# Chapter 3 Trigonometric Functions Solutions – NCERT Class 11 Mathematics

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## Exercise 3.1

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

Relationship between angle in radians and degrees is
Angle in Radians = Angle in degrees × 𝛑/180
Using this simple formula all of these angles in degrees can be converted to radians

2 . Find the degree measures corresponding to the following radian measures
(Use 𝛑 = 22/7).
(i) 11/16
(ii) – 4
(iii) 5π/3
(iv) 7π/6

Relationship between angle in radians and in degrees is
Angle in degrees = Angle in radians × 180/𝛑

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

Wheel is making 360 revolutions in one minute
1 minute → 360 revolutions
As 1 revolutions means 360°
Hence
1 minute → 360 × 360 degrees
So within 1 minute wheel is revolving 360 × 360 degrees
Let’s now convert degrees to radians using formula Angle in Radians = Angle in degrees × 𝛑/180
So
1 minute → 360 × 360 × 𝛑/180 radians
1 minute → 720𝛑 radians
As 1 minute = 60 seconds
Hence
60 seconds → 720𝛑 radians
1 second → 12𝛑 radians

Thus wheel is taking 12𝛑 radians of revolutions in 1 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).

It is given that
Length of arc = 22 cm
Radius of circle = 100 cm

Relationship between Length of arc, Radius of circle and angle subtended at the centre of circle is
Angle subtended at centre of circle in Radians = Length of arc/Radius of circle

Plugging in values as given in the question
Angle subtended at centre of circle in Radians = 22/100 radians

But as per question we need to find out degree measure of this angle?
So we need to convert this angle from radians to degrees.
Which can be done using formula
Angle in degrees = Angle in radians × 180/𝛑

Hence
Angle subtended at centre of circle in degrees = 22/100 × 180/𝛑 = 12° 36′

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.

This question can be depicted as following image.

As per the question we’re given that
Diameter = 40 cm ⇒ Radius(AC or BC) = 20 cm
Length of chord = 20 cm ⇒ AB = 20 cm

And we need to find out Length of minor arc which is AB.

As is clear from the above diagram AC = BC = AB = 20 cm hence triangle ABC is an Equilateral Triangle ⇒ All of its internal angles are of 60 degrees.

Thus ∠ACB = 60° or 𝛑/3 radians

We know that angle subtended by an arc at centre in radians is length of arc divided by radius of circle.
Thus
∠ACB in radians = Length of arc AB/Radius of circle
Plugging in the values
𝛑/3 = Length of arc AB/20
Length of arc AB = 20𝛑/3 cm

Hence if length of chord is 20 cm and circle have 40 cm diameter then length of minor arc of the chord will be 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.

This question can be depicted as following image.

Converting both of given angles to radians?
Angle in Radians = Angle in degrees × 𝛑/180

Angle subtended by arc at the centre of first circle in radians = 60 × 𝛑/180
= 𝛑/3

Angle subtended by arc at the centre of second circle in radians = 75 × 𝛑/180
= 5𝛑/12

Angle subtended by arc at the centre of circle in radians = Length of arc/Radius of circle

Angle subtended by arc at the centre of first circle in radians
= Length of arc of first circle/Radius of first circle
(Equation 1) ⇒ 𝛑/3 = Length of arc of first circle/Radius of first circle

Angle subtended by arc at the centre of second circle in radians
= Length of arc of second circle/Radius of second circle
(Equation 2) ⇒ 5𝛑/12 = Length of arc of second circle/Radius of second circle

Dividing (Equation 2) by (Equation 1)
5𝛑/12 × 3/𝛑 = Length of arc of second circle/Radius of second circle × Radius of first circle/Length of arc of first circle

15/12 = Length of arc of second circle/Length of arc of first circle × Radius of first circle/Radius of second circle

As per question it’s given that Radius of second circle = Length of arc of first circle

Hence above equation becomes
⇒ 15/12 = Radius of first circle/Radius of second circle
⇒ Radius of first circle/Radius of second circle = 5/4

Thus if for two circles it’s given that same length arc subtend angles 60° and 75° at the centre then ratio of their radii will be 5:4.

