JEE Advanced Maths compound angle previous year questions with solutions are available on this page. A compound angle is defined as the algebraic sum of two or more angles. Students can expect one or two questions from this topic for the JEE Advanced exam. Those who are preparing for JEE Advanced should practise a lot of sample question papers and previous years’ question papers. Students can also download chapter-wise JEE Advanced important questions provided in PDF format.
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JEE Advanced Maths Compound Angle Previous Year Questions With Solutions
Question 1: If (sin4x)/2 + (cos4x)/3 = ⅕, then
(a) tan2x = ⅔
(b) (sin8x)/8 + (cos8x)/27 = 1/125
(c) tan2x = ⅓
(d) (sin8x)/8 + (cos8x)/27 = 2/125
Solution:
Given (sin4x)/2 + (cos4x)/3 = ⅕
=> (sin4x)/2 + (1 – sin2x)2/3 = ⅕
=> (sin4x)/2 + (1 – 2 sin2x + sin4x)/3 = 1/5
Put t = sin2x
=> t2/2 + (1-2t+t2)/3 = ⅕
=> 3t2 + 2 – 4t + 2t2 = 6/5
=> 5t2 – 4t + 2 = 6/5
=> 25t2 – 20t + 4 = 0
=> (5t-2)2 = 0
=> t = 2/5
sin2x = 2/5
cos2x = 1-sin2x
= 1 – 2/5
= 3/5
tan2x = 2/3
(sin8x)/8 + (cos8x)/27 = (2/5)4/8 + (3/5)4/27
= (2/54) + (3/54)
= 5/54
= 1/125
Hence option a and b are correct.
Question 2: The maximum value of the expression 1/(sin2x + 3 sin x cos x + 5 cos2x) is
(a) 2
(b) 1
(c) 3
(d) 4
Solution:
Let f(x) = sin2x + 3 sin x cos x + 5 cos2x
= (sin2x+ cos2x) + (3/2) (2 sin x cos x) + 4 cos2x
= 1 + (3/2) sin 2x + 2(2 cos2x)
= 1 + (3/2) sin 2x + 2(1 + cos 2x)
= 3 + (3/2) sin 2x + 2 cos 2x
= 3 + (3 sin 2x + 4 cos 2x)/2 …(i)
Use -√(a2+b2) ≤ a sin x + b cos x ≤ √(a2+b2)
=> – (5/2) ≤ (3 sin 2x + 4 cos 2x)/2 ≤ (5/2)
(i) => 3 – (5/2) ≤ f(x) ≤ 3 + 5/2
=> ½ ≤ f(x) ≤ 11/2
=> 2/11 ≤ 1/f(x) ≤ 2
Hence the maximum value of the expression is 2.
Hence option a is the answer.
Question 3: Let α and β be non zero real numbers such that 2(cos β – cos α) + cos α cos β = 1. Then which of the following is/are true.
(a) tan (α/2) – √3 tan (β/2) = 0
(b) √3 tan (α/2) – tan (β/2) = 0
(c) tan (α/2) + √3 tan (β/2) = 0
(d) √3tan (α/2) + tan (β/2) = 0
Solution:
Given that 2(cos β – cos α) + cos α cos β = 1
=> 2(cos β – cos α) = 1 – cos α cos β
=> (cos β – cos α) = (1 – cos α cos β)/2
cos 2x = (1 – tan2x)/(1 + tan2x)
Put tan (α/2) = x and tan (β/2) = y
So cos α = (1 – x2)/(1 + x2)
And cos β = (1-y2)/(1+y2)
=> 2[(1-y2)/(1+y2) – (1-x2)/(1+x2)] = 1 – [(1-x2)(1-y2)/(1+x2)(1+y2)]
=> 2[(1+x2)(1-y2) – (1-x2)(1+y2)] = (1+x2)(1+y2) – (1-x2)(1-y2)
=> 4(x2 – y2)= 2(x2+y2)
=> x2 = 3y2
=> x= ±√3y
=> tan (α/2) = ±√3 tan (β/2)
=> tan (α/2) ±√3 tan (β/2) = 0
Hence option a and c are correct.
