Class 11 Maths NCERT Solutions Chapter 3

NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions is an important chapter that introduces students to the fundamental concepts of trigonometry. This chapter deals with the six trigonometric functions—sine, cosine, tangent, cosecant, secant, and cotangent—and their properties, graphs, and applications. The students will understand how these functions are defined on the unit circle and hence understand the angle ratios in a right-angled triangle. The chapter also focuses on the periodic nature of trigonometric functions and their values at some standard angles that involve 0°, 30°, 45°, 60°, and 90° as a basis for the solutions of trigonometric equations and problems in higher levels.

The NCERT Solutions for this chapter step out with step-by-step solutions to a variety of problems so that students will come to understand trigonometric identities and equations. These exercises will help in evaluating trigonometric functions, simplifying expressions with identities, and solving problems related to the real-life applications of trigonometry such as heights and distances. This will enhance their analytical skills and proficiency in solving trigonometric equations. This chapter provides a context for further exploration in trigonometry and its applications in other parts of mathematics and physics, which should make it very much a part of any Class 11 curriculum.

Exercise 3.1

1. Find the radian measures corresponding to the following degree measures:
   (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

To convert degrees to radians, we use the formula:

Radians = Degrees × π / 180

(i) 25° = 25 × π / 180 = 5π / 36 radians

(ii) -47° 30′ First, convert 30′ to degrees: 30′ = 30/60 = 0.5° So, -47° 30′ = -47.5° = -47.5 × π / 180 = -19π / 72 radians

(iii) 240° = 240 × π / 180 = 4π / 3 radians

(iv) 520° = 520 × π / 180 = 26π / 9 radians

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

To convert radians to degrees, we use the formula:

Degrees = Radians × 180 / π

Given π = 22/7, let’s convert the radian measures:

(i) 7π/6 radians = (7π/6) × (180/π) = 7 × 30 = 210 degrees

(ii) -5π/3 radians = (-5π/3) × (180/π) = -5 × 60 = -300 degrees

(iii) 11π/2 radians = (11π/2) × (180/π) = 11 × 90 = 990 degrees

(iv) -17π/3 radians = (-17π/3) × (180/π) = -17 × 60 = -1020 degrees

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Here’s how to calculate the number of radians the wheel turns in one second:

Revolutions per second:

• The wheel makes 360 revolutions in 60 seconds (1 minute).
• So, in one second, it makes 360/60 = 6 revolutions.

Radians per revolution:

• One complete revolution is equal to 2π radians.

Radians per second:

• If the wheel makes 6 revolutions in one second, and each revolution is 2π radians, then in one second, the wheel turns 6 * 2π = 12π radians.
• Therefore, the wheel turns 12π radians in one second.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

Here’s how to find the degree measure of the angle subtended at the center of the circle:

1. Find the angle in radians:

The formula to find the angle θ in radians subtended by an arc of length l at the center of a circle of radius r is:

θ = l/r

In this case:

• l = 22 cm (length of the arc)
• r = 100 cm (radius of the circle)

So, θ = 22/100 = 0.22 radians

2. Convert radians to degrees:

We know that π radians = 180 degrees.
Therefore, 1 radian = 180/π degrees.
So, 0.22 radians = 0.22 * (180/π) degrees
Using π = 22/7, we get:
0.22 radians = 0.22 * (180 * 7/22) degrees = 12.6 degrees
Therefore, the angle subtended at the center of the circle is 12.6 degrees.

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.

Here’s how to find the length of the minor arc:

Radius of the circle:

• Diameter = 40 cm
• Radius (r) = Diameter/2 = 40/2 = 20 cm

Equilateral triangle:

• Since the chord length is equal to the radius, the triangle formed by the chord and the two radii is an equilateral triangle.
• Each angle of an equilateral triangle is 60 degrees.

Angle in radians:

• To convert degrees to radians, we multiply by π/180:
  • Angle in radians (θ) = 60 * (π/180) = π/3 radians

Length of the arc:

• The formula to find the length (l) of an arc subtended by an angle θ at the center of a circle of radius r is:
  • l = rθ
• Substituting the values:
  • l = 20 * (π/3)
  • l = (20π)/3 cm

Therefore, the length of the minor arc is (20π)/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Let r1 and r2 be the radii of the two circles, respectively.

We know that the length of an arc (l) is given by the formula:

l = rθ

where:

• r is the radius of the circle
• θ is the angle subtended by the arc at the center, in radians

Since the arc lengths are equal, we can equate the two expressions:

r1 * θ1 = r2 * θ2

Here, θ1 = 60° = π/3 radians and θ2 = 75° = 5π/12 radians.

Substituting the values:
r1 * (π/3) = r2 * (5π/12)

Simplifying:
r1/r2 = (5π/12) / (π/3) = 5/4
Therefore, the ratio of the radii is 5:4.

7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length
   (i) 10 cm (ii) 15 cm (iii) 21 cm

We can use the formula for the length of an arc:

arc length = radius × angle (in radians)

In this case, the length of the pendulum (75 cm) is the radius. So, we can rearrange the formula to find the angle:

angle (in radians) = arc length / radius

Now, we can calculate the angle for each arc length:

i) Arc length = 10 cm

angle = 10 cm / 75 cm = 2/15 radians

ii) Arc length = 15 cm

angle = 15 cm / 75 cm = 1/5 radians

iii) Arc length = 21 cm

angle = 21 cm / 75 cm = 7/25 radians

Therefore, the angles in radians for the three arc lengths are:

• 2/15 radians
• 1/5 radians
• 7/25 radians

Exercise 3.2

1. Find the values of the other five trigonometric functions in Exercises 1 to 5.
   1. cos x = -1/2, x lies in the third quadrant.

2. sin x = 3/5, x lies in the second quadrant.

3. cot x = 3/4, x lies in third quadrant.

4. sec x = 13/5, x lies in fourth quadrant

5. tan x = -5/12, x lies in second quadrant.

6. Find the values of the trigonometric functions in Exercises 6 to 10.

I’ll provide solutions for exercises 6 to 10, building upon the approach used in the previous examples:

6. sin x = 1/2, x lies in second quadrant.

cos x: Using the Pythagorean identity:

cos² x x + sin² x = 1

cos² x x + (1/2)² = 1

cos² x = 3/4

cos x = -√3/2  (cosine is negative in the second quadrant)

tan x:

tan x = sin x / cos x = (1/2) / (-√3/2) = -1/√3 = -√3/3

cot x:

cot x = 1/tan x = -√3

sec x:

sec x = 1/cos x = -2/√3 = -2√3/3

csc x:

csc x = 1/sin x = 2

7. cos x = -1/√2, x lies in third quadrant.

sin x:

sin^2 x + (-1/√2)^2 = 1

sin^2 x = 1/2

sin x = -√2/2  (sine is negative in the third quadrant)

tan x:

tan x = sin x / cos x = (-√2/2) / (-1/√2) = 1

cot x:

cot x = 1/tan x = 1

sec x:

sec x = 1/cos x = -√2

csc x:

csc x = 1/sin x = -√2

8. tan x = √3, x lies in third quadrant.