7. Find the angle in radian through 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

Angle in radians through which a pendulum swings = Length of arc/Pendulum length

As per question it’s given that
Pendulum length = 75 cm

If Length of arc = 10 cm
Angle in radians through which a pendulum swings = 10/75 = 2/15 radians

If Length of arc = 15 cm
Angle in radians through which a pendulum swings = 15/75 = 3/15 radians

If Length of arc = 21 cm
Angle in radians through which a pendulum swings = 21/75 = 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

Using formulas
Sin2x + Cos2x = 1
Sec2x = 1 + Tan2x
Cosec2x = 1 + Cot2x

Keeping these three formulas and +/- table in mind all of values of other trigonometric functions can be calculated.

1. Cosx = -1/2, x lies in third quadrant
Using formula
Sin2x + Cos2x = 1
Sin2x + (-1/2)2 = 1
Sin2x + 1/4 = 1
Sin2x = 1 – 1/4 = 3/4
Sinx = ± √3/2
But as per the question itself, x lies in third quadrant ⇒ Sinx is negative
⇒ Sinx = – √3/2

We know that Tanx is just ratio of Sinx and Cosx
⇒ Tanx = Sinx/Cosx = – √3/2/ -1/2 = √3
⇒ Tanx = √3

As Cotx is just inverse of Tanx ⇒ Cotx = 1/Tanx
⇒ Cotx = 1/√3

As Cosecx is just inverse of Sinx ⇒ Cosecx = 1/Sinx
⇒ Cosecx = 1/ – √3/2 = – 2/√3
⇒ Cosecx = – 2/√3

As Secx is just inverse of Cosx ⇒ Secx = 1/Cosx
⇒ Secx = 1/ – 1/2 = – 2
⇒ Secx = -2

Summarising all this in below table

2. Sin x = 3/5, x lies in second quadrant
Using formula
Sin2x + Cos2x = 1
(3/5)2 + Cos2x = 1
9/25 + Cos2x = 1
Cos2x = 1 – 9/25 = 16/25
Cos2x = 16/25
Cosx = ± 4/5
But as x lies in second quadrant ⇒ Cosx is negative
⇒ Cosx = – 4/5

We know that Tanx is just ratio of Sinx and Cosx
⇒ Tanx = Sinx/Cosx = 3/5 / – 4/5 = – 3/4
⇒ Tanx = – 3/4

As Cotx is just inverse of Tanx ⇒ Cotx = 1/Tanx
⇒ Cotx = – 4/3

As Cosecx is just inverse of Sinx ⇒ Cosecx = 1/Sinx
⇒ Cosecx = 1/ 3/5 = 5/3
⇒ Cosecx = 5/3

As Secx is just inverse of Cosx ⇒ Secx = 1/Cosx
⇒ Secx = 1/ – 4/5 = – 5/4
⇒ Secx = – 5/4

Summarising all this in below table

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

We know that Tanx is just inverse of Cotx
⇒ Tanx = 1/Cotx = 1/ 4/3 = 3/4
⇒ Tanx = 3/4

Using formula
Cosec2x = 1 + Cot2x
Cosec2x = 1 + (4/3)2
Cosec2x = 1 + 16/9 = 25/9
Cosec2x = 25/9
Cosecx = ± √25/9 = ± 5/3
Cosecx = ± 5/3
But as per the question itself, x lies in third quadrant ⇒ Cosecx is negative
⇒ Cosecx = – 5/3

We know that Cosecx is just inverse of Sinx
⇒ Cosecx = 1/Sinx
⇒ Sinx = 1/Cosecx = 1/ – 5/3 = – 3/5
⇒ Sinx = – 3/5

Using formula
Sin2x + Cos2x = 1
(-3/5)2 + Cos2x = 1
9/25 + Cos2x = 1
Cos2x = 1 – 9/25 = 16/25
Cos2x = 16/25
Cosx = ± √16/25 = ± 4/5
Cosx = ± 4/5
But as per the question itself, x lies in third quadrant ⇒ Cosx is negative
⇒ Cosx = – 4/5