Question 4: Let P = {x: sin x – cos x = √2 cos x} and Q = {x: sin x + cos x = √2 sin x} be two sets. Then
(a) P⊂Q and Q-P ≠ Ø
(b) Q ⊄ P
(c) P ⊄ Q
(d) P = Q
Solution:
Given P = sin x – cos x = √2 cos x
Divide by cos x
=> tan x – 1 = √2
=> tan x = √2+1 …(i)
Given Q = sin x + cos x = √2 sin x
Divide by cos x
tan x + 1 = √2 tan x
tan x (√2-1) = 1
tan x = 1/(√2-1)
Multiply numerator and denominator by (√2+1)
=> tan x = (√2+1)/(√2-1)(√2+1)
= (√2+1)/1
= (√2+1) …(ii)
From (i) and (ii), we get P = Q.
Hence option d is the answer.
Question 5: If tan A = n/(n+1) and tan B = 1/(2n+1), 0 < A+B < 2 π, then A+B equals
(a) π/4
(b) π/2
(c) π/3
(d) π/4
Solution:
Given tan A = n/(n+1) and tan B = 1/(2n+1)
We know tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
= [n/(n+1) + 1/(2n+1)]/(1 – n/(n+1)(2n+1)]
= (2n2 + n + n+1)/(2n2+2n+n+1 – n)
= (2n2 + 2n+1)/(2n2 + 2n+1)
= 1
=> A+B = tan-1 1
= π/4
Hence option a is the answer.
Question 6: The value of cos (π/4 – A) cos (π/4 – B) – sin (π/4 – A) sin (π/4 – B) is
(a) sin (A+B)
(b) cos (A+B)
(c) 0
(d) 1
Solution:
We know that cos X cos Y – sin X sin Y = cos (X+Y)
(take X = π/4 – A and Y = π/4 – B)
So cos (π/4 – A) cos (π/4 – B) – sin (π/4 – A) sin (π/4 – B) = cos (π/4 – A + π/4 – B)
= cos (π/2 – (A+B))
= sin (A+B) (since cos (90-x) = sin x)
Hence option a is the answer.
Question 7: If sin (A-B) = 1/√10, cos (A+B) = 2/√29, find the value of tan 2A where A and B lie between 0 and π/4.
(a) 1/17
(b) 17
(c) 17/6
(d) 1
Solution:
tan 2A = tan (A+B + A-B)
= [tan (A+B) + tan (A-B)]/(1 – tan (A+B) tan (A-B)) …(i)
Given that sin (A-B) = 1/√10
=> tan (A-B) = ⅓ (Use Pythagoras theorem)
Given that cos (A+B) = 2/√29
=> tan (A+B) = 5/2 (Use Pythagoras theorem)
Substitute tan (A-B) and tan (A+B) in (i)
tan 2A = [ (5/2) + ⅓]/(1-(5/2)(⅓))
= (17/6)/(⅙)
= 17
Hence option b is the answer.
Question 8: The value of tan(π/4 + x).tan (3π/4 + x) is
(a) 1
(b) 0
(c) 2
(d) -1
Solution:
We know tan (A+B) = (tan A + tan B)/(1-tan A tan B)
tan(π/4 + x) = (1 + tan x)/(1 – tan x) (since tan π/4 = 1)
tan (3π/4 + x) = (-1 + tan x)/(1 + tan x) (since tan 3π/4 = -1)
tan(π/4 + x)tan (3π/4 + x) = (1 + tan x)(-1 + tan x)/(1 – tan x)(1 + tan x)
= (-1 + tan x)/(1 – tan x)
= -1(1 – tan x)/(1 – tan x)
= -1
Hence option d is the answer.
Question 9: If cos (x-y) + cos (y-z) + cos (z-x) = -3/2, then find the value of sin x + sin y + sin z
(a) 0
(b) 1
(c) 3
(d) -1
Solution:
We know cos (A-B) = cos A cos B + sin A sin B
Given that cos (x-y) + cos (y-z) + cos (z-x) = -3/2
=> cos x cos y + sin x sin y + cos y cos z + sin y sin z + cos z cos x + sin z sin x = -3/2
=> 2 cos x cos y + 2 sin x sin y + 2 cos y cos z + 2 sin y sin z + 2 cos z cos x + 2 sin z sin x = -3
=> 2 cos x cos y + 2 sin x sin y + 2 cos y cos z + 2 sin y sin z + 2 cos z cos x + 2 sin z sin x + 3 = 0
=> 2 cos x cos y + 2 sin x sin y + 1 + 2 cos y cos z + 2 sin y sin z + 1 + 2 cos z cos x + 2 sin z sin x + 1 = 0
Instead of 1, put sin2x + cos2x, sin 2y + cos2y, and sin2z + cos2z.
