**Chapter 3.1: Introduction to Trigonometry**

**Definition:**The study of relationships between the sides and angles of a triangle.**Trigonometric Ratios:**Sine, cosine, tangent, cotangent, secant, cosecant.**Trigonometric Identities:**Fundamental relationships between trigonometric functions.

**Chapter 3.2: Trigonometric Functions**

**Domain and Range:**The set of possible input and output values for trigonometric functions.**Graphs of Trigonometric Functions:**Visual representation of the behavior of trigonometric functions.**Periodic Functions:**Functions that repeat their values after a fixed interval (period).

**Chapter 3.3: Trigonometric Identities**

**Sum and Difference Formulas:**Formulas for trigonometric functions of the sum or difference of angles.**Multiple Angle Formulas:**Formulas for trigonometric functions of multiples of an angle.

**Chapter 3.4: Trigonometric Equations**

**Solving Trigonometric Equations:**Finding the values of x that satisfy a given trigonometric equation.**General Solution:**Expressing the solution of a trigonometric equation in terms of the principal solution and the period of the function.

**Key Concepts:**

- Trigonometric ratios and their relationships
- Trigonometric functions and their graphs
- Trigonometric identities
- Solving trigonometric equations

This chapter provides a foundational understanding of trigonometry, which is essential for various mathematical and scientific applications.

**Exercise 3.1**

**1. Find the radian measures corresponding to the following degree measures: **

**(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°**

**Ans : **

o convert from degrees to radians, we multiply by π/180.

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

(ii) -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. **

**Ans : **

To convert from radians to degrees, we multiply by 180/π.

Given: π = 22/7

(i) 11π/16 = 11 * (22/7) * (180/π) = 11 * 22 * 180 / 7 = 5820/7 ≈ 831.43°

(ii) -4 = -4 * (180/π) ≈ -4 * 180 * 7/22 ≈ -229.09°

(iii) 5π/3 = 5 * (22/7) * (180/π) = 5 * 22 * 180 / 7 = 2828.57° ≈ 300°

(iv) 7π/6 = 7 * (22/7) * (180/π) = 7 * 22 * 180 / 7 = 2520° ≈ 210°

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

**Ans : **

**Given:**

- The wheel makes 360 revolutions in one minute.

**Steps:**

**Convert minutes to seconds:**- 1 minute = 60 seconds

**Calculate revolutions per second:**- Revolutions per second = 360 revolutions / 60 seconds = 6 revolutions/second

**Convert one revolution to radians:**- 1 revolution = 2π radians

**Calculate radians per second:**- Radians per second = 6 revolutions/second * 2π radians/revolution =
**12π radians/second**

- Radians per second = 6 revolutions/second * 2π radians/revolution =

**Therefore, the wheel turns through 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).**

**Ans : **

- Radius of the circle (r) = 100 cm
- Arc length (s) = 22 cm

**Formula:**

- θ (in radians) = s / r

**Calculation:**

- θ = 22 cm / 100 cm = 0.22 radians

**Converting radians to degrees:**

- 1 radian = 180° / π
- θ (in degrees) = 0.22 * (180° / π) ≈ 0.22 * (180° * 7/22) ≈ 12.63°

**Therefore, the angle subtended at the center of the circle is approximately 12.63 degrees.**

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

**Ans : **

**Given:**

- Diameter of the circle = 40 cm
- Length of the chord = 20 cm

**Steps:**

**Find the radius of the circle:**- Radius = Diameter / 2 = 40 cm / 2 = 20 cm

**Determine the type of triangle formed:**- Since the chord is equal to the radius, the triangle formed by the chord and the radii to its endpoints is an equilateral triangle.

**Find the central angle θ:**- Convert 60 degrees to radians: θ = 60 * (π/180) = π/3 radians

**Use the formula for arc length:**- Arc length (s) = r * θ
- s = 20 cm * (π/3) ≈ 20.94 cm

**Therefore, the length of the minor arc of the chord is approximately 20.94 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.**

**Ans : **

**Given:**

- Two circles with arcs of equal length.
- One arc subtends an angle of 60° at the center.
- The other arc subtends an angle of 75° at the center.

