RD Sharma Class 11 Solutions for Chapter 5: Trigonometric Functions

Step 1: Understand what a trigonometric function is.

A trigonometric function relates the angles of a triangle to the lengths of its sides, particularly used in right-angled triangles.

Step 2: Identify the three primary trigonometric functions.

The three primary functions are based on the ratio of sides in a right triangle:

• sin(θ) = Opposite / Hypotenuse
• cos(θ) = Adjacent / Hypotenuse
• tan(θ) = Opposite / Adjacent

Step 3: Derive the three reciprocal trigonometric functions.

• cosec(θ) = 1 / sin(θ) = Hypotenuse / Opposite
• sec(θ) = 1 / cos(θ) = Hypotenuse / Adjacent
• cot(θ) = 1 / tan(θ) = Adjacent / Opposite

Answer: The six standard trigonometric functions are: sin, cos, tan, cosec, sec, and cot.

Step 1: Determine the domain of sine and cosine functions.

The sine and cosine functions are defined for all real numbers as they can represent angles of any magnitude.

Domain of sin(θ) and cos(θ): All real numbers (ℝ)

Step 2: Determine the range of sine and cosine functions.

For a right triangle, these ratios represent sides/hypotenuse, which can never exceed 1 in absolute value.

Using the unit circle representation, both functions oscillate between -1 and 1.

Answer: Domain: ℝ (all real numbers), Range: [-1, 1]

Step 1: Understand the concept of principal value.

Since trigonometric functions are periodic, they can have multiple solutions for a single value.

The principal value is the unique solution within a designated range for inverse trigonometric functions.

Step 2: Identify the principal ranges.

• sin⁻¹(x): Range [-π/2, π/2]
• cos⁻¹(x): Range [0, π]
• tan⁻¹(x): Range [-π/2, π/2]

Step 3: Provide an example.

If sin⁻¹(1/2) = θ, then sin(θ) = 1/2

Possible values: 30°, 150°, 390°, etc.

Principal value: 30° or π/6 radians (within range [-π/2, π/2])

Answer: The principal value of sin⁻¹(1/2) is π/6 radians (30°).

Step 1: Use the conversion formula.

Step 2: Substitute 225° into the formula.

Radians = 225 × (π/180)

= 225π/180

= 5π/4 (simplifying by dividing by 45)

Step 3: Determine the quadrant.

Angle 225° = 180° + 45°

This places the angle in the third quadrant (angles between 180° and 270°)

In the third quadrant: sin < 0, cos < 0, tan > 0

Answer: 225° = 5π/4 radians; Third quadrant

Step 1: Recall the definition of cosecant.

cosec θ is the reciprocal of sin θ

Step 2: Write the mathematical relationship.

cosec θ = 1/sin θ

Step 3: Verify this relationship.

In a right triangle: cosec θ = Hypotenuse/Opposite

sin θ = Opposite/Hypotenuse

Therefore: cosec θ = 1/sin θ

Answer: cosec θ = 1/sin θ

Step 1: Recall the standard trigonometric values.

sin 60° = √3/2

cos 30° = √3/2

Step 2: Add the values.

sin 60° + cos 30° = √3/2 + √3/2

= (√3 + √3)/2

= 2√3/2

= √3

Answer: √3

Step 1: Recall the standard values for 45°.

tan 45° = 1

Since cot θ = 1/tan θ, we have cot 45° = 1/tan 45° = 1/1 = 1

Step 2: Add the values.

tan 45° + cot 45° = 1 + 1 = 2

Answer: 2

Step 1: Recall the fundamental identity.

We know that sec²θ = 1 + tan²θ

Step 2: Apply the identity to the left side.

LHS: sec²θ - tan²θ

Substitute: (1 + tan²θ) - tan²θ

= 1 + tan²θ - tan²θ

= 1

Step 3: Compare with the right side.

RHS: 1

LHS = RHS

Therefore, sec²θ - tan²θ = 1 is proven.

Step 1: Rewrite the left side using powers.

sin⁴A + cos⁴A = (sin²A)² + (cos²A)²

Step 2: Use the algebraic identity (a² + b²) = (a + b)² - 2ab.

(sin²A)² + (cos²A)² = (sin²A + cos²A)² - 2sin²Acos²A

Step 3: Apply the fundamental identity sin²A + cos²A = 1.

= 1² - 2sin²Acos²A

= 1 - 2sin²Acos²A

Step 4: Relate to sin 2A.