=> 2 cos x cos y + 2 sin x sin y + sin2x + cos2x + 2 cos y cos z + 2 sin y sin z + sin 2y + cos2y + 2 cos z cos x + 2 sin z sin x + sin2z + cos2z = 0
=> (sin x + sin y + sin z)2 + (cos x + cos y + cos z)2 = 0
Sum of perfect squares = 0 implies both quantities will be zero.
=> sin x + sin y + sin z = 0
Hence option a is the answer.
Question 10: If tan a and tan b are the roots of the equation x2 + px + q = 0, p ≠ 0, then
(a) sin2(a+b) + p.sin(a+b). cos(a+b) + q cos2(a+b) = q
(b) tan (a+b) = p/(q-1)
(c) cos (a+b) = 1-q
(d) sin (a+b) = -p
Solution:
Given that tan a and tan b are the roots of the equation x2 + px + q = 0.
Sum of roots = tan a + tan b = -p
Product of roots = tan a tan b = q
tan (a+b) = (tan a + tan b)/(1 – tan a tan b)
= -p/(1-q)
= p/(q-1) …(i)
Option b is correct.
Check option a.
sin2(a+b) + p. sin (a+b). cos (a+b) + q cos2(a+b)
Multiply and divide by cos2(a+b)
=> cos2(a+b)[ tan2 (a+b) + p.tan (a+b) + q]
= [ tan2 (a+b) + p.tan (a+b) + q]/sec2(a+b)
= [ tan2 (a+b) + p.tan (a+b) + q] /(1+tan2(a+b))
Put (i) in above equation
= [p2/(q-1)2 + p2/(q-1) + q)]/(1 + p2/(q-1)2)
= q(p2+q2-2q+1)/(p2+q2-2q+1)
= q
Option a is also correct.
Hence option a and b is the answer.
Question 11: Let a, b, c be three non zero real numbers such that the equation √3a cos x + 2b sin x = c, x belongs to [-π/2, π/2] has two distinct roots α and β with α + β = π/3. Then the value of b/a is
(a) 1
(b) 4/3
(c) 1/2
(d) 0
Solution:
Given that √3a cos x + 2b sin x = c
Divide by a, we get
√3 cos x + (2b/a) sin x = c/a …(i)
Since α and β are the roots of (i)
√3 cos α + (2b/a) sin α = c/a …(ii)
√3 cos β + (2b/a) sin β = c/a …(iii)
Subtract (iii) from (ii)
=> √3 (cos α – cos β) + (2b/a) (sin α – sin β) = 0
We know cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
Also sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
=> √3 (-2 sin ((α+β)/2) sin ((α-β)/2)) + (2b/a) (2 cos ((α+β)/2) sin ((α-β)/2)) = 0
Given α + β = π/3
=> √3 (-2 sin (π/6) sin (α-β)/2) + (2b/a) (2 cos (π/6) sin (α-β)/2) = 0
=> √3 (-2 ×(½) × sin (α-β)/2) + (2b/a) (2×(√3/2) sin (α-β)/2) = 0
=> √3 ( sin (α-β)/2) = (2b/a) (√3 sin (α-β)/2)
=> 1 = 2b/a
=> b/a = 1/2
Hence option c is the answer.
Question 12: Suppose θ and φ (≠ 0) are such that sec (θ + φ), sec θ, and sec (θ – φ) are in AP. If cos θ = k cos (φ/2) for some k, then k is equal to
(a) ±√2
(b) ± 1
(c) ±1/√2
(d) ±2
Solution:
Given that sec (θ + φ), sec θ, and sec (θ – φ) are in AP.
So 2 sec θ = sec (θ + φ) + sec (θ – φ)
=> 2/cos θ = [1/cos (θ + φ)] + [1/cos (θ – φ)]
Take LCM and add
2/cos θ = [cos (θ – φ) + cos (θ + φ)]/cos (θ – φ).cos (θ + φ)
2/cos θ = 2 cos θ cos φ/(cos 2θ – sin2φ)
(cos 2θ – sin2φ) = cos2 θ cos φ
(cos 2θ – cos2 θ cos φ) = sin2φ
cos 2θ (1 – cos φ) = sin2φ
cos 2θ (1 – cos φ) = 1 – cos2φ
cos 2θ (1 – cos φ) = (1 – cos φ)(1 + cos φ)
cos 2θ = (1 + cos φ)
cos 2θ = 2 cos2 (φ/2)
So cos θ = ±√2 cos (φ/2)
Given that cos θ = k cos (φ/2)
=> k = ±√2
Hence option a is the answer.
Question 13: In a triangle PQR, let ∠PQR = 300 and the sides PQ and QR have length 10√3 and 10, respectively. Then which of the following statement(s) is (are) true?
(a) ∠QPR = 450
(b) Area of the triangle PQR is 25√3 and ∠QRP = 1200
(c) The radius of the incircle of the triangle PQR is 10√3 – 15
(d) The area of the circumcircle of the triangle is PQR is 100π
Solution:
Given that ∠PQR = 300
PQ = 10√3
QR = 10
Area of triangle PQR = ½ ab sin C
= ½ ×10√3×10×sin 30
= ½ ×10√3×10×1/2
= 25√3 …(i)
Let PR = x
Use cosine rule
cos 30 = ((10√3)2 + (10)2 – x2)/2×10√3×10
√3/2 = (300 + 100 – x2)/200√3
(300 + 100 – x2) = 300
100 – x2 = 0
=> x = 10
So triangle PQR is isosceles.
So ∠QPR = 300
∠QRP = 180 – (30+30)
∠QRP = 1200 …(ii)
Radius of incircle = ∆/s
= 25√3/(10+5√3) (semi perimeter = 10+5√3)
= 25√3(10-5√3)/(10+5√3)(10-5√3)
= (250√3 – 375)/(100 – 75)
= (250√3 – 375)/25
= 10√3 – 15 …(iii)
Area of circumcircle = πR2, where R is the circum radius
(sin A)/a = 1/2R
(½)/10 = 1/2R
=> R = 10
So area = π(10)2
= 100π …(iv)
From (i), (ii), (iii) and (iv), we can say that options b, c, d are correct.
Question 14: If tan A and tan B are the roots of the quadratic equation x2 – px + q = 0, then sin2(A+B) equals
(a) p2/[p2 + (1- q)2]
(b) p2/[p2 – (1- q)2]
(c) q2/[p2 + (1- q)2]
(d) none of these
Solution:
Given tan A and tan B are the roots of the quadratic equation x2 – px + q = 0.
Sum of roots = tan A + tan B = p
Product of roots = tan A tan B = q
We know
tan(A + B) = (tan A + tan B)/ (1 – tan A tan B)
= p/(1 – q)
tan2(A + B) = p2/(1 – q)2
cot2(A + B) = (1 – q)2/p2
Using the identity cosec2x – cot2x = 1,
cosec2(A + B) = 1 + cot2(A + B)
= 1 + [(1 – q)2/p2]
= [p2 +(1- q)2]/p2
So, sin2(A + B) = 1/cosec2(A + B) = p2/[p2 +(1- q)2]
Hence option a is the answer.
Question 15: If 5(tan2 x – cos2 x) = 2cos 2x + 9, then the value of cos 4x is :
(a) 1/3
(b) -7/9
(c) 2/9
(d) -3/5
Solution:
Given that 5(tan2 x – cos2 x) = 2cos 2x + 9
(We use tan2x = sec2x – 1 and cos 2x = 2 cos2x – 1)
5(sec2x – 1 – cos2 x) = 2(2 cos2x – 1) + 9
5 sec2 x – 5 = 9 cos2 x + 7
Put cos2 x = t
(5 / t) = 9t + 12
9t2 + 12t – 5 = 0
t = 1/3 or – 5/3
t = 1/3 as t ≠ – 5/3
cos2 x = 1/3, cos 2x = 2cos2 x – 1 = -1/3
cos 4x = 2cos2 2x – 1
= (2/9) – 1
= -7/ 9
Hence option b is the answer.
Applying Concepts: Trigonometric Ratios of Compound Angles for JEE
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JEE Main Maths Trigonometry Previous Year Questions With Solutions
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