**Formula:**

- Arc length (s) = r * θ
- (where r is the radius and θ is the angle in radians)

**Approach:**

- Since the arc lengths are equal, we can set up an equation using the formula for arc length.
- Let r1 and r2 be the radii of the two circles, respectively.
- Then, we can write:
- r1 * (60° * π/180) = r2 * (75° * π/180)

**Simplifying the equation:**

- r1 * (π/3) = r2 * (5π/12)
- r1 / r2 = (5π/12) / (π/3)
- r1 / r2 = 5/4

**radii of the two circles is 5:4.**

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

**Ans : **

θ = s / r

where:

- θ is the angle in radians
- s is the arc length
- r is the radius (length of the pendulum)

In this case, the radius (r) is 75 cm.

(i) When the arc length (s) is 10 cm: θ = 10 cm / 75 cm = 2/15 radians

(ii) When the arc length (s) is 15 cm: θ = 15 cm / 75 cm = 1/5 radians

(iii) When the arc length (s) is 21 cm: θ = 21 cm / 75 cm = 7/25 radians

Therefore, the angles in radians through which the pendulum swings for the given arc lengths are:

(i) 2/15 radians

(ii) 1/5 radians

(iii) 7/25 radians

**Exercise 3.2**

**Find the values of other five trigonometric functions in Exercises 1 to 5. **

**1. cos x = – 1/ 2 , x lies in third quadrant. **

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

**3. cot x = 4 /3 , 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.**

**Ans : **

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

**sin x:**Since x is in the third quadrant, sin x is negative. Using the Pythagorean identity:- sin^2 x + cos^2 x = 1
- sin^2 x + (-1/2)^2 = 1
- sin^2 x = 3/4
- sin x = -√3/2

**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 = 1 / (-1/2) = -2**csc x:**csc x = 1 / sin x = 1 / (-√3/2) = -2√3/3

**2. **

**cos x:**Since x is in the second quadrant, cos x is negative. Using the Pythagorean identity:- sin^2 x + cos^2 x = 1
- (3/5)^2 + cos^2 x = 1
- cos^2 x = 16/25
- cos x = -4/5

**tan x:**tan x = sin x / cos x = (3/5) / (-4/5) = -3/4**cot x:**cot x = 1 / tan x = 1 / (-3/4) = -4/3**sec x:**sec x = 1 / cos x = 1 / (-4/5) = -5/4**csc x:**csc x = 1 / sin x = 1 / (3/5) = 5/3

**3.**

**tan x:**tan x = 1 / cot x = 1 / (4/3) = 3/4**sin x:**Since x is in the third quadrant, sin x is negative. Using the Pythagorean identity:- 1 + tan^2 x = sec^2 x
- 1 + (3/4)^2 = sec^2 x
- sec^2 x = 25/16
- sec x = -5/4
- cos x = 1 / sec x = -4/5
- sin x = tan x * cos x = (3/4) * (-4/5) = -3/5

**csc x:**csc x = 1 / sin x = 1 / (-3/5) = -5/3

**4. **

**cos x:**cos x = 1 / sec x = 1 / (13/5) = 5/13**sin x:**Since x is in the fourth quadrant, sin x is negative. Using the Pythagorean identity:- sin^2 x + cos^2 x = 1
- sin^2 x + (5/13)^2 = 1
- sin^2 x = 144/169
- sin x = -12/13

**tan x:**tan x = sin x / cos x = (-12/13) / (5/13) = -12/5**cot x:**cot x = 1 / tan x = 1 / (-12/5) = -5/12**csc x:**csc x = 1 / sin x = 1 / (-12/13) = -13/12

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

**cot x:**cot x = 1 / tan x = 1 / (-5/12) = -12/5**sin x:**Since x is in the second quadrant, sin x is positive. Using the Pythagorean identity:- 1 + tan^2 x = sec^2 x
- 1 + (-5/12)^2 = sec^2 x
- sec^2 x = 169/144
- sec x = -13/12
- cos x = 1 / sec x = -12/13
- sin x = tan x * cos x = (-5/12) * (-12/13) = 5/13

