theta plus 2ab sin theta cos theta. The square of the second is a squared sin squared theta plus b squared cos squared theta minus 2ab sin theta cos theta. The cross terms cancel and the remaining terms pair up: a squared times (cos squared plus sin squared) plus b squared times (sin squared plus cos squared). Both brackets equal 1, so the sum is a squared plus b squared. Source: NCERT Exemplar Class 11 Mathematics, Chapter 3, Exercise 3.3, Short Answer Type, Q7.

Q9 (NCERT Exemplar, 3 marks)

Find the value of tan 22 degrees 30 minutes.

Answer: The angle is half of 45 degrees. Use the half angle form for tangent, writing tan of half an angle as (1 minus cos) divided by sin with the full angle 45 degrees. Then cos 45 is 1 by root 2 and sin 45 is also 1 by root 2, so the expression is (1 minus 1 by root 2) divided by 1 by root 2. Multiplying numerator and denominator by root 2 gives (root 2 minus 1). The angle is acute, so the positive value is correct, and tan 22 degrees 30 minutes equals root 2 minus 1. Source: NCERT Exemplar Class 11 Mathematics, Chapter 3, Exercise 3.3, Short Answer Type, Q8.

Q10 (NCERT Exemplar, 3 marks)

Solve 2 sin squared theta equals 3 cos theta for theta between 0 and 2 pi.

Answer: Replace sin squared theta by 1 minus cos squared theta. The equation becomes 2 minus 2 cos squared theta equal to 3 cos theta, that is 2 cos squared theta plus 3 cos theta minus 2 equal to 0. Factor as (2 cos theta minus 1)(cos theta plus 2) equal to 0. The second factor gives cos theta equal to minus 2, which is impossible. The first gives cos theta equal to 1 by 2, whose solutions in the interval are theta equal to pi by 3 and theta equal to 5 pi by 3. Both values satisfy the original equation. Source: NCERT Exemplar Class 11 Mathematics, Chapter 3, Exercise 3.3, Short Answer Type, Q18.