We know sin 2A = 2sin A cos A

Therefore, sin²2A = 4sin²Acos²A

So, sin²Acos²A = sin²2A/4

Thus, 2sin²Acos²A = 2 × sin²2A/4 = sin²2A/2 = (1/2)sin²2A

Step 5: Substitute back.

sin⁴A + cos⁴A = 1 - 2sin²Acos²A = 1 - (1/2)sin²2A

Therefore, sin⁴A + cos⁴A = 1 - (1/2)sin²2A is proven.

Step 1: Write the definitions of tan and cot.

tan θ = sin θ / cos θ

cot θ = cos θ / sin θ

Step 2: Substitute into the left side.

LHS: tan θ + cot θ

= sin θ / cos θ + cos θ / sin θ

Step 3: Compare with the right side.

RHS: sin θ / cos θ + cos θ / sin θ

LHS = RHS

Therefore, tan θ + cot θ = sin θ / cos θ + cos θ / sin θ is proven.

Step 1: Apply the fundamental identity.

We know: sin²A + cos²A = 1

Therefore: 1 - cos²A = sin²A

Step 2: Substitute into the expression.

(1 - cos²A) / sin²A = sin²A / sin²A

Step 3: Simplify the fraction.

sin²A / sin²A = 1

Answer: 1

Step 1: Apply the cofunction identity.

The cofunction identity states: cos(90° - θ) = sin θ

Step 2: Verify with an example.

Test with θ = 30°: cos(90° - 30°) = cos 60° = 1/2

sin 30° = 1/2

Both values match, confirming our identity.

Answer: sin θ

Step 1: Recognize the addition formula pattern.

The expression matches the form: sin A cos B + cos A sin B

Step 2: Apply the addition formula.

sin A cos B + cos A sin B = sin(A + B)

Step 3: Substitute the values.

sin 18° cos 72° + cos 18° sin 72° = sin(18° + 72°)

= sin 90°

Step 4: Evaluate sin 90°.

sin 90° = 1

Answer: 1

Step 1: Replace sec θ and tan θ with their definitions.

sec θ = 1/cos θ

tan θ = sin θ/cos θ

Step 2: Substitute into the expression.

sec θ (1 - sin θ)(sec θ + tan θ)

= (1/cos θ)(1 - sin θ)(1/cos θ + sin θ/cos θ)

Step 3: Simplify the second parenthesis.

(1/cos θ + sin θ/cos θ) = (1 + sin θ)/cos θ

Step 4: Multiply the terms.

= (1/cos θ)(1 - sin θ)((1 + sin θ)/cos θ)

= (1 - sin θ)(1 + sin θ)/cos²θ

Step 5: Apply the difference of squares formula.

(1 - sin θ)(1 + sin θ) = 1 - sin²θ

= cos²θ (using sin²θ + cos²θ = 1)

Step 6: Final simplification.

cos²θ/cos²θ = 1

Answer: 1

Step 1: Set up a right triangle using tan θ = 3/4.

tan θ = Opposite/Adjacent = 3/4

Let Opposite = 3k and Adjacent = 4k (for some constant k)

Step 2: Find the hypotenuse using Pythagorean theorem.

Hypotenuse² = Opposite² + Adjacent²

Hypotenuse² = (3k)² + (4k)² = 9k² + 16k² = 25k²

Hypotenuse = 5k

Step 3: Calculate all trigonometric functions.

• sin θ = Opposite/Hypotenuse = 3k/5k = 3/5
• cos θ = Adjacent/Hypotenuse = 4k/5k = 4/5
• tan θ = 3/4 (given)
• cosec θ = 1/sin θ = 5/3
• sec θ = 1/cos θ = 5/4
• cot θ = 1/tan θ = 4/3

sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, cosec θ = 5/3, sec θ = 5/4, cot θ = 4/3

Step 1: Define the period of a function.

The period is the smallest positive value T such that f(x + T) = f(x) for all x in the domain.

Step 2: For sine function.

sin(x + 2π) = sin x (verified by unit circle)

The sine function completes one full cycle every 2π units.

Step 3: For cosine function.

cos(x + 2π) = cos x (verified by unit circle)

The cosine function also completes one full cycle every 2π units.

Step 4: Graphical verification.

Both sine and cosine graphs show one complete wave cycle from 0 to 2π:

• Sine: Starts at (0,0), peaks at (π/2,1), zero at (π,0), trough at (3π/2,-1), back to (2π,0)
• Cosine: Starts at (0,1), zero at (π/2,0), trough at (π,-1), zero at (3π/2,0), back to (2π,1)

Period of both sin x and cos x is 2π

Step 1: Define the four quadrants.