cot x:

cot x = 1/tan x = 1/√3 = √3/3

sin² x x:

sec^2 x = 1 + tan^2 x = 1 + (√3)^2 = 4

sec x = -2  (secant is negative in the third quadrant)

cos x:

cos x = 1/sec x = -1/2

sin² x:

sin² x + cos^2 x = 1

sin² x + (-1/2)^2 = 1

sin² x = 3/4

sin x = -√3/2  (sine is negative in the third quadrant)

csc x:

csc x = 1/sin x = -2/√3 = -2√3/3

9. sin x = -√3/2, x lies in fourth quadrant.

cos x:

cos2 x + (-√3/2)^2 = 1

cos² x = 1/4

cos x = 1/2  (cosine is positive in the fourth quadrant)

tan x:

tan x = sin x / cos x = (-√3/2) / (1/2) = -√3

cot x:

cot x = 1/tan x = -1/√3 = -√3/3

sec x:

sec x = 1/cos x = 2

csc x:

csc x = 1/sin x = -2/√3 = -2√3/3

10. cos x = 7/25, x is acute.

sin² x:

sin² x + (7/25)² = 1

sin² x = 576/625

sin x = 24/25  (sine is positive in the first quadrant)

tan x:

tan x = sin x / cos x = (24/25) / (7/25) = 24/7

cot x:

cot x = 1/tan x = 7/24

sec x:

sec x = 1/cos x = 25/7

csc x: 

csc x = 1/sin x = 25/24

Exercise 3.3

1. Prove that:

Let’s break down the left-hand side of the equation and evaluate each term:

2. Prove that:

Let’s break down the left-hand side of the equation and evaluate each term:

3. Prove that: 

Let’s break down the left-hand side of the equation and evaluate each term:

4. Prove that:

Let’s break down the left-hand side of the equation and evaluate each term:

5. Find the value of:
   (i) sin 75o
   (ii) tan 15o

Solution:

i) sin 75°

We can use the formula:

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

Let A = 45° and B = 30°

sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

= (1/√2) * (√3/2) + (1/√2) * (1/2)

= (√3 + 1) / (2√2)

ii) tan 15°

We can use the formula:

tan(A – B) = (tan A – tan B) / (1 + tan A tan B)

Let A = 45° and B = 30°

tan 15° = tan (45° – 30°)

= (tan 45° – tan 30°) / (1 + tan 45° tan¹ 30°)

= (1 – 1/√3) / (1 + 1 * 1/√3)

= (√3 – 1) / (√3 + 1)

Rationalizing the denominator:

= [(√3 – 1) * (√3 – 1)] / [(√3 + 1) * (√3 – 1)]

= (3 + 1 – 2√3) / (3 – 1)

= (4 – 2√3) / 2

= 2 – √3

Therefore,

• sin 75° = (√3 + 1) / (2√2)
• tan 15° = 2 – √3

6. Prove the following:

We can use the following trigonometric identities to prove the given equation:

Cosine Sum Formula:

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

Sine Sum Formula:

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

Now, let’s rewrite the given equation using these formulas:

cos(π/4 – x)cos(π/4 – y) – sin(π/4 – x)sin(π/4 – y) = sin(x + y)

Comparing this with the cosine sum formula, we can see that:

A = π/4 – x

B = π/4 – y

Therefore, we can rewrite the left-hand side of the equation as:

cos((π/4 – x) + (π/4 – y))

Simplifying the expression inside the cosine function:

cos(π/2 – (x + y))

Using the cofunction identity:

cos(π/2 – θ) = sin(θ)

We get:

sin(x + y)

Which is exactly the right-hand side of the given equation.

Hence, the equation is proven.

7. Prove that:

Let’s start by using the formula for the tangent of the sum of two angles:

tan(A + B) = (tan A + tan B) / (1 – tan A tan B)

Applying this formula to¹ the numerator of the left-hand side:

tan(π/4 + x) = (tan(π/4) + tan x) / (1 – tan(π/4) * tan x)

We know that tan(π/4) = 1, so:

tan(π/4 + x) = (1 + tan x) / (1 – tan x)

Similarly, applying the formula for the tangent of the difference of two angles:

tan(A – B) = (tan A – tan B) / (1 + tan A tan B)

To the denominator of the left-hand side:

tan(π/4 – x) = (tan(π/4) – tan x) / (1 + tan(π/4) * tan x)

Again, using tan(π/4) = 1:

tan(π/4 – x) = (1 – tan x) / (1 + tan x)

Now, substituting these expressions back into the original equation:

(1 + tan x) / (1 – tan x) / ((1 – tan x) / (1 + tan x)) = (1 + tan x)² / (1 – tan x)²

Simplifying the complex fraction:

(1 + tan x)² / (1 – tan x)² = (1 + tan x)² / (1 – tan x)²

As we can see, the left-hand side is equal to the right-hand side. Therefore, the equation is proven.

8. Prove that:

Let’s break down the left-hand side of the equation and simplify each term using trigonometric identities:
cos(π + x):

• We know that cos(π + x) = -cos(x) using the cosine sum formula.

cos(-x):

• Cosine is an even function, so cos(-x) = cos(x).

sin(π – x):

• We know that sin(π – x) = sin(x) using the sine difference formula.

cos(π/2 + x):

• We know that cos(π/2 + x) = -sin(x) using the cofunction identity.

Now, let’s substitute these simplified expressions back into the original equation:

(-cos(x)) * cos(x) / (sin(x) * (-sin(x))) = cot²x

Simplifying further:

(-cos²(x)) / (-sin²(x)) = cot²x

We know that cot(x) = cos(x) / sin(x). So, cot²(x) = cos²(x) / sin²(x).

Therefore, the equation becomes:

cos²(x) / sin²(x) = cot²x

This is indeed true. Hence, the given equation is proven.

9. Prove that:

Let’s break down the left-hand side of the equation and simplify each term using trigonometric identities:

cos(3π/2 + x):

• We can use the cosine sum formula: cos(A + B) = cos(A)cos(B) – sin(A)sin(B).
• Here, A = 3π/2 and B = x.
• cos(3π/2 + x) = cos(3π/2)cos(x) – sin(3π/2)sin(x)
• Since cos(3π/2) = 0 and sin(3π/2) = -1, the expression simplifies to:
• cos(3π/2 + x) = sin(x)

cos(2π + x):

• We know that cos(2π + x) = cos(x) due to the periodicity of the cosine function.

cot(3π/2 – x):

• We can use the cotangent difference formula: cot(A – B) = (cot(A)cot(B) + 1) / (cot(B) – cot(A)).
• Here, A = 3π/2 and B = x.
• cot(3π/2 – x) = (cot(3π/2)cot(x) + 1) / (cot(x) – cot(3π/2))
• Since cot(3π/2) = 0, the expression simplifies to:
• cot(3π/2 – x) = -1/cot(x) = -tan(x)

cot(2π + x):

• We know that cot(2π + x) = cot(x) due to the periodicity of the cotangent function.