**csc x:**csc x = 1 / sin x = 1 / (5/13) = 13/5

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

**6. sin 765° **

**7. cosec (– 1410°) **

**8. tan 19π /3 **

**9. sin (– 11π /3 ) **

**10. cot (– 15π/ 4 )**

**Ans : **

**6. sin 765°**

- 765° = 2 * 360° + 45°
- Since sin(360° + θ) = sin(θ), we have:
- sin 765° = sin 45° = √2/2

**7. cosec (-1410°)**

- csc(-1410°) = -csc(1410°) (since cosecant is odd)
- 1410° = 3 * 360° + 330°
- csc(-1410°) = -csc(330°) = -csc(360° – 30°) = -csc(30°) = -2

**8. tan (19π/3)**

- 19π/3 = 6π + π/3
- Since tan(2π + θ) = tan(θ), we have:
- tan (19π/3) = tan (π/3) = √3

**9. sin (-11π/3)**

- sin(-11π/3) = -sin(11π/3) (since sine is odd)
- 11π/3 = 3π + 2π/3
- sin(-11π/3) = -sin(2π/3) = -√3/2

**10. cot (-15π/4)**

- cot(-15π/4) = -cot(15π/4) (since cotangent is odd)
- 15π/4 = 4π + 3π/4
- cot(-15π/4) = -cot(3π/4) = 1

**Exercise 3.3**

**Prove that: **

**1. ****sin2 π/6 + cos2 π/3 – tan2 π/4 = – ½**

**Ans : **

**Given:** sin²(π/6) + cos²(π/3) – tan²(π/4) = -1/2

**Solution:**

Let’s calculate each term separately:

**sin²(π/6):**The sine of π/6 (30 degrees) is 1/2. So, sin²(π/6) = (1/2)² = 1/4.**cos²(π/3):**The cosine of π/3 (60 degrees) is also 1/2. So, cos²(π/3) = (1/2)² = 1/4.**tan²(π/4):**The tangent of π/4 (45 degrees) is 1. So, tan²(π/4) = 1².

Substituting these values into the original equation:

(1/4) + (1/4) – 1 = -1/2 1/2 – 1

= -1/2 -1/2 = -1/2

**Conclusion:** The left side of the equation equals the right side. Therefore, the identity is proven:

**sin²(π/6) + cos²(π/3) – tan²(π/4) **

**= -1/2**

**2. sin2 π/6 + cosec2 7π/6 cos2 π/3 = 3/2**

**Ans : **

**3. cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6**

**Ans : **

**1. cot²(π/6):**

- cot(π/6) = 1 / tan(π/6)
- = 1 / (1/√3)
- = √3
- cot²(π/6) = (√3)² = 3

**2. cosec(5π/6):**

- 5π/6 is in the second quadrant, where cosecant is positive.
- cosec(5π/6) = 1 / sin(5π/6) = 1 / (1/2) = 2

**3. tan²(π/6):**

- tan(π/6) = 1/√3
- tan²(π/6) = (1/√3)² = 1/3

3 + 2 + 3 * (1/3) = 6

Simplifying:

3 + 2 + 1 = 6 6 = 6

**Therefore, the given identity is true.**

**4. 2 sin2 3π/4 + 2 cos2 π/4 + 2 sec2 π/3 = 10**

**Ans : **

**1. sin²(3π/4):**

- sin(3π/4) = 1/√2
- sin²(3π/4) = (1/√2)² = 1/2

**2. cos²(π/4):**

- cos(π/4) = 1/√2
- cos²(π/4) = (1/√2)² = 1/2

**3. sec²(π/3):**

- sec(π/3) = 1 / cos(π/3) = 1 / (1/2) = 2
- sec²(π/3) = 2² = 4

2 * (1/2) + 2 * (1/2) + 2 * 4 = 10

Simplifying:

1 + 1 + 8 = 10 10 = 10

**Therefore, the given identity is true.**

**5. Find the value of: **

**(i) sin 75° (ii) tan 15° **

**Ans : **

**1. sin 75°**

- 75° = 150°/2
- cos 150° = -√3/2 (since 150° is in the second quadrant)