• Quadrant I: 0° to 90° (x > 0, y > 0)
• Quadrant II: 90° to 180° (x < 0, y > 0)
• Quadrant III: 180° to 270° (x < 0, y < 0)
• Quadrant IV: 270° to 360° (x > 0, y < 0)

Step 2: Determine the signs based on definitions.

In the unit circle: x = cos θ, y = sin θ

Step 3: Analyze each quadrant.

• Quadrant I: All positive (sin > 0, cos > 0, hence all functions positive)
• Quadrant II: sin > 0, cos < 0 → sin, cosec > 0; cos, sec, tan, cot < 0
• Quadrant III: sin < 0, cos < 0 → tan, cot > 0; sin, cosec, cos, sec < 0
• Quadrant IV: sin < 0, cos > 0 → cos, sec > 0; sin, cosec, tan, cot < 0

Quadrant I: All positive
Quadrant II: sin, cosec positive
Quadrant III: tan, cot positive
Quadrant IV: cos, sec positive

Step 1: Use addition formulas for sine.

sin(A + B) = sin A cos B + cos A sin B

Step 2: Apply with A = π and B = x.

sin(π + x) = sin π cos x + cos π sin x

Since sin π = 0 and cos π = -1:

= 0 · cos x + (-1) · sin x

= 0 - sin x = -sin x

Step 3: Use addition formulas for cosine.

cos(A + B) = cos A cos B - sin A sin B

Step 4: Apply with A = π and B = x.

cos(π + x) = cos π cos x - sin π sin x

Since sin π = 0 and cos π = -1:

= (-1) · cos x - 0 · sin x

= -cos x

sin(π + x) = -sin x and cos(π + x) = -cos x are proven.

Step 1: Use the definition of tangent.

tan x = sin x / cos x

Step 2: Express tan(-x).

tan(-x) = sin(-x) / cos(-x)

Step 3: Apply the even-odd properties.

sin(-x) = -sin x (sine is an odd function)

cos(-x) = cos x (cosine is an even function)

Step 4: Substitute and simplify.

tan(-x) = -sin x / cos x = -tan x

Therefore, tan(-x) = -tan x is proven, confirming tan is an odd function.

Step 1: Evaluate sin 150°.

150° = 180° - 30°

sin(180° - θ) = sin θ (supplementary angle property)

sin 150° = sin 30° = 1/2

Step 2: Evaluate cos 270°.

270° = 360° - 90° = 3 × 90°

On the unit circle, at 270°, the coordinates are (0, -1)

cos 270° = 0

Step 3: Evaluate tan 315°.

315° = 360° - 45°

tan(360° - θ) = -tan θ

tan 315° = -tan 45° = -1

sin 150° = 1/2, cos 270° = 0, tan 315° = -1

Step 1: Identify the key points of the sine function.

The sine function starts at the origin for x = 0.

• At x = 0: sin 0 = 0 → Point (0, 0)
• At x = π/2: sin(π/2) = 1 → Point (π/2, 1)
• At x = π: sin π = 0 → Point (π, 0)
• At x = 3π/2: sin(3π/2) = -1 → Point (3π/2, -1)
• At x = 2π: sin(2π) = 0 → Point (2π, 0)

Step 2: Describe the shape and properties.

The graph is a smooth wave that:

• Oscillates between -1 and 1
• Has a period of 2π
• Is symmetric about the origin (odd function)
• Crosses the x-axis at multiples of π

Key points: (0,0), (π/2,1), (π,0), (3π/2,-1), (2π,0)

Step 1: Find the principal value.

tan x = √3

x = tan⁻¹(√3)

x = π/3 or 60° (since tan 60° = √3)

Step 2: Consider the periodicity of tangent.

tan function has a period of π.

If tan α = tan β, then α = β + nπ (where n ∈ ℤ)

Step 3: Write the general solution.

x = π/3 + nπ, where n ∈ ℤ

This gives solutions: ..., -2π/3, π/3, 4π/3, 7π/3, ...

General solution: x = π/3 + nπ, where n ∈ ℤ

Step 1: Find principal values where cos θ = -1/2.

Since cos θ is negative, θ must be in quadrant II or III.

In quadrant II: θ = π - π/3 = 2π/3 (since cos π/3 = 1/2)

In quadrant III: θ = π + π/3 = 4π/3

Step 2: Apply periodicity of cosine.

cos function has a period of 2π.

If cos α = cos β, then α = ±β + 2nπ (where n ∈ ℤ)

Step 3: Write the general solution.

θ = 2π/3 + 2nπ or θ = 4π/3 + 2nπ

This can be combined as: θ = 2nπ ±2π/3, where n ∈ ℤ

General solution: θ = 2nπ ± 2π/3, where n ∈ ℤ