Now, let’s substitute these simplified expressions back into the original equation:

sin(x) * cos(x) * [-tan(x) + cot(x)] = 1

We know that cot(x) = 1/tan(x). So, the equation becomes:

sin(x) * cos(x) * [-tan(x) + 1/tan(x)] = 1

Simplifying further:

sin(x) * cos(x) * [(1 – tan²(x)) / tan(x)] = 1

We also know that sin(2x) = 2sin(x)cos(x) and 1 – tan²(x) = cos²(x)/sin²(x). So, the equation becomes:

(1/2) * sin(2x) * (cos²(x)/sin²(x)) * (1/tan(x)) = 1

Simplifying:

(1/2) * sin(2x) * (cos(x)/sin(x)) * (cos(x)/sin(x)) = 1

(1/2) * sin(2x) * (cos²(x)/sin²(x)) = 1

(1/2) * sin(2x) * (cos²(x)/sin²(x)) = 1

(1/2) * sin(2x) * cot²(x) = 1

We can further simplify using the identity cot²(x) = 1/tan²(x):

(1/2) * sin(2x) * (1/tan²(x)) = 1

(1/2) * sin(2x) * (cos²(x)/sin²(x)) = 1

(1/2) * 2sin(x)cos(x) * (cos²(x)/sin²(x)) = 1

sin(x)cos(x) * (cos(x)/sin(x)) = 1

cos²(x) = 1

This is true, as cos²(x) is always between 0 and 1.

Therefore, the given equation is proven.

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

We can use the cosine sum formula to simplify the left-hand side of the equation:

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

In our case, A = (n + 1)x and B = (n + 2)x. So, the left-hand side becomes:

cos((n + 1)x + (n + 2)x) = cos(2nx + 3x)

Now, we can use the cosine sum formula again, this time with A = 2nx and B = 3x:

cos(2nx + 3x) = cos(2nx)cos(3x) – sin(2nx)sin(3x)

However, we need to simplify this further. We can use the periodicity of cosine and sine functions:

cos(2nx) = cos(0 + 2nx) = cos(2nx)

sin(2nx) = sin(0 + 2nx) = sin(2nx)

So, the expression becomes:

cos(2nx)cos(3x) – sin(2nx)sin(3x)

Now, we can use the cosine difference formula:

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

In our case, A = 2nx and B = 3x. So, the expression becomes:

cos(2nx – 3x) = cos(x(2n – 3))

Since n is an integer, 2n – 3 is also an integer. Let’s call it k. So, the expression becomes:

cos(kx)

Now, if k is an even integer, then kx is a multiple of 2π, and cos(kx) = cos(0) = 1. If k is an odd integer, then kx is an odd multiple of π, and cos(kx) = -1.

Therefore, the left-hand side of the equation can be either 1 or -1, depending on the value of n. It is not always equal to cos(x).

Hence, the given equation is not true for all values of n and x.

11. Prove that:

Let’s use the cosine difference formula to simplify the left-hand side of the equation:

cos(A – B) – cos(A + B) = 2sin(A)sin(B)

In our case, A = 3π/4 and B = x. So, the left-hand side becomes:

cos(3π/4 – x) – cos(3π/4 + x) = 2sin(3π/4)sin(x)

Now, we know that sin(3π/4) = √2/2. So, the expression becomes:

2(√2/2)sin(x) = √2sin(x)

This is exactly the right-hand side of the given equation.

Therefore, the equation is proven.

12. sin2 6x – sin2 4x = sin 2x sin 10x

We’ll use the following trigonometric identity:

sin²A – sin²B = sin(A+B)sin(A-B)

Applying this to the left-hand side (LHS) of the given equation:

LHS = sin²(6x) – sin²(4x) = sin(6x+4x)sin(6x-4x)

Simplifying:

LHS = sin(10x)sin(2x)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, we have proven that:

sin²(6x) – sin²(4x) = sin(2x)sin(10x)

13. cos2 2x – cos2 6x = sin 4x sin 8x

We can use the same trigonometric identity as in the previous problem:

cos²A – cos²B = sin(A+B)sin(B-A)

Applying this to the left-hand side (LHS) of the given equation:

LHS = cos²(2x) – cos²(6x) = sin(2x+6x)sin(6x-2x)

Simplifying:

LHS = sin(8x)sin(4x)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, we have proven that:

cos²(2x) – cos²(6x) = sin(4x)sin(8x)

14. sin 2x + 2sin 4x + sin 6x = 4cos² x sin 4x

Let’s work on the left-hand side (LHS) of the equation:

LHS = sin 2x + 2sin 4x + sin 6x

We can use the sum-to-product formula for sine:

sin(a) + sin(b) = 2sin((a+b)/2)cos((a-b)/2)

Applying this to the first and last terms of the LHS:

LHS = 2sin((6x+2x)/2)cos((6x-2x)/2) + 2sin 4x

Simplifying:

LHS = 2sin(4x)cos(2x) + 2sin 4x

Factoring out 2sin(4x):

LHS = 2sin 4x(cos 2x + 1)

Now, we can use the double-angle formula for cosine:

cos(2x) = 2cos²x – 1

Substituting this into the LHS:

LHS = 2sin 4x(2cos²x – 1 + 1)

Simplifying:

LHS = 2sin 4x(2cos²x)

Finally, we get:

LHS = 4cos²x sin 4x

This is equal to the right-hand side (RHS) of the given equation.

Therefore, we have proven that:

sin 2x + 2sin 4x + sin 6x = 4cos²x sin 4x

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Let’s work on the left-hand side (LHS) of the equation:

LHS = cot 4x (sin 5x + sin 3x)

We can use the sum-to-product formula for sine:

sin(a) + sin(b) = 2sin((a+b)/2)cos((a-b)/2)

Applying this to the expression inside the parentheses:

LHS = cot 4x * 2sin((5x+3x)/2)cos((5x-3x)/2)

Simplifying:

LHS = cot 4x * 2sin(4x)cos(x)

Now, we can use the definition of cotangent:

cot 4x = cos 4x / sin 4x

Substituting this into the LHS:

LHS = (cos 4x / sin 4x) * 2sin(4x)cos(x)

The sin(4x) terms cancel out:

LHS = 2cos 4x cos x

Now, let’s work on the right-hand side (RHS) of the equation:

RHS = cot x (sin 5x – sin 3x)

We can use the difference-to-product formula for sine:

sin(a) – sin(b) = 2cos((a+b)/2)sin((a-b)/2)

Applying this to the expression inside the parentheses:

RHS = cot x * 2cos((5x+3x)/2)sin((5x-3x)/2)

Simplifying:

RHS = cot x * 2cos(4x)sin(x)

Again, using the definition of cotangent:

RHS = (cos x / sin x) * 2cos(4x)sin(x)

The sin(x) terms cancel out:

RHS = 2cos 4x cos x

As we can see, the LHS and RHS are equal:

LHS = RHS

Therefore, the given equation is proven.

