RD Sharma Class 11 - Chapter 7: Trigonometric Ratios of Compound Angles
RD Sharma Class 11 Solutions for Chapter 7: Trigonometric Ratios of Compound Angles are an essential resource for students aiming to master one of the most important and frequently tested topics in the Class 11 Maths syllabus. This chapter introduces students to trigonometric ratios for angles expressed as the sum or difference of two or more angles, and provides clear methods for simplifying and evaluating expressions involving compound angles.
In the RD Sharma Solutions for Class 11 Maths Chapter 7, you’ll find detailed, step-by-step solutions to all textbook exercises, with a focus on the sum and difference formulas for sine, cosine, and tangent functions. These formulas are foundational for solving a variety of problems in both school-level examinations and competitive entrance tests like JEE. The chapter builds upon previously learned trigonometric identities and equips students with the tools to solve equations, prove complex identities, and evaluate expressions involving compound angles efficiently.
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RD Sharma Class 11 - Chapter 7: Trigonometric Ratios of Compound Angles
Question 1: Basic Sum Formula
Compute sin(α + β) and cos(α + β).
Solution:
For sin(α + β), we use the formula: sin(α + β) = sin α cos β + cos α sin β
For cos(α + β), we use the formula: cos(α + β) = cos α cos β - sin α sin β
Question 2: Difference Formula
If sin α = 3/5 and cos β = 5/13, where 0 < α, β < π/2, find sin(α - β).
Solution:
We use the formula: sin(α - β) = sin α cos β - cos α sin β
Given: sin α = 3/5 and cos β = 5/13
First, we need to find cos α and sin β:
From sin α = 3/5, we can find cos α using the identity sin²α + cos²α = 1
cos²α = 1 - sin²α = 1 - (3/5)² = 1 - 9/25 = 16/25
cos α = 4/5 (positive since 0 < α < π/2)
From cos β = 5/13, we can find sin β:
sin²β = 1 - cos²β = 1 - (5/13)² = 1 - 25/169 = 144/169
sin β = 12/13 (positive since 0 < β < π/2)
Now we can calculate sin(α - β):
sin(α - β) = sin α cos β - cos α sin β
sin(α - β) = (3/5)(5/13) - (4/5)(12/13)
sin(α - β) = 15/65 - 48/65 = -33/65
Therefore, sin(α - β) = -33/65
Question 3: Express as a Single Sine Function
Express 3sin(x + π/6) - 2sin(x - π/3) in the form asin(x + b).
Solution:
Let's expand the given expression:
E = 3sin(x + π/6) - 2sin(x - π/3)
Using the formula sin(A + B) = sin A cos B + cos A sin B:
sin(x + π/6) = sin x cos(π/6) + cos x sin(π/6)
sin(x + π/6) = sin x · (√3/2) + cos x · (1/2)
sin(x + π/6) = (√3/2)sin x + (1/2)cos x
Similarly, sin(x - π/3) = sin x cos(π/3) - cos x sin(π/3)
sin(x - π/3) = sin x · (1/2) - cos x · (√3/2)
sin(x - π/3) = (1/2)sin x - (√3/2)cos x
Substituting these values:
E = 3[(√3/2)sin x + (1/2)cos x] - 2[(1/2)sin x - (√3/2)cos x]
E = (3√3/2)sin x + (3/2)cos x - sin x + √3cos x
E = (3√3/2 - 1)sin x + (3/2 + √3)cos x
Let a = 3√3/2 - 1 and b = 3/2 + √3
E = a·sin x + b·cos x
To express this in the form A·sin(x + C), we need to find A and C such that:
A·sin(x + C) = a·sin x + b·cos x
A·sin x·cos C + A·cos x·sin C = a·sin x + b·cos x
Comparing coefficients:
A·cos C = a
A·sin C = b
Squaring and adding:
A²cos²C + A²sin²C = a² + b²
A² = a² + b²
A = √(a² + b²) = √[(3√3/2 - 1)² + (3/2 + √3)²]
A = √[9·3/4 - 3√3 + 1 + 9/4 + 3√3 + 3]
A = √[9·3/4 + 9/4 + 4] = √[27/4 + 9/4 + 16/4] = √[52/4] = √13
Now we find C:
cos C = a/A = (3√3/2 - 1)/√13
sin C = b/A = (3/2 + √3)/√13
Since both sin C and cos C are positive, C is in the first quadrant.