We know that Secx is just inverse of Cosx
⇒ Secx = 1/Cosx
⇒ Secx = 1/ – 4/5 = – 5/4
⇒ Secx = – 5/4

Summarising all this in below table

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

As Secx is just inverse of Cosx
⇒ Secx = 1/Cosx
⇒ Cosx = 1/Sinx = 1/ 13/5 = 5/13
⇒ Cosx = 5/13

Using formula
Sin2x + Cos2x = 1
Sin2x + (5/13)2 = 1
Sin2x + 25/169 = 1
Sin2x = 1 – 25/169 = 144/169
Sin2x = 144/169
Sinx = ± √144/169 = ± 12/13
Sinx = ± 12/13
As per question itself, x lies in fourth quadrant ⇒ Sinx is negative
⇒ Sinx = – 12/13

As Cosecx is just inverse of Sinx
⇒ Cosecx = 1/Sinx = 1/ – 12/13 = – 13/12
⇒ Cosecx = – 13/12

As Tanx is just ratio of Sinx and Cosx
⇒ Tanx = Sinx/Cosx
⇒ Tanx = – 12/13 / 5/13 = – 12/5
⇒ Tanx = – 12/5

As Cotx is just inverse of Tanx
⇒ Cotx = 1/Tanx
⇒ Cotx = 1/ – 12/5 = – 5/12
⇒ Cotx = – 5/12

Summarising all this in below table

5. Tan x = –5/12 , x lies in second quadrant

As Cotx is just inverse of Tanx
⇒ Cotx = 1/Tanx
⇒ Cotx = 1/ – 5/12 = – 12/5
⇒ Cotx = – 12/5

Using formula
Cosec2x = 1 + Cot2x
Cosec2x = 1 + (- 12/5)2 = 1 + 144/25
Cosec2x = 169/25
Cosecx = ± √169/25 = ± 13/5
Cosecx = ± 13/5
But as per the question itself, x lies in second quadrant ⇒ Cosecx is positive
⇒ Cosecx = 13/5

As Cosecx is just inverse of Sinx
⇒ Cosecx = 1/Sinx
⇒ Sinx = 1/Cosecx
⇒ Sinx = 1/ 13/5 = 5/13
⇒ Sinx = 5/13

Using formula
Sin2x + Cos2x = 1
(5/13)2 + Cos2x = 1
25/169 + Cos2x = 1
Cos2x = 1 – 25/169 = 144/169
Cos2x = 144/169
Cosx = ± √144/169 = ± 12/13
Cosx = ± 12/13
But as per the question itself, x lies in second quadrant ⇒ Cosx is negative
⇒ Cosx = – 12/13

As Secx is just inverse of Cosx
Secx = 1/Cosx = 1/ – 12/13 = – 13/12
⇒ Secx = – 13/12

Summarising all this in below table

Find the values of the trigonometric functions in Exercises 6 to 10.
6. Sin765°
7. Cosec(- 1410°)
8. Tan(19𝛑/3)
9. Sin(- 11𝛑/3)
10. Cot(- 15𝛑/4)

Let’s solve each of this one by one

6. Sin765°
Sin765° = Sin(360° × 2 + 45°)
We know that Sin(360° × n + θ) = Sinθ where n ∈ I and 0 ≤ θ ≤ 90°
⇒ Sin765° = Sin(360° × 2 + 45°) = Sin45° = 1/√2
Hence
Value of Sin765° = 1/√2

7. Cosec(- 1410°)
We know that Cosec(- θ) = – Cosecθ for any angle θ
⇒ Cosec(- 1410°) = – Cosec1410° = – Cosec(360° × 4 – 30°)
We know that Cosec(360° × n – θ) = – Cosecθ where n ∈ I and 0 ≤ θ ≤ 90°
⇒ Cosec(- 1410°) = – Cosec(360° × 4 – 30°) = – (- Cosec30°) = Cosec30° = 2
Thus
Value of Cosec(-1410°) = 2