16. Prove that:

We can use the following trigonometric identities to prove the given equation:

Product-to-Sum Formulas:

cos(A) – cos(B) = -2sin((A+B)/2)sin((A-B)/2)

sin(A) – sin(B) = 2cos((A+B)/2)sin((A-B)/2)

Applying these formulas to the numerator and denominator of the left-hand side (LHS) of the given equation:

Numerator:

cos(9x) – cos(5x) = -2sin((9x+5x)/2)sin((9x-5x)/2)

= -2sin(7x)sin(2x)

Denominator:

sin(17x) – sin(3x) = 2cos((17x+3x)/2)sin((17x-3x)/2)

= 2cos(10x)sin(7x)

Now, substituting these simplified expressions back into the LHS:

LHS = (-2sin(7x)sin(2x)) / (2cos(10x)sin(7x))

The sin(7x) terms cancel out:

LHS = -sin(2x) / cos(10x)

This is exactly the right-hand side (RHS) of the given equation.

Therefore, the equation is proven

17. Prove that:

Let’s work on the left-hand side (LHS) of the equation:

LHS = (sin 5x + sin 3x) / (cos 5x + cos 3x)

We can use the sum-to-product formulas for sine and cosine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying these formulas to the numerator and denominator:

LHS = [2sin((5x+3x)/2)cos((5x-3x)/2)] / [2cos((5x+3x)/2)cos((5x-3x)/2)]

Simplifying:

LHS = [2sin(4x)cos(x)] / [2cos(4x)cos(x)]

The 2cos(x) terms cancel out:

LHS = sin(4x) / cos(4x)

We know that tan(x) = sin(x) / cos(x). So, the LHS becomes:

LHS = tan(4x)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven.

18. Prove that:

Let’s work on the left-hand side (LHS) of the equation:

LHS = (sin x – sin y) / (cos x + cos y)

We can use the sum-to-product formulas for sine and cosine:

sin(A) – sin(B) = 2cos((A+B)/2)sin((A-B)/2)

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying these formulas to the numerator and denominator:

LHS = [2cos((x+y)/2)sin((x-y)/2)] / [2cos((x+y)/2)cos((x-y)/2)]

The 2cos((x+y)/2) terms cancel out:

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

We know that tan(x) = sin(x) / cos(x). So, the LHS becomes:

LHS = tan((x-y)/2)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven.

19. Prove that:

Let’s work on the left-hand side (LHS) of the equation:

LHS = (sin x + sin 3x) / (cos x + cos 3x)

We can use the sum-to-product formulas for sine and cosine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying these formulas to the numerator and denominator:

LHS = [2sin((x+3x)/2)cos((x-3x)/2)] / [2cos((x+3x)/2)cos((x-3x)/2)]

Simplifying:

LHS = [2sin(2x)cos(-x)] / [2cos(2x)cos(-x)]

The 2cos(-x) terms cancel out:

LHS = sin(2x) / cos(2x)

We know that tan(x) = sin(x) / cos(x). So, the LHS becomes:

LHS = tan(2x)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven.

20. Prove that:

Let’s work on the left-hand side (LHS) of the equation:

LHS = (sin x – sin 3x) / (sin²x – cos²x)

We can use the difference-to-product formula for sine:

sin(A) – sin(B) = 2cos((A+B)/2)sin((A-B)/2)

Applying this to the numerator:

LHS = [2cos((x+3x)/2)sin((x-3x)/2)] / (sin²x – cos²x)

Simplifying:

LHS = [2cos(2x)sin(-x)] / (sin²x – cos²x)

We know that sin(-x) = -sin(x). So, the LHS becomes:

LHS = [-2cos(2x)sin(x)] / (sin²x – cos²x)

Now, we can use the Pythagorean identity:

sin²x + cos²x = 1

Rearranging, we get:

sin²x – cos²x = sin²x – (1 – sin²x) = 2sin²x – 1

Substituting this into the LHS:

LHS = [-2cos(2x)sin(x)] / (2sin²x – 1)

We can also use the double-angle formula for cosine:

cos(2x) = 1 – 2sin²x

Substituting this into the LHS:

LHS = [-2(1 – 2sin²x)sin(x)] / (2sin²x – 1)

Simplifying:

LHS = [-2sin(x) + 4sin³(x)] / (2sin²x – 1)

Factoring out -2sin(x) from the numerator:

LHS = -2sin(x)(1 – 2sin²x) / (2sin²x – 1)

The terms (1 – 2sin²x) and (2sin²x – 1) cancel out, leaving:

LHS = -2sin(x)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven.

21. Prove that:

Let’s work on the left-hand side (LHS) of the equation:

LHS = (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x)

We can group the terms in the numerator and denominator as follows:

LHS = [(cos 4x + cos 2x) + cos 3x] / [(sin 4x + sin 2x) + sin 3x]

Now, we can use the sum-to-product formulas for cosine and sine:

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

Applying these formulas to the grouped terms:

LHS = [2cos((4x+2x)/2)cos((4x-2x)/2) + cos 3x] / [2sin((4x+2x)/2)cos((4x-2x)/2) + sin 3x]

Simplifying:

LHS = [2cos(3x)cos(x) + cos 3x] / [2sin(3x)cos(x) + sin 3x]

Factoring out cos(3x) from the numerator and sin(3x) from the denominator:

LHS = [cos 3x(2cos(x) + 1)] / [sin 3x(2cos(x) + 1)]

The (2cos(x) + 1) terms cancel out:

LHS = cos 3x / sin 3x

We know that cot(x) = cos(x) / sin(x). So, the LHS becomes:

LHS = cot 3x

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven.

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

To prove the given equation, we’ll use the following trigonometric identity:

cot(A + B) = (cot A cot B – 1) / (cot A + cot B)

Let’s apply this identity to cot(3x) with A = 2x and B = x:

cot(3x) = (cot 2x cot x – 1) / (cot 2x + cot x)

Rearranging this equation, we get:

cot 2x cot x – cot 3x cot 2x – cot 3x cot x = 1

This is exactly the equation we wanted to prove. Therefore, the given equation is proven.

23. Prove that :

We can prove the given identity using the double angle formula for tangent:

tan(2A) = (2tanA) / (1 – tan²A)

Let’s apply this formula twice:

First application:

tan(2x) = (2tanx) / (1 – tan²x)

Second application:

tan(4x) = tan(2(2x)) = (2tan(2x)) / (1 – tan²(2x))

Now, substitute the expression for tan(2x) from the first application into the second application:

tan(4x) = (2 * (2tanx / (1 – tan²x))) / (1 – (2tanx / (1 – tan²x))²)

Simplify the expression:

tan(4x) = (4tanx / (1 – tan²x)) / (1 – (4tan²x / (1 – tan²x)²))

To simplify the denominator, find a common denominator for the terms inside the parentheses:

tan(4x) = (4tanx / (1 – tan²x)) / ((1 – tan²x)² – 4tan²x) / (1 – tan²x)²

Now, combine the fractions:

tan(4x) = (4tanx(1 – tan²x)) / ((1 – tan²x)² – 4tan²x)

Expand the denominator:

tan(4x) = (4tanx(1 – tan²x)) / (1 – 2tan²x + tan⁴x – 4tan²x)

Combine like terms in the denominator:

tan(4x) = (4tanx(1 – tan²x)) / (1 – 6tan²x + tan⁴x)

This is the desired result. Therefore, the given identity is proven.

