Trigonometric Functions Ex 3.3 Class 11 NCERT: Concepts and Solutions

Core Identities for Exercise 3.3

Sum and Difference Identities

sin(x + y) = sin x cos y + cos x sin y sin(x - y) = sin x cos y - cos x sin y cos(x + y) = cos x cos y - sin x sin y cos(x - y) = cos x cos y + sin x sin y tan(x + y) = (tan x + tan y) / (1 - tan x tan y) tan(x - y) = (tan x - tan y) / (1 + tan x tan y)

Double Angle Identities

sin(2x) = 2 sin x cos x = (2 tan x) / (1 + tan²x) cos(2x) = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x = (1 - tan²x) / (1 + tan²x) tan(2x) = (2 tan x) / (1 - tan²x)

Triple Angle Identities

sin(3x) = 3sin x - 4sin³x cos(3x) = 4cos³x - 3cos x tan(3x) = (3tan x - tan³x) / (1 - 3tan²x)

Sum-to-Product Identities

sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2) sin x - sin y = 2 cos((x+y)/2) sin((x-y)/2) cos x + cos y = 2 cos((x+y)/2) cos((x-y)/2) cos x - cos y = -2 sin((x+y)/2) sin((x-y)/2)

How and Why to Use These Identities

Have you ever wondered how to find the exact value of sin(75°)? You know the values for 45° and 30°, but you can't simply add sin(45°) + sin(30°). This is precisely where the sum and difference identities shine. By rewriting 75° as 45° + 30°, you can use the formula sin(x+y) = sin x cos y + cos x sin y to find the exact value. These identities are the fundamental 'rules of grammar' for trigonometric functions. They allow us to break down, transform, and simplify complex expressions. When you're asked to prove an identity, think of it as a puzzle. Your goal is to use these formulas as legal moves to transform the Left-Hand Side (LHS) of the equation until it looks exactly like the Right-Hand Side (RHS). The key is to recognize patterns: does the expression involve a sum of angles like (π/4 + x)? Or a multiple angle like 3x? Recognizing the pattern tells you which identity to use.

Worked Examples for Trigonometric Functions Ex 3.3

• Problem 1: Find the value of sin(15°) Step 1: Express the angle as a sum or difference of known angles. We can write 15° as (45° - 30°) or (60° - 45°). Let's use 45° and 30° since their trigonometric values are well known. Step 2: Apply the appropriate difference identity. The identity for sin(x - y) is: sin(x - y) = sin x cos y - cos x sin y. Here, x = 45° and y = 30°. So, sin(15°) = sin(45° - 30°) = sin(45°)cos(30°) - cos(45°)sin(30°). Step 3: Substitute the known values and simplify. We know: sin(45°) = 1/√2 cos(30°) = √3/2 cos(45°) = 1/√2 sin(30°) = 1/2 Substituting these values: sin(15°) = (1/√2)(√3/2) - (1/√2)(1/2) Step 4: Combine the terms. sin(15°) = √3 / (2√2) - 1 / (2√2) = (√3 - 1) / (2√2). Final Answer: The exact value of sin(15°) is (√3 - 1) / (2√2).
• Problem 2: Prove that (sin 5x + sin 3x) / (cos 5x + cos 3x) = tan 4x Step 1: Identify the structure of the expression. The numerator is a sum of sines (sin x + sin y) and the denominator is a sum of cosines (cos x + cos y). This indicates we should use the sum-to-product formulas. Step 2: Apply the sum-to-product formulas. For the numerator: sin 5x + sin 3x = 2 sin((5x+3x)/2) cos((5x-3x)/2) = 2 sin(8x/2) cos(2x/2) = 2 sin(4x) cos(x). For the denominator: cos 5x + cos 3x = 2 cos((5x+3x)/2) cos((5x-3x)/2) = 2 cos(8x/2) cos(2x/2) = 2 cos(4x) cos(x). Step 3: Substitute the expanded forms back into the fraction. LHS = (2 sin(4x) cos(x)) / (2 cos(4x) cos(x)). Step 4: Simplify the expression. The terms 2 and cos(x) are common in the numerator and denominator, so they cancel out. LHS = sin(4x) / cos(4x). Step 5: Use the definition of tangent. We know that tan θ = sin θ / cos θ. Therefore, sin(4x) / cos(4x) = tan(4x). LHS = tan(4x) = RHS. Final Answer: Hence, the identity is proved.

Exam Traps and Key Strategies

A major pitfall for students is mixing up the signs in the sum and difference formulas. Remember this trick: For cos(A±B), the sign in the formula is opposite to the operation in the bracket. For sin(A±B), the sign in the formula matches the operation. Another critical point is choosing the correct form of cos(2x). There are three versions: cos²x - sin²x, 2cos²x - 1, and 1 - 2sin²x. Look at the RHS of your proof! If it only contains cos x, use the 2cos²x - 1 form to eliminate sin x quickly. If it only contains sin x, use 1 - 2sin²x. This strategic choice can drastically simplify your proof and save valuable time in an exam.