8. Tan(19𝛑/3)
Tan(19𝛑/3) = Tan(2𝛑 × 3 + 𝛑/3)
We know that Tan(2𝛑 × n + θ) = Tanθ where n ∈ I and 0 ≤ θ ≤ 90°
⇒ Tan(19𝛑/3) = Tan(2𝛑 × 3 + 𝛑/3) = Tan(𝛑/3) = √3
Thus
Value of Tan(19𝛑/3) = √3

9. Sin(- 11𝛑/3)
We know that Sin(- θ) = – Sinθ for any value of angle θ
⇒ Sin(- 11𝛑/3) = – Sin(11𝛑/3) = – Sin(2𝛑 × 2 – 𝛑/3)
⇒ Sin(- 11𝛑/3) = – Sin(2𝛑 × 2 – 𝛑/3)
We know that Sin(2𝛑 × n – θ) = – Sinθ where n ∈ I and 0 ≤ θ ≤ 90°
⇒ Sin(- 11𝛑/3) = – (- Sin(𝛑/3)) = Sin(𝛑/3) = √3/2
⇒ Sin(- 11𝛑/3) = √3/2
Thus
Value of Sin(- 11𝛑/3) = √3/2

10. Cot(- 15𝛑/4)
We know that Cot(- θ) = – Cotθ for any value of angle θ
⇒ Cot(- 15𝛑/4) = – Cot(15𝛑/4) = – Cot(2𝛑 × 2 – 𝛑/4)
⇒ Cot(- 15𝛑/4) = – Cot(2𝛑 × 2 – 𝛑/4)
We know that Cot(2𝛑 × n – θ) = – Cotθ where n ∈ I and 0 ≤ θ ≤ 90°
⇒ Cot(- 15𝛑/4) = – (- Cot(𝛑/4)) = Cot(𝛑/4) = 1
⇒ Cot(- 15𝛑/4) = 1
Thus
Value of Cot(- 15𝛑/4) = 1

## Exercise 3.3

Prove that
1. Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) = – 1/2
2. 2Sin2(𝛑/6) + Cosec2(7𝛑/6) Cos2(𝛑/3) = 3/2
3. Cot2(𝛑/6) + Cosec(5𝛑/6) + 3Tan2(𝛑/6) = 6
4. 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 10

Let’s solve each of these one by one

1. Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) = 1/2
Let’s simplify left hand side of this equation
We know that
• Sin(𝛑/6) = 1/2
• Cos(𝛑/3) = 1/2
• Tan(𝛑/4) = 1

⇒ Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) = (1/2)2 + (1/2)2 – 12 = 1/4 + 1/4 – 1
⇒ Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) = 2/4 – 1 = 1/2 – 1 = – 1/2
⇒ Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) = – 1/2
Hence Proved that Sin2(𝛑/6) + Cos2(𝛑/3) – Tan2(𝛑/4) is equals to – 1/2

2. 2Sin2(𝛑/6) + Cosec2(7𝛑/6)Cos2(𝛑/3) = 3/2
Let’s simplify left hand side of this equation
We know that
• Sin(𝛑/6) = 1/2
• Cos(𝛑/3) = 1/2

Cosec(7𝛑/6) = Cosec(𝛑 + 𝛑/6)
We know that Cosec(𝛑 + θ) = – Cosecθ where 0 ≤ θ ≤ 90°
⇒ Cosec(7𝛑/6) = Cosec(𝛑 + 𝛑/6) = – Cosec(𝛑/6) = – 2
⇒ Cosec(7𝛑/6) = – 2

Let’s now plug in all these values into Left Hand Side of given equation
⇒ 2Sin2(𝛑/6) + Cosec2(7𝛑/6) Cos2(𝛑/3) = 2(1/2)2 + (- 2)2(1/2)2
⇒ 2Sin2(𝛑/6) + Cosec2(7𝛑/6) Cos2(𝛑/3) = 2(1/4) + 4(1/4) = 1/2 + 1 = 3/2
⇒ 2Sin2(𝛑/6) + Cosec2(7𝛑/6) Cos2(𝛑/3) = 3/2
Hence Proved that 2Sin2(𝛑/6) + Cosec2(7𝛑/6) Cos2(𝛑/3) is equals to 3/2