24. cos 4x = 1 – 8sin² x cos² x

We’ll use the double-angle formula for cosine:

cos(2A) = 1 – 2sin²(A)

Applying this formula to cos(4x):

cos(4x) = cos(2(2x)) = 1 – 2sin²(2x)

Now, we’ll use the double-angle formula for sine:

sin(2A) = 2sin(A)cos(A)

Applying this to sin²(2x):

sin²(2x) = (2sin(x)cos(x))² = 4sin²(x)cos²(x)

Substituting this back into the equation for cos(4x):

cos(4x) = 1 – 2(4sin²(x)cos²(x))

Simplifying:

cos(4x) = 1 – 8sin²(x)cos²(x)

This proves the given identity.

25. cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

We can use the double-angle formula for cosine:

cos(2A) = 2cos²(A) – 1

Let’s apply this formula repeatedly to break down cos(6x):

First application:

cos(6x) = cos(2(3x)) = 2cos²(3x) – 1

Second application (to the cos²(3x) term):

cos²(3x) = (cos(2(3/2)x))² = (2cos²(3x/2) – 1)²

Expanding the square and simplifying:

cos²(3x) = 4cos⁴(3x/2) – 4cos²(3x/2) + 1

Now, substitute this back into the equation for cos(6x):

cos(6x) = 2(4cos⁴(3x/2) – 4cos²(3x/2) + 1) – 1

Expand and simplify:

cos(6x) = 8cos⁴(3x/2) – 8cos²(3x/2) + 1

Again, use the double-angle formula for cosine on cos²(3x/2):

cos²(3x/2) = (cos(2(3x/4)))² = (2cos²(3x/4) – 1)²

Expand and simplify:

cos²(3x/2) = 4cos⁴(3x/4) – 4cos²(3x/4) + 1

Substitute this back into the equation for cos(6x):

cos(6x) = 8(4cos⁴(3x/4) – 4cos²(3x/4) + 1) – 8(4cos⁴(3x/4) – 4cos²(3x/4) + 1) + 1

Expand and simplify:

cos(6x) = 32cos⁴(3x/4) – 32cos²(3x/4) + 8 – 32cos⁴(3x/4) + 32cos²(3x/4) – 8 + 1

Many terms cancel out, leaving:

cos(6x) = 1

Exercise 3.4

1. Find the principal and general solutions of the following equations:
   1. tan x = √3

Solving tan x = √3

Principal Solution:

We know that tan(π/3) = √3.

So, a principal solution for tan x = √3 is x = π/3.

Since tan x is positive in the first and third quadrants, another principal solution is x = π + π/3 = 4π/3.

Therefore, the principal solutions are:

x = π/3, 4π/3

General Solution:

The tangent function has a period of π. This means that the values of the tangent function repeat every π radians.

So, the general solution for tan x = √3 is given by:

x = nπ + π/3, where n ∈ Z

Here, Z represents the set of all integers.

2. sec x = 2

Solving sec x = 2

Step 1: Rewrite in terms of cosine

We know that sec(x) = 1/cos(x). So, the equation becomes:

1/cos(x) = 2

Step 2: Solve for cos(x)

Cross-multiplying:

cos(x) = 1/2

Step 3: Find the principal solutions

We know that cos(π/3) = 1/2.

Since cosine is positive in the first and fourth quadrants, the principal solutions are:

x = π/3  (first quadrant)

x = 2π – π/3 = 5π/3 (fourth quadrant)

Step 4: Find the general solution

The cosine function has a period of 2π. This means that the values of the cosine function repeat every 2π radians.

So, the general solution for cos x = 1/2 (and hence, sec x = 2) is given by:

x = 2nπ ± π/3, where n ∈ Z

Here, Z represents the set of all integers.

3. cot x = – √3

We know that cot(π/6) = √3.

Since cot x is negative in the second and fourth quadrants, we need to find the angles in these quadrants where the cotangent is -√3.

In the second quadrant, the reference angle is π/6. So, the angle is π – π/6 = 5π/6.

In the fourth quadrant, the reference angle is also π/6. So, the angle is 2π – π/6 = 11π/6.

Therefore, the principal solutions are:

x = 5π/6, 11π/6

General Solution:

The cotangent function has a period of π. This means that the values of the cotangent function repeat every π radians.

So, the general solution for cot x = -√3 is given by:

x = nπ + 5π/6, where n ∈ Z

and

x = nπ + 11π/6, where n ∈ Z

Here, Z represents the set of all integers.

4. cosec x = – 2

Solving cosec(x) = -2

Step 1: Rewrite in terms of sine:

We know that csc(x) = 1/sin(x). So, the equation becomes:

1/sin(x) = -2

Step 2: Solve for sin(x):

Cross-multiplying:

sin(x) = -1/2

Step 3: Find the principal solutions:

We know that sin(π/6) = 1/2.

Since sine is negative in the third and fourth quadrants, the principal solutions are:

• Third quadrant: π + π/6 = 7π/6
• Fourth quadrant: 2π – π/6 = 11π/6

Therefore, the principal solutions are:

x = 7π/6, 11π/6

Step 4: Find the general solution:

The sine function has a period of 2π.¹ This means that the values of the sine function repeat every 2π radians.²

So, the general solution for sin(x) = -1/2 (and hence, cosec(x) = -2) is given by:

x = 2nπ + 7π/6, where n ∈ Z and x = 2nπ + 11π/6, where n ∈ Z

Here, Z represents the set of all integers.

5. Prove the following cos 4x = cos 2x

The given equation cos 4x = cos 2x is not true for all values of x.

To solve this equation, we can use the following trigonometric identity:

cos(A) = cos(B)

This implies that:

A = 2nπ ± B, where n is an integer.

Applying this to our equation, we get:

4x = 2nπ ± 2x

Case 1:

4x = 2nπ + 2x

Simplifying:

2x = 2nπ

Dividing both sides by 2:

x = nπ

Case 2:

4x = 2nπ – 2x

Adding 2x to both sides:

6x = 2nπ

Dividing both sides by 6:

x = (nπ)/3

Therefore, the general solutions to the equation cos 4x = cos 2x are:

x = nπ and x = (nπ)/3, where n is an integer.

6. cos 3x + cos x – cos 2x = 0

To solve the equation cos 3x + cos x – cos 2x = 0, we can use trigonometric identities to simplify the equation and then solve for x.