Practice Questions with Solutions

• Q: Prove the identity: cos(π/4 - x)cos(π/4 - y) - sin(π/4 - x)sin(π/4 - y) = sin(x+y). A: Step 1: Recognize the structure of the Left-Hand Side (LHS). The LHS is in the form cos A cos B - sin A sin B, where A = (π/4 - x) and B = (π/4 - y). Step 2: Apply the sum identity for cosine, which is cos(A + B) = cos A cos B - sin A sin B. LHS = cos(A + B) = cos((π/4 - x) + (π/4 - y)). Step 3: Simplify the expression inside the cosine function. LHS = cos(π/4 + π/4 - x - y) = cos(π/2 - (x+y)). Step 4: Use the co-function identity cos(π/2 - θ) = sin θ. Here, θ = (x+y). Therefore, cos(π/2 - (x+y)) = sin(x+y). Step 5: Conclude the proof. LHS = sin(x+y), which is equal to the Right-Hand Side (RHS). Final answer: Hence, the identity is proved.
• Q: Find the value of tan(105°). A: Step 1: Express 105° as a sum of two standard angles. We can write 105° = 60° + 45°. Step 2: Apply the sum identity for tangent: tan(x + y) = (tan x + tan y) / (1 - tan x tan y). Here, x = 60° and y = 45°. tan(105°) = (tan 60° + tan 45°) / (1 - tan 60° tan 45°). Step 3: Substitute the known values of tan 60° and tan 45°. We know tan 60° = √3 and tan 45° = 1. tan(105°) = (√3 + 1) / (1 - (√3)(1)) = (1 + √3) / (1 - √3). Step 4: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is (1 + √3). tan(105°) = [(1 + √3)(1 + √3)] / [(1 - √3)(1 + √3)] = (1 + 2√3 + 3) / (1 - 3) = (4 + 2√3) / (-2). Step 5: Simplify the final expression. tan(105°) = - (2 + √3). Final answer: The value of tan(105°) is -(2 + √3).
• Q: Prove that (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) = cot 3x. A: Step 1: Rearrange the terms on the LHS to group terms that can be simplified using sum-to-product formulas. Group cos 4x with cos 2x and sin 4x with sin 2x. LHS = (cos 4x + cos 2x + cos 3x) / (sin 4x + sin 2x + sin 3x). Step 2: Apply cos A + cos B = 2 cos((A+B)/2)cos((A-B)/2) to the numerator and sin A + sin B = 2 sin((A+B)/2)cos((A-B)/2) to the denominator. Numerator: 2 cos((4x+2x)/2)cos((4x-2x)/2) + cos 3x = 2 cos 3x cos x + cos 3x. Denominator: 2 sin((4x+2x)/2)cos((4x-2x)/2) + sin 3x = 2 sin 3x cos x + sin 3x. Step 3: Factor out the common terms in the numerator and denominator. Numerator: cos 3x (2 cos x + 1). Denominator: sin 3x (2 cos x + 1). Step 4: Substitute the factored forms back into the fraction. LHS = [cos 3x (2 cos x + 1)] / [sin 3x (2 cos x + 1)]. Step 5: Cancel the common factor (2 cos x + 1). LHS = cos 3x / sin 3x. Step 6: Use the identity cot θ = cos θ / sin θ. LHS = cot 3x, which is equal to the RHS. Final answer: Hence, the identity is proved.
• Q: Prove that sin 3x + sin 2x - sin x = 4 sin x cos(x/2) cos(3x/2). A: Step 1: Rearrange the terms on the LHS to apply the difference-to-product formula. Group sin 3x and sin x. LHS = (sin 3x - sin x) + sin 2x. Step 2: Apply the identity sin A - sin B = 2 cos((A+B)/2)sin((A-B)/2) to the grouped terms. sin 3x - sin x = 2 cos((3x+x)/2)sin((3x-x)/2) = 2 cos(2x) sin(x). LHS = 2 cos(2x) sin(x) + sin 2x. Step 3: Expand the sin 2x term using the double angle formula sin 2x = 2 sin x cos x. LHS = 2 cos(2x) sin(x) + 2 sin(x) cos(x). Step 4: Factor out the common term 2 sin(x). LHS = 2 sin(x) [cos(2x) + cos(x)]. Step 5: Apply the sum-to-product formula cos A + cos B = 2 cos((A+B)/2)cos((A-B)/2) to the term in the brackets. cos(2x) + cos(x) = 2 cos((2x+x)/2)cos((2x-x)/2) = 2 cos(3x/2) cos(x/2). Step 6: Substitute this back into the expression for the LHS. LHS = 2 sin(x) [2 cos(3x/2) cos(x/2)] = 4 sin(x) cos(x/2) cos(3x/2). This matches the RHS. Final answer: Hence, the identity is proved.

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