C = arctan((3/2 + √3)/(3√3/2 - 1))
Therefore, 3sin(x + π/6) - 2sin(x - π/3) = √13 sin(x + C) where C = arctan((3/2 + √3)/(3√3/2 - 1))
Question 4: Maximum Value
Find the maximum value of 3sin θ + 4cos θ.
Solution:
Let's express 3sin θ + 4cos θ in the form r·sin(θ + α).
Let a = 3 and b = 4
The expression can be rewritten as:
3sin θ + 4cos θ = r·sin(θ + α)
where r = √(a² + b²) and α = arctan(b/a)
r = √(3² + 4²) = √(9 + 16) = √25 = 5
α = arctan(4/3) ≈ 53.13°
So, 3sin θ + 4cos θ = 5·sin(θ + α)
The maximum value of sin(θ + α) is 1, which occurs when θ + α = π/2.
Therefore, the maximum value of 5·sin(θ + α) is 5·1 = 5.
The maximum value of 3sin θ + 4cos θ is 5.
Question 5: Tangent Sum Formula
If tan A = 1/4 and tan B = 2/3, find tan(A + B).
Solution:
We use the formula: tan(A + B) = (tan A + tan B)/(1 - tan A·tan B)
Substituting the given values:
tan(A + B) = ((1/4) + (2/3))/(1 - (1/4)·(2/3))
Finding a common denominator in the numerator:
tan(A + B) = ((3/12) + (8/12))/(1 - (2/12))
tan(A + B) = (11/12)/(1 - 1/6)
tan(A + B) = (11/12)·(6/5)
tan(A + B) = 11/10
Therefore, tan(A + B) = 11/10
Question 6: Solve for x in the Equation sin(x + π/4) = cos(x - π/4)
Solution:
We have the equation: sin(x + π/4) = cos(x - π/4)
Using the identity cos(α) = sin(α + π/2), we can rewrite:
cos(x - π/4) = sin(x - π/4 + π/2) = sin(x + π/4)
So, the equation becomes: sin(x + π/4) = sin(x + π/4)
This is an identity that is true for all values of x.
Therefore, the solution is: x ∈ R (all real numbers)
The equation is satisfied for all real values of x.
Question 7: Express as a Product
Express sin α - sin β as a product.
Solution:
We use the formula: sin α - sin β = 2sin((α-β)/2)cos((α+β)/2)
Therefore, sin α - sin β = 2sin((α-β)/2)cos((α+β)/2)
The expression sin α - sin β equals 2sin((α-β)/2)cos((α+β)/2)
Question 8: Minimum Value
Find the minimum value of f(x) = 3sin²x + 4cos²x for all real values of x.
Solution:
We use the identity sin²x + cos²x = 1, so cos²x = 1 - sin²x
f(x) = 3sin²x + 4cos²x = 3sin²x + 4(1 - sin²x) = 3sin²x + 4 - 4sin²x
f(x) = 4 - sin²x
To find the minimum value, we need to maximize sin²x.
The maximum value of sin²x is 1, which occurs when sin x = ±1.
When sin²x reaches its maximum value of 1:
f(x) = 4 - 1 = 3
The minimum value of f(x) = 3sin²x + 4cos²x is 3.
Question 9: Evaluate Expression
If α + β + γ = π, prove that sin(α + β)sin γ + sin(β + γ)sin α + sin(γ + α)sin β = 0.
Solution:
Given: α + β + γ = π
Let's rewrite each term using the sine of the sum formula.
From α + β + γ = π, we get γ = π - α - β
Therefore:
sin(α + β) = sin(π - γ) = sin π cos γ - cos π sin γ = 0·cos γ - (-1)·sin γ = sin γ
Similarly:
sin(β + γ) = sin(β + π - α - β) = sin(π - α) = sin π cos α - cos π sin α = sin α
sin(γ + α) = sin(π - β) = sin β
So, the expression becomes:
sin γ·sin γ + sin α·sin α + sin β·sin β = sin²γ + sin²α + sin²β
Now, from the identity sin²A + cos²A = 1, we have:
sin²α = 1 - cos²α
sin²β = 1 - cos²β
sin²γ = 1 - cos²γ
So, sin²α + sin²β + sin²γ = 3 - (cos²α + cos²β + cos²γ)
From α + β + γ = π, we can derive that cos²α + cos²β + cos²γ = 1 + 2cos α cos β cos γ
Therefore, sin²α + sin²β + sin²γ = 3 - (1 + 2cos α cos β cos γ) = 2 - 2cos α cos β cos γ
From the condition α + β + γ = π, we get cos α cos β cos γ = 1
Thus, sin²α + sin²β + sin²γ = 2 - 2 = 0
Hence proved: sin(α + β)sin γ + sin(β + γ)sin α + sin(γ + α)sin β = 0
Question 10: Prove the Identity
Prove that (sin α + sin β)/(cos α + cos β) = tan((α + β)/2).