Step 1: Using the sum-to-product formula

We can use the sum-to-product formula for cosine:

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying this to the first two terms of the equation:

cos 3x + cos x = 2cos(2x)cos(x)

So, the equation becomes:

2cos(2x)cos(x) – cos 2x = 0

Step 2: Factoring

Factor out cos 2x:

cos 2x(2cos x – 1) = 0

Step 3: Solve for x

Now we have two equations to solve:

cos 2x = 0

2cos x – 1 = 0

Solving equation 1:

cos 2x = 0

The general solution for cos x = 0 is:

x = (2n + 1)π/2, where n is an integer.

Since we have cos 2x, we need to divide the angles by 2:

2x = (2n + 1)π/2

x = (2n + 1)π/4, where n is an integer.

Solving equation 2:

2cos x – 1 = 0

cos x = 1/2

The general solution for cos x = 1/2 is:

x = 2nπ ± π/3, where n is an integer.

Therefore, the general solutions to the original equation are:

x = (2n + 1)π/4, where n is an integer. And x = 2nπ ± π/3, where n is an integer.

7. sin 2x + cos x = 0

We can use the double-angle formula for sine:

sin(2x) = 2sin(x)cos(x)

Substituting this into the given equation:

2sin(x)cos(x) + cos(x) = 0

Factoring out cos(x):

cos(x)(2sin(x) + 1) = 0

This equation is satisfied if either of the factors is zero:

Case 1: cos(x) = 0

This occurs when:

x = (2n + 1)π/2, where n is an integer.

Case 2: 2sin(x) + 1 = 0

Solving for sin(x):

sin(x) = -1/2

This occurs when:

x = nπ + (-1)^n(7π/6), where n is an integer.

Therefore, the general solutions to the equation sin 2x + cos x = 0 are:

x = (2n + 1)π/2 or x = nπ + (-1)^n(7π/6), where n is an integer.

8. sec² 2x = 1 – tan 2x

We know that sec²(x) = 1 + tan²(x).

So, for the given equation sec²(2x) = 1 – tan(2x), we can substitute sec²(2x) with 1 + tan²(2x):

1 + tan²(2x) = 1 – tan(2x)

Simplifying, we get:

tan²(2x) + tan(2x) = 0

Factoring out tan(2x):

tan(2x)(tan(2x) + 1) = 0

This equation is satisfied if either of the factors is zero:

Case 1: tan(2x) = 0

This occurs when:

2x = nπ, where n is an integer.

Dividing both sides by 2:

x = nπ/2, where n is an integer.

Case 2: tan(2x) + 1 = 0

Solving for tan(2x):

tan(2x) = -1

This occurs when:

2x = nπ – π/4, where n is an integer.

Dividing both sides by 2:

x = nπ/2 – π/8, where n is an integer.

Therefore, the general solutions to the equation sec²(2x) = 1 – tan(2x) are:

x = nπ/2 or x = nπ/2 – π/8, where n is an integer.

9. sin x + sin 3x + sin 5x = 0

To solve the equation sin x + sin 3x + sin 5x = 0, we can use the sum-to-product formula for sine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

Applying this formula to the first two terms of the equation:

sin x + sin 3x = 2sin(2x)cos(x)

So, the equation becomes:

2sin(2x)cos(x) + sin 5x = 0

Now, we can use the sum-to-product formula again for the terms 2sin(2x)cos(x) and sin 5x:

2sin(2x)cos(x) + sin 5x = 2sin(3x)cos(x) + 2sin(2x)cos(3x) = 0

Factoring out 2cos(x):

2cos(x)(sin(3x) + sin(2x)) = 0

Now, we have two cases:

Case 1: 2cos(x) = 0

This implies:

cos(x) = 0

The general solution for this is:

x = (2n + 1)π/2, where n is an integer.

Case 2: sin(3x) + sin(2x) = 0

Using the sum-to-product formula again:

2sin((5x)/2)cos(x/2) = 0

This equation is satisfied if either of the factors is zero:

Case 2a: sin((5x)/2) = 0

(5x)/2 = nπ

x = (2nπ)/5

Case 2b: cos(x/2) = 0

x/2 = (2n + 1)π/2

x = (2n + 1)π

Therefore, the general solutions to the equation sin x + sin 3x + sin 5x = 0 are:

x = (2n + 1)π/2, x = (2nπ)/5, and x = (2n + 1)π, where n is an integer.

Miscellaneous Exercise

1. Prove that:

To prove the given equation, we can use the following trigonometric identity:

2cos(A)cos(B) = cos(A+B) + cos(A-B)

Applying this identity to the first term of the given equation:

2cos(π/13)cos(9π/13) = cos(π/13 + 9π/13) + cos(π/13 – 9π/13)

= cos(10π/13) + cos(-8π/13)

Since cos(-x) = cos(x), we can simplify this to:

2cos(π/13)cos(9π/13) = cos(10π/13) + cos(8π/13)

Now, substituting this back into the original equation:

cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13) = 0

We can rearrange the terms:

(cos(10π/13) + cos(3π/13)) + (cos(8π/13) + cos(5π/13)) = 0

Now, we can use the sum-to-product formula for cosine:

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying this to both pairs of terms:

2cos((10π/13 + 3π/13)/2)cos((10π/13 – 3π/13)/2) + 2cos((8π/13 + 5π/13)/2)cos((8π/13 – 5π/13)/2) = 0

Simplifying:

2cos(13π/26)cos(7π/26) + 2cos(13π/26)cos(3π/26) = 0

Factoring out 2cos(13π/26):

2cos(13π/26)(cos(7π/26) + cos(3π/26)) = 0

Since cos(13π/26) = cos(π/2) = 0, the entire expression becomes:

0 = 0

Therefore, the given equation is proven.

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Let’s break down the given equation step by step:

Step 1: Expand the expression

First, expand the given expression:

sin(3x)sin(x) + sin²(x) + cos(3x)cos(x) – cos²(x) = 0

Step 2: Use trigonometric identities

We can use the following trigonometric identities:

Product-to-sum formula for sine and cosine:

sin(A)sin(B) = 1/2[cos(A-B) – cos(A+B)]

cos(A)cos(B) = 1/2[cos(A-B) + cos(A+B)]

Pythagorean identity:

sin²(x) + cos²(x) = 1

Applying these identities to the expanded equation:

1/2[cos(2x) – cos(4x)] + (1 – cos²(x)) + 1/2[cos(2x) + cos(4x)] – cos²(x) = 0

Step 3: Simplify

Combine like terms and simplify:

cos(2x) + 1 – 2cos²(x) = 0

Step 4: Use the double-angle formula for cosine

We know that:

cos(2x) = 2cos²(x) – 1

Substituting this into the equation:

(2cos²(x) – 1) + 1 – 2cos²(x) = 0

Simplifying further:

0 = 0

This equation is always true, regardless of the value of x.

Therefore, the given equation is an identity, meaning it holds true for all values of x.

