Trigonometry is one of the most important topics in JEE Main exam. Every year about 1-3 questions are asked from this topic. The overall weightage of Trigonometry in JEE Main Question Paper is 7%. Some of the topics in Trignometry include Identities of Trignometry and Trigonometric Equations, Functions of Trignometry, Properties of Inverse trigonometric functions, and Problems of Heights and Distances. Check JEE Main Mathematics Syllabus
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JEE Main aspirants can refer to the previous years’ questions listed below, to get an idea of the type of questions asked about Trigonometry.
In ∆PQR, if 3sin P + 4cos Q = 6 and 4sin Q + 3cos P = 1, then the angle R is equal to?
56
6
4
34
Answer: Option b: 6
Solution:
Let, 3sin P + 4cos Q = 6 … (1)
4sin Q + 3 cos P = 1 … (2)
(1)2 + (2)2 = (9) (1) + 16 (1) + 24
Sin P Cos Q + Cos P Sin Q = 36 + 1
25 + 24 sin (P + Q) = 37
24 sin (P + Q) = 12
Sin (P + Q) = ½
P + Q = 6, R = 56
If P + Q = π, then equation (1) and (2) is not satisfied.
The sides of a triangle are 3x + 4y, 4x + 3y, and 5x + 5y, where x, y > 0, then the triangle is?
Right angled
Obtuse angled
Equilateral
None of these
Answer: Option b: Obtuse angled
Solution:
Let a = 3x + 4y, b = 4x + 3y, and c = 5x + 5y
As x, y > 0, c = 5x + 5y is the largest side
Hence, C is the largest angle
Now, cos C = (3x +4y)2 + (4x + 3y)2 + (5x + 5y)2 2 3x + 4y (4x + 3y)
= - 2xy 2 3x + 4y (4x + 3y) < 0
So, C is an obtuse angle
The number of solutions for tan x + sec x = 2cos x in [0, 2π) is?
2
3
0
1
Answer: Option b: 3
Solution:
The given equation is tan x + sec x = 2 cos x
Sin x + 1 = 2cos2 x
2sin2 x + sin x – 1 = 0
(2 sin x – 1) (sin x + 1) = 0
Sin x = ½, -1
x = 30˚, 150˚, 270˚
In a triangle ABC, let ∠C = 2. If r is in radius and R is the circumradius of the triangle ABC, then 2(r + R) is equal to?
b + c
a + b
a + b + c
c + a
Answer: Option b: a + b
Solution:
2r + 2R = c + 2ab(a + b + c)
c2 + c a + b + 2 aba + b + c
(a + b)2 + c a + ba + b + c
(a + b + c) a + ba + b + c
a+ b
If A = sin2 x + cos4 x = then for all real x:
1 ≤ A ≤ 2
34 ≤ A ≤ 1316
34 ≤ A ≤ 1
1316 ≤ A ≤ 1
Answer: Option c: 34 ≤ A ≤ 1
Solution:
A = 1 – cos2 x + cos4 x
1 - cos2 x (1 - cos2 x)
1 - cos2 x sin2 x
1 – ¼ (sin2 2x)
1 – ¼ (1) ≤ A ≤ 1 – ¼ (0)
¾ ≤ A ≤ 1
The equation esin x- e-sin x-4=0 has?