3. Cot2(𝛑/6) + Cosec(5𝛑/6) + 3Tan2(𝛑/6) = 6
Let’s simplify left hand side of this equation
We know that
• Cot(𝛑/6) = √3
• Tan(𝛑/6) = 1/√3

Cosec(5𝛑/6) = Cosec(𝛑 – 𝛑/6)
We know that Cosec(𝛑 – θ) = Cosecθ where 0 ≤ θ ≤ 90°
⇒ Cosec(5𝛑/6) = Cosec(𝛑 – 𝛑/6) = Cosec(𝛑/6) = 2
⇒ Cosec(5𝛑/6) = 2

Let’s now plug in all these values into Left Hand Side of given equation
⇒ Cot2(𝛑/6) + Cosec(5𝛑/6) + 3Tan2(𝛑/6) = (√3)2 + 2 + 3(1/√3)2
⇒ Cot2(𝛑/6) + Cosec(5𝛑/6) + 3Tan2(𝛑/6) = 3 + 2 + 3(1/3) = 3 + 2 + 1 = 6
⇒ Cot2(𝛑/6) + Cosec(5𝛑/6) + 3Tan2(𝛑/6) = 6
Hence Proved that Cot2(𝛑/6) + Cosec(5𝛑/6) + 3 Tan2(𝛑/6) is equals to 6

4. 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 10
Let’s simplify left hand side of this equation
We know that
• Cos(𝛑/4) = 1/√2
• Sec(𝛑/3) = 2

Sin(3𝛑/4) = Sin(𝛑/2 + 𝛑/4)
We know that Sin(𝛑/2 + θ) = Sinθ where 0 ≤ θ ≤ 90°
⇒ Sin(3𝛑/4) = Sin(𝛑/2 + 𝛑/4) = Sin(𝛑/4) = 1/√2
⇒ Sin(3𝛑/4) = 1/√2

Let’s now plug in all these values into Left Hand Side of given equation
⇒ 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 2(1/√2)2 + 2(1/√2)2 + 2(2)2
⇒ 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 2(1/2) + 2(1/2) + 2(4)
⇒ 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 1 + 1 + 8 = 10
⇒ 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) = 10
Hence Proved that 2Sin2(3𝛑/4) + 2Cos2(𝛑/4) + 2Sec2(𝛑/3) is equals to 10

5. Find the value of
(i) Sin75°
(ii) Tan15°

Let’s find out values of each of these one by one
(i) Sin75° = Sin(90° – 15°)
We know that Sin(90° – θ) = Sinθ where where 0 ≤ θ ≤ 90°
⇒ Sin75° = Sin(90° – 15°) = Sin15° = Sin(45° – 30°)
⇒ Sin75° = Sin(45° – 30°)

We know that Sin(A – B) = SinA CosB – CosA SinB
⇒ Sin75° = Sin(45° – 30°) = Sin45° Cos30° – Cos45° Sin30°
⇒ Sin75° = Sin45° Cos30° – Cos45° Sin30°

Values of
• Sin45° = 1/√2
• Cos30° = √3/2
• Cos45° = 1/√2
• Sin30° = 1/2

Putting in these values into equation Sin75° = Sin45° Cos30° – Cos45° Sin30°
⇒ Sin75° = (1/√2)(√3/2) – (1/√2)(1/2) = √3/2√2 – 1/2√2 = (√3 – 1)/2√2
⇒ Sin75° = (√3 – 1)/2√2

Hence value of Sin75° is equals to (√3 – 1)/2√2.