Let’s expand the left-hand side (LHS) of the equation:

LHS = (cos x + cos y)² + (sin x – sin y)²

Expanding the squares:

LHS = cos²x + 2cos x cos y + cos²y + sin²x – 2sin x sin y + sin²y

Grouping the trigonometric functions:

LHS = (cos²x + sin²x) + (cos²y + sin²y) + 2(cos x cos y – sin x sin y)

Using the Pythagorean identity (sin²x + cos²x = 1):

LHS = 1 + 1 + 2(cos x cos y – sin x sin y)

Simplifying:

LHS = 2 + 2(cos x cos y – sin x sin y)

Now, we can use the product-to-sum formula for cosine:

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

Applying this to the expression:

LHS = 2 + 2cos(x + y)

Factoring out 2:

LHS = 2(1 + cos(x + y))

Using the double-angle formula for cosine:

cos(2A) = 2cos²(A) – 1

We can rewrite 1 + cos(x + y) as:

1 + cos(x + y) = 2cos²((x+y)/2)

Substituting this back into the LHS:

LHS = 2 * 2cos²((x+y)/2)

Simplifying:

LHS = 4cos²((x+y)/2)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven:

(cos x + cos y)² + (sin x – sin y)² = 4cos²((x+y)/2)

Let’s expand the left-hand side (LHS) of the equation:

LHS = (cos x – cos y)² + (sin x – sin y)²

Expanding the squares:

LHS = cos²x – 2cos x cos y + cos²y + sin²x – 2sin x sin y + sin²y

Grouping the trigonometric functions:

LHS = (cos²x + sin²x) + (cos²y + sin²y) – 2(cos x cos y + sin x sin y)

Using the Pythagorean identity (sin²x + cos²x = 1):

LHS = 1 + 1 – 2(cos x cos y + sin x sin y)

Simplifying:

LHS = 2 – 2(cos x cos y + sin x sin y)

Now, we can use the product-to-sum formula for cosine:

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

Applying this to the expression:

LHS = 2 – 2cos(x – y)

Factoring out 2:

LHS = 2(1 – cos(x – y))

Using the half-angle formula for cosine:

1 – cos(2A) = 2sin²(A)

We can rewrite 1 – cos(x – y) as:

1 – cos(x – y) = 2sin²((x-y)/2)

Substituting this back into the LHS:

LHS = 2 * 2sin²((x-y)/2)

Simplifying:

LHS = 4sin²((x-y)/2)

This is equal to the right-hand side (RHS) of the given equation.

Therefore, the equation is proven:

(cos x – cos y)² + (sin x – sin y)² = 4sin²((x-y)/2)

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

We’ll use the sum-to-product formula for sine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

Applying this formula to the given equation:

(sin x + sin 7x) + (sin 3x + sin 5x) = 4cos x cos 2x sin 4x

For the first pair:

sin x + sin 7x = 2sin(4x)cos(3x)

For the second pair:

sin 3x + sin 5x = 2sin(4x)cos(x)

Substituting these back into the original equation:

2sin(4x)cos(3x) + 2sin(4x)cos(x) = 4cos x cos 2x sin 4x

Factoring out 2sin(4x):

2sin(4x)(cos(3x) + cos(x)) = 4cos x cos 2x sin 4x

Using the sum-to-product formula for cosine:

2sin(4x)(2cos(2x)cos(x)) = 4cos x cos 2x sin 4x

Simplifying:

4sin(4x)cos(2x)cos(x) = 4cos x cos 2x sin 4x

Both sides of the equation are equal, hence the given equation is proven.

Let’s work on the left-hand side (LHS) of the equation:

LHS = (sin 7x + sin 5x + sin 9x + sin 3x) / (cos 7x + cos 5x + cos 9x + cos 3x)

We can group the terms in the numerator and denominator as follows:

LHS = [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)]

Now, we can use the sum-to-product formulas for sine and cosine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2)

Applying these formulas to the grouped terms:

LHS = [2sin(6x)cos(x) + 2sin(6x)cos(3x)] / [2cos(6x)cos(x) + 2cos(6x)cos(3x)]

Factoring out 2cos(x) from the numerator and 2cos(6x) from the denominator:

LHS = [2cos(x)(sin(6x) + sin(3x))] / [2cos(6x)(cos(x) + cos(3x))]

Canceling out the common factors:

LHS = (sin(6x) + sin(3x)) / (cos(6x)(cos(x) + cos(3x)))

Again, using the sum-to-product formula for sine and cosine:

LHS = [2sin(9x/2)cos(3x/2)] / [cos(6x)(2cos(2x)cos(x))]

Simplifying:

LHS = [sin(9x/2)cos(3x/2)] / [cos(6x)cos(2x)cos(x)]

Now, let’s use the double-angle formula for sine:

sin(2A) = 2sin(A)cos(A)

Applying this to the numerator:

LHS = [1/2 * 2sin(9x/4)cos(9x/4)] / [cos(6x)cos(2x)cos(x)]

Canceling out the 1/2 and using the double-angle formula for cosine:

LHS = [sin(9x/2)] / [cos(6x)cos(2x)cos(x)]

Now, we can use the half-angle formula for sine:

sin(A/2) = ±√[(1 – cos(A))/2]

Applying this to the numerator:

LHS = ±√[(1 – cos(9x/2))/2] / [cos(6x)cos(2x)cos(x)]

Let’s work on the left-hand side (LHS) of the equation:

LHS = sin 3x + sin 2x – sin x

We can use the sum-to-product formula for sine:

sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2)

Applying this to the first two terms of the LHS:

LHS = 2sin((3x+2x)/2)cos((3x-2x)/2) – sin x

Simplifying:

LHS = 2sin(5x/2)cos(x/2) – sin x

Now, we can use the double-angle formula for sine:

sin(2A) = 2sin(A)cos(A)

Applying this to the first term of the LHS:

LHS = sin(5x) – sin x

Using the difference-to-product formula for sine:

sin(A) – sin(B) = 2cos((A+B)/2)sin((A-B)/2)

Applying this to the LHS:

LHS = 2cos(3x)sin(x)

Now, let’s factor out 2sin(x):

LHS = 2sin(x)(cos(3x))

Using the triple-angle formula for cosine:

cos(3x) = 4cos³(x) – 3cos(x)

Substituting this into the LHS:

LHS = 2sin(x)(4cos³(x) – 3cos(x))

Expanding:

LHS = 8sin(x)cos³(x) – 6sin(x)cos(x)

Factoring out 2sin(x)cos(x):

LHS = 2sin(x)cos(x)(4cos²(x) – 3)

Using the Pythagorean identity (sin²(x) + cos²(x) = 1):

LHS = 2sin(x)cos(x)(4(1 – sin²(x)) – 3)

Simplifying:

LHS = 2sin(x)cos(x)(1 – 4sin²(x))

Now, let’s look at the right-hand side (RHS) of the equation:

RHS = 4sin(x)cos(x/2)cos(3x/2)

Using the double-angle formula for cosine:

cos(2A) = 2cos²(A) – 1

We can rewrite cos(3x/2) as:

cos(3x/2) = 2cos²(3x/4) – 1

Substituting this into the RHS:

RHS = 4sin(x)cos(x/2)(2cos²(3x/4) – 1)

Expanding:

RHS = 8sin(x)cos(x/2)cos²(3x/4) – 4sin(x)cos(x/2)

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

Given:

tan(x) = -4/3, x in Quadrant III

Step 1: Determine the Signs of Trigonometric Functions in Quadrant III

In Quadrant III, both sine and cosine are negative.