Infinite number of real roots
No real roots
Exactly one real root
Exactly four real roots
Answer: Option b: No real roots
Solution: esin x- e-sin x-4=0
Let esin x=t
t – 1t = 4
t2 – 4t – 1 = 0
t = 4 ± √202 = 2 ± √5
esin x= 2 ± √5
Let A and B denote statements
A: cos α + cos β + cos γ = 0 and
B: sin α + sin β + sin γ = 0
If cos (β – γ) + cos (γ – α) + cos (α - β) = - 32
A is true and B is false
A is false and B is true
Both A and B are true
Both A and B are false
Answer: Option c: Both A and B are true
Solution:
cos (β – γ) + cos (γ – α) + cos (α - β) = - 32
2 [cos (β – γ) + cos (γ – α) + cos (α - β)] + 3 = 0
2 [cos (β – γ) + cos (γ – α) + cos (α - β)] + sin2 α + cos2 α + sin2 β + cos2 β + sin2 γ + cos2 γ = 0
(sin α + sin β + sin γ)2 + (cos α + cos β + cos γ)2 = 0
If 0 ≤ x < 2π then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0 is:
9
5
7
3
Answer: Option c: 7
Solution:
2 cos 2x cos x + 2 cos 3x cos x = 0
2 cos x (cos 2x + cos 3x) = 0
2 cos x 2 cos 5x/2 cos x/2 = 0
x = 2, 3π2,, 5 , 3π5, 7π5, 9π5
7 solutions
If in a triangle ABC, the altitudes from the vertices A, B, C on the opposite sides are in H. P, then sin A, sin B, sin C are in:
G.P
A.P
Arithmetic – Geometric Progression
H.P
Answer: Option b: A.P
Solution: ∆ ½ p1a = ½ p2b = ½ p3c
p1, p2, p3 are in H.P
2∆a, 2∆b, 2∆c are in H.P
1a, 1b, 1c are in H.P
a, b, c is in A.P
Sin A, sin B, sin C are in A.P
If sum of all solutions of the equation 8 cos x. (cos (6+x) . cos 6+x – ½ ) = 1, in [0, ] is k, then k equals to?
139
89
209
23
Answer: Option a: 139
Solution:
8 cos x (cos2 6 - sin2 x – ½ = 1
8 cos x (¼ - (1 - cos2 x)) = 1
8 cos x (cos2 x – ¾) = 1
2 cos 3x = 1
cos 3x = ½
3x + 2n ± 3, n I
x = 2nπ3 ± 9 , in x [0, ]
x = 9 , 2π3 + 9 , 2π3 - 9
k = 139
Check JEE Main Practice Papers
Most of the questions asked on Trigonometry topic in JEE Mains exam can be solved by applying the formulas directly. Hence, students need to remember all the major formulae from this chapter. Here’s a list of the important formulae for JEE Main Trigonometry:
Trigonometric ratios of acute angles:
Sin = ph cos = bh tan = pb
cosec = hp sec = hb cot = bp
Trigonometric identities
Sin2 + cos2 = 1
Sec2 = 1 + tan2
cosec2 = 1 + cot2
Trigonometric Ratios of compound angles
Sin (A ± B) = sin A Cos B ± cos A Sin B
cos (A ± B) = cos A Cos B ∓ sin A Sin B
tan (A ± B) = tan A B 1+Atan B
cot (A ± B) = cot Acot B 1 B A
sin (A + B) sin (A - B) = sin2 A – Sin2 B = cos2 B – cos2 A
cos (A + B) cos (A - B) = cos2 A - Sin2 B = cos2 B - sin2 A
sin (A + B + C) = sin A Cos B Cos C + sin B CosA Cos C + sin C Cos A Cos B – sin A sin B sin C
cos (A + B + C) = cos A Cos B Cos C – cos A Sin B Sin C – cos B Sin A Sin C – cos C sin A sin B
tan (A + B + C) = tan A +B +C -tan Atan Btan C 1 -Atan B -tan Btan C-tan Ctan A
Transformation formulae in Trigonometry:
2sin A Cos B = sin (A + B) + Sin (A – B)
2sin B Cos A = sin (A + B) - Sin (A – B)
2 cos A Cos B = cos (A + B) + cos (A – B)
2 sin A Sin B = cos (A – B) – cos (A + B)
Sin A + Sin B = 2 Sin (A + B2) Cos (A- B2)
Sin A - Sin B = 2 Cos (A + B2) Sin (A- B2)
Cos A + Cos B = 2 Cos (A + B2) Cos (A- B2)
Cos A - Cos B = 2 Sin (A + B2) Sin (A- B2)
*The article might have information for the previous academic years, please refer the official website of the exam.