(ii) Tan15°
⇒ Tan15° = Tan(45° – 30°)

We know that Tan(A – B) = TanA – TanB/(1 + TanA TanB)
⇒ Tan15° = Tan(45° – 30°) = Tan45° – Tan30°/(1 + Tan45° Tan30°)
⇒ Tan15° = Tan45° – Tan30°/(1 + Tan45° Tan30°)

Values of
• Tan45° = 1
• Tan30° = 1/√3

Putting in these values into equation Tan15° = Tan45° – Tan30°/(1 + Tan45° Tan30°)
⇒ Tan15° = 1 – 1/√3/(1 + 1 × 1/√3) = (√3 – 1)/(√3 + 1)
⇒ Tan15° = (√3 – 1)/(√3 + 1)

Hence value of Tan15° is equals to (√3 – 1)/(√3 + 1)

6. Prove Cos(𝛑/4 – x) Cos(𝛑/4 – y) – Sin(𝛑/4 – x) Sin(𝛑/4 – y) = Sin(x + y)
We know that
Cos(A – B) = CosA CosB + SinA SinB
Sin(A – B) = SinA CosB – CosA SinB

Using these formulas
Cos(𝛑/4 – x) = Cos(𝛑/4) Cosx + Sin(𝛑/4) Sinx

Cos(𝛑/4 – y) = Cos(𝛑/4) Cosy + Sin(𝛑/4) Siny

Sin(𝛑/4 – x) = Sin(𝛑/4) Cosx – Cos(𝛑/4) Sinx

Sin(𝛑/4 – y) = Sin(𝛑/4) Cosy – Cos(𝛑/4) Siny

Value of Sin(𝛑/4) = Cos(𝛑/4) = 1/√2
Thus above four equations can be rewritten as

Cos(𝛑/4 – x) = Cos(𝛑/4) Cosx + Sin(𝛑/4) Sinx = 1/√2(Cosx + Sinx)

Cos(𝛑/4 – y) = Cos(𝛑/4) Cosy + Sin(𝛑/4) Siny = 1/√2(Cosy + Siny)

Sin(𝛑/4 – x) = Sin(𝛑/4) Cosx – Cos(𝛑/4) Sinx = 1/√2(Cosx – Sinx)

Sin(𝛑/4 – y) = Sin(𝛑/4) Cosy – Cos(𝛑/4) Siny = 1/√2(Cosy – Siny)

Plugging in these values into Left Hand Side of given equation Cos(𝛑/4 – x) Cos(𝛑/4 – y) – Sin(𝛑/4 – x) Sin(𝛑/4 – y)

⇒ Cos(𝛑/4 – x) Cos(𝛑/4 – y) – Sin(𝛑/4 – x) Sin(𝛑/4 – y) = 1/√2(Cosx + Sinx) × 1/√2(Cosy + Siny) – 1/√2(Cosx – Sinx) × 1/√2(Cosy – Siny)

= 1/2(Cosx + Sinx)(Cosy + Siny) – 1/2(Cosx – Sinx)(Cosy – Siny)

= 1/2[(Cosx + Sinx)(Cosy + Siny) – (Cosx – Sinx)(Cosy – Siny)]
Simplifying

= 1/2[Cosx(Cosy + Siny) + Sinx(Cosy + Siny) – Cosx(Cosy – Siny) + Sinx(Cosy – Siny)]

= 1/2[Cosx Cosy + Cosx Siny + Sinx Cosy + Sinx Siny – Cosx Cosy + Cosx Siny + Sinx Cosy – Sinx Siny]

= 1/2[Cosx Siny + Sinx Cosy + Cosx Siny + Sinx Cosy]

= 1/2[2 Cosx Siny + 2 Sinx Cosy] = 2/2[Cosx Siny + Sinx Cosy]

= Cosx Siny + Sinx Cosy

We know that SinA CosB + CosA SinB = Sin(A + B)
Thus above equation can be rewritten as

= Sin(x + y)

Thus
⇒ Cos(𝛑/4 – x) Cos(𝛑/4 – y) – Sin(𝛑/4 – x) Sin(𝛑/4 – y) = Sin(x + y)
Hence proved that Cos(𝛑/4 – x) Cos(