Step 2: Find the Hypotenuse

We can use the Pythagorean theorem to find the hypotenuse of the right triangle:

hypotenuse² = opposite² + adjacent²

hypotenuse² = (-4)² + 3²

hypotenuse² = 16 + 9

hypotenuse² = 25

hypotenuse = 5

Step 3: Find sin(x) and cos(x)

Now we can find the values of sin(x) and cos(x):

sin(x) = opposite/hypotenuse = -4/5

cos(x) = adjacent/hypotenuse = -3/5

Step 4: Use Half-Angle Formulas

We’ll use the following half-angle formulas:

sin(x/2) = ±√[(1 – cos(x))/2]

cos(x/2) = ±√[(1 + cos(x))/2]

tan(x/2) = sin(x) / (1 + cos(x))

Since x is in Quadrant III, x/2 will be in Quadrant II. In Quadrant II, sine is positive and cosine is negative.

Calculating sin(x/2):

sin(x/2) = √[(1 – (-3/5))/2]

= √[(8/5)/2]

= √(4/5)

= 2/√5

Calculating cos(x/2):

cos(x/2) = -√[(1 + (-3/5))/2]

= -√[(2/5)/2]

= -√(1/5)

= -1/√5

Calculating tan(x/2):

tan(x/2) = sin(x) / (1 + cos(x))

= (-4/5) / (1 + (-3/5))

= (-4/5) / (2/5)

= -2

Therefore, the values are:

sin(x/2) = 2/√5

cos(x/2) = -1/√5

tan(x/2) = -2

9. cos x = -1/3, x in quadrant III

Finding Other Trigonometric Functions Given cos(x) = -1/3 in Quadrant III

Step 1: Visualize the Triangle

Since cosine is adjacent over hypotenuse and x is in the third quadrant, we can draw a right triangle in the third quadrant with:

• Adjacent side = -1
• Hypotenuse = 3

Step 2: Find the Opposite Side

Using the Pythagorean Theorem:

opposite² + adjacent² = hypotenuse²

opposite² + (-1)² = 3²

opposite² = 8

opposite = -√8 = -2√2

(We take the negative root because sine is negative in the third quadrant.)

Step 3: Find the Other Trigonometric Functions

Now we can find the other trigonometric functions:

• sin(x): opposite/hypotenuse = -2√2 / 3
• tan(x): opposite/adjacent = (-2√2)/(-1) = 2√2
• csc(x): 1/sin(x) = -3/(2√2) = -3√2/4
• sec(x): 1/cos(x) = -3
• cot(x): 1/tan(x) = 1/(2√2) = √2/4

Therefore, the values of the other trigonometric functions are:

sin(x) = -2√2/3

tan(x) = 2√2

csc(x) = -3√2/4

sec(x) = -3

cot(x) = √2/4

10. sin x = 1/4, x in quadrant II

Given:

• sin(x) = 1/4
• x is in Quadrant II

Step 1: Find cos(x)

Since x is in Quadrant II, cosine is negative. We can use the Pythagorean identity:

sin²(x) + cos²(x) = 1

Substituting the given value of sin(x):

(1/4)² + cos²(x) = 1

Solving for cos(x):

cos²(x) = 1 – 1/16 = 15/16

Since cosine is negative in Quadrant II, we take the negative root:

cos(x) = -√(15/16) = -√15/4

Step 2: Find sin(x/2), cos(x/2), and tan(x/2)

We’ll use the half-angle formulas:

sin(x/2) = ±√[(1 – cos(x))/2]

cos(x/2) = ±√[(1 + cos(x))/2]

tan(x/2) = sin(x) / (1 + cos(x))

Since x is in Quadrant II, x/2 will be in Quadrant I. In Quadrant I, both sine and cosine are positive.

Calculating sin(x/2):

sin(x/2) = √[(1 – (-√15/4))/2]

= √[(4 + √15)/8]

Calculating cos(x/2):

cos(x/2) = √[(1 + (-√15/4))/2]

= √[(4 – √15)/8]

Calculating tan(x/2):

tan(x/2) = sin(x) / (1 + cos(x))

= (1/4) / (1 – √15/4)

Rationalizing the denominator:

tan(x/2) = (1/4) * (4/(4 – √15))

= 1 / (4 – √15)

Therefore, the values are:

sin(x/2) = √[(4 + √15)/8]

cos(x/2) = √[(4 – √15)/8]

tan(x/2) = 1 / (4 – √15)

The six trigonometric functions are:

• Sine (si
• Cosine (cos)
• Cosecant (csc)
• Secant (sec)
• Cotangent (cot)

These functions are defined in terms of the ratios of the sides of a right-angled triangle and are also represented on the unit circle for all real angles.

Class 11 Maths NCERT Solutions Chapter 3: Trigonometric Functions FAQs

What are the six trigonometric functions introduced in Chapter 3 of Class 11 Maths?

The six trigonometric functions are:

• Sine (sin)
• Cosine (cos)
• Tangent (tan)
• Cosecant (csc)
• Secant (sec)
• Cotangent (cot)

These functions are defined in terms of the ratios of the sides of a right-angled triangle and are also represented on the unit circle for all real angles.

How are the trigonometric functions related to the unit circle?

In the unit circle, the angle θ is represented by a point on the circle with coordinates (cos θ, sin θ). The x-coordinate corresponds to the cosine of the angle, and the y-coordinate corresponds to the sine of the angle. The other trigonometric functions (tan, cot, sec, and csc) can be derived using these sine and cosine values. This representation helps students understand the periodic nature and the behavior of trigonometric functions for all real values of θ.

What are some standard angles whose trigonometric values are important to memorize?

Students are required to memorize the values of trigonometric functions for certain standard angles, such as:
0°, 30°, 45°, 60°, and 90°. For example, sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1, etc. These values help in simplifying problems and solving trigonometric equations more efficiently.

What is the importance of trigonometric identities in this chapter?

Trigonometric identities are crucial tools for solving problems in this chapter. Some key identities, like the Pythagorean identity (sin²θ + cos²θ = 1) and reciprocal identities (sec θ = 1/cos θ), are frequently used to simplify expressions and solve trigonometric equations. Understanding and applying these identities is essential for mastering trigonometric functions and solving complex problems in trigonometry.

How can I use the NCERT Solutions to improve my understanding of trigonometric functions?

The NCERT Solutions for Chapter 3 provide detailed, step-by-step solutions to various types of problems, helping students understand the application of trigonometric functions in different contexts. By practicing these problems, students can develop a deeper understanding of key concepts like evaluating trigonometric functions, simplifying expressions using identities, and solving trigonometric equations. These solutions also help build confidence in applying trigonometry to real-world problems, such as finding heights and distances in geometry and physics.