Therefore,

sin 75° = √[(1 – (-√3/2))/2] = √[(2 + √3)/4] = **(√6 + √2)/4**

**2. tan 15°**

- 15° = 30°/2
- cos 30° = √3/2

Therefore,

tan 15° = √[(1 – √3/2)/(1 + √3/2)] = √[(2 – √3)/(2 + √3)] = **2 – √3**

So, the values are:

(i) sin 75° = (√6 + √2)/4 (ii) tan 15° = 2 – √3

**Prove the following: **

**6. ****cos(****π/4****−x)cos(****π/4****−y)−sin(****π/4****−x)sin(****π/4****−y)**** = sin (x + y)**

**Ans : **

**Given:** cos(π/4−x)cos(π/4−y)−sin(π/4−x)sin(π/4−y) = sin (x + y)

**Proof:**

We will use the following trigonometric identity:*cos(A + B) = cosAcosB – sinAsinB*

In our case, we have:

- A = π/4 – x
- B = π/4 – y
- cos((π/4 – x) + (π/4 – y))

Simplifying:

- cos(π/2 – (x + y))

Using the identity cos(π/2 – θ) = sin θ, we get:

- sin(x + y)

**7. tan(π/4+x)/tan(π/4−x)=(1+tanx1−tanx)2**

**Ans : **

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

In this case, A = π/4 and B = x.

Substituting into the formula:

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

Since tan(π/4) = 1, we get:

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

Now, we can rewrite the original identity as:

(tan(π/4 + x) / tan(π/4 – x)) = (1 + tan x / 1 – tan x)²

Substituting the expression for tan(π/4 + x):

((1 + tan x) / (1 – tan x)) / (tan(π/4 – x)) = (1 + tan x / 1 – tan x)²

Since tan(π/4 – x) = (1 – tan x) / (1 + tan x) (using the tangent subtraction formula), we can substitute this into the equation:

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

Simplifying the left side:

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

Therefore, the identity is proven.

**8. cos(π+x)cos(−x)/sin(π−x)cos(π/2+x) = cot2 x**

**Ans : **

To prove the given identity, we will use the following trigonometric identities:

- cos(π + x) = -cos(x)
- cos(-x) = cos(x)
- sin(π – x) = sin(x)
- cos(π/2 + x) = -sin(x)

Substituting these identities into the given equation, we get:

(-cos(x)) * cos(x) / (sin(x) * (-sin(x))) = cot² x

Simplifying the expression:

cos²(x) / sin²(x) = cot² x

Since cot x = cos x / sin x, we can rewrite the left side as:

(cos x / sin x)² = cot² x

Therefore, the given identity is proven.

**9.**

**Ans : **

**=1= R.H.S**

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

**Ans : **

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

In this case, let A = (n + 1)x and B = (n + 2)x.

Substituting into the identity:

cos((n + 1)x + (n + 2)x) = cos(n + 1)x cos(n + 2)x – sin(n + 1)x sin(n + 2)x

Simplifying the left side:

cos(2n + 3)x = cos(n + 1)x cos(n + 2)x – sin(n + 1)x sin(n + 2)x

Now, we can rearrange the terms to match the given identity:

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

Since cos(2n + 3)x is equal to cos(x) for all integer values of n (due to the periodicity of the cosine function), we have:

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

Therefore, the identity is proven.

**11. cos(3π/4+x)−cos(3π/4−x)=−2–√sinx**

**Ans : **

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

let A = 3π/4 and B = x.

Substituting into the identity:

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

Since sin(3π/4) = √2/2, we can substitute this value into the equation:

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

Simplifying:

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

Therefore, the given identity is proven.

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

**Ans : **

To prove this identity, we will use the following trigonometric identity:

sin² A – sin² B = sin(A + B)sin(A – B)

Substituting into the identity:

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

Simplifying:

sin²(6x) – sin²(4x) = sin(10x)sin(2x)

Therefore, the given identity is proven.