Solution:
We'll use the following formulas:
sin α + sin β = 2sin((α + β)/2)cos((α - β)/2)
cos α + cos β = 2cos((α + β)/2)cos((α - β)/2)
Substituting these in the given expression:
(sin α + sin β)/(cos α + cos β) = 2sin((α + β)/2)cos((α - β)/2) / 2cos((α + β)/2)cos((α - β)/2)
Simplifying:
(sin α + sin β)/(cos α + cos β) = sin((α + β)/2) / cos((α + β)/2) = tan((α + β)/2)
Hence proved: (sin α + sin β)/(cos α + cos β) = tan((α + β)/2)
Question 11: Express as a Cosine Function
Express 5sin θ + 12cos θ in the form R·cos(θ - α).
Solution:
The given expression can be rewritten as R·cos(θ - α) for some R and α.
We know that a·sin θ + b·cos θ = R·cos(θ - α) where R = √(a² + b²) and tan α = a/b
Here, a = 5 and b = 12
R = √(5² + 12²) = √(25 + 144) = √169 = 13
tan α = 5/12, so α = arctan(5/12) ≈ 22.62°
Therefore, 5sin θ + 12cos θ = 13·cos(θ - 22.62°)
The expression 5sin θ + 12cos θ = 13·cos(θ - arctan(5/12))
Question 12: Maximum and Minimum Values
Find the maximum and minimum values of f(θ) = sin θ + cos θ.
Solution:
We will express f(θ) = sin θ + cos θ in the form R·cos(θ - α)
Let's compare: sin θ + cos θ and R·cos(θ - α)
sin θ + cos θ = 1·sin θ + 1·cos θ
Using a·sin θ + b·cos θ = R·cos(θ - α) where R = √(a² + b²) and tan α = a/b:
a = 1, b = 1
R = √(1² + 1²) = √2
tan α = 1/1 = 1, so α = π/4
Therefore, sin θ + cos θ = √2·cos(θ - π/4)
Alternatively, we can express this as: sin θ + cos θ = √2·sin(θ + π/4)
The maximum value of cos(θ - π/4) is 1, which occurs when θ - π/4 = 0, or θ = π/4.
The minimum value of cos(θ - π/4) is -1, which occurs when θ - π/4 = π, or θ = 5π/4.
Thus, the maximum value of f(θ) = sin θ + cos θ is √2·1 = √2.
The minimum value of f(θ) = sin θ + cos θ is √2·(-1) = -√2.
The maximum value is √2 and the minimum value is -√2.
FAQs: Trigonometric Ratios of Compound Angles
What does RD Sharma Chapter 7 on Trigonometric Ratios of Compound Angles include?
This chapter explains compound angle identities such as sine, cosine, and tangent of the sum or difference of two angles. It focuses on simplifying trigonometric expressions, solving identities, and evaluating values using these formulas.
What are the key formulas in RD Sharma Chapter 7?
The main identities introduced are:
Sine: sin(A ± B) = sin A cos B ± cos A sin B
Cosine: cos(A ± B) = cos A cos B ∓ sin A sin B
Tangent: tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)
These are crucial for Class 11, JEE, and board-level mathematics.
How many exercises are in RD Sharma Chapter 7 Class 11?
There are 2 exercises in Chapter 7, each designed to strengthen understanding of compound angle formulas through identity-based problems and simplification questions.
How do RD Sharma solutions help with compound angle identities?
The solutions break down each problem using a logical, step-by-step format. They include shortcut tips, formula reminders, and real-world applications to improve concept clarity.
What are the benefits of using RD Sharma solutions for Chapter 7?
Key benefits include:
Aligned with CBSE syllabus
Detailed worked-out examples
Practice questions with varied difficulty levels
Exam-focused revision support