**13. cos****2**** 2x cos****2**** 6x = sin 4x sin 8x**

**Ans : **

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

We can rewrite the given identity as:

2sin(4x)cos(4x) = sin(8x)sin(2x)

Now, using the identity sin(2A) = 2sin(A)cos(A) for both terms:

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

Simplifying:

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

Dividing both sides by 2sin(2x):

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

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

the given identity is proven.

**14. sin 2x + 2 sin 4x + sin 6x = 4 cos****2**** x sin 4x**

**Ans : **

L.H.S.= sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= [2sin(2x+6×2)cos(2x−6×2)] + 2 sin 4x

[∵ sin A + sin B = 2 sin(A+B2)cos(A−B2)

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

= 2 sin 4x cos 2x + 2 sin 4x

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

= 2 sin 4x (2 cos2 x – 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4 cos2 x sin 4x

= R.H.S.

Hence proved.

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

**Ans : **

**16. Cos9x−cos5x / sin17x−sin3x =−sin2x / cos10x**

**Ans : **

**17. Sin5x+sin3x / cos5x+cos3x = tan 4x**

**Ans : **

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

**cos(A) + cos(B) = 2cos((A + B)/2)cos((A – B)/2)**

Substituting A = 5x and B = 3x into these formulas:

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

cos(5x) + cos(3x) = 2cos((5x + 3x)/2)cos((5x – 3x)/2) = 2cos(4x)cos(x)

Now, we can substitute these expressions into the given identity:

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

Simplifying:

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

Since tan(x) = sin(x) / cos(x), we have:

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

Therefore, the given identity is proven.

**18. sinx−siny/cosx+cosy =tanx−y/2**

**Ans : **

.

**19. Sinx+sin3x / cosx+cos3x = tan 2x**

**Ans : **

**20. Sinx−sin3x / sin2x−cos2x = 2 sin x**

**Ans : **

It is known that

sin A – sin B = 2 cos(A+B /2)

sin(A−B/ 2)cos2 A – sin2 A = cos 2A

∴ L.H.S. = sinx−sin3x / sin2x−cos2x

= 2cos(x+3x/2)sin(x−3x/2)/ 2cos2x

= – 2 × (- sin x) = 2 sin x

= R.H.S

Hence proved.

**21. Cos4x+cos3x+cos2x / sin4x+sin3x+sin2x= cot 3x**

**Ans : **

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

**Ans : **

L.H.S.= cot x cot 2x – cot 2x cot 3x – cot 3x cot x

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

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – [cot2xcotx−1 / cotx+cot2x] (cot 2x + cot x)

[∵ cot (A + B) = cotAcotB−1 / cotA+cotB]

= cot x cot 2x – (cot 2x cot x – 1)

= 1 = R.H.S.

Hence proved.

**23. tan 4x = 4tanx(1−tan2x) / 1−6tan2x+tan4x**

**Ans : **

.

**24. cos 4x = 1 – 8sin2 x cos2 x**

**Ans : **

To prove the given identity, we will use the following trigonometric identities:

cos(2A) = 1 – 2sin²(A) sin(2A) = 2sin(A)cos(A)

Substituting A = 2x into the first identity:

cos(4x) = 1 – 2sin²(2x)

Now, using the identity sin(2A) = 2sin(A)cos(A) with A = x:

sin²(2x) = (2sin(x)cos(x))²

Substituting this into the previous equation:

cos(4x) = 1 – 2(2sin(x)cos(x))²

Simplifying:

cos(4x) = 1 – 8sin²(x)cos²(x)

Therefore, the given identity is proven.

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

**Ans : **

To prove this identity, we will use the double angle formula for cosine:

cos(2A) = 2cos²(A) – 1

We can start by expressing cos(6x) in terms of cos(2x) using this formula:

cos(6x) = cos(2(3x)) = 2cos²(3x) – 1

Now, we can use the double angle formula again to express cos²(3x) in terms of cos(6x):

cos²(3x) = (1 + cos(6x))/2

Substituting this into the previous equation:

cos(6x) = 2((1 + cos(6x))/2) – 1

Simplifying:

cos(6x) = 1 + cos(6x) – 1

cos(6x) = cos(6x)

This is a tautology, meaning it is always true. Therefore, the given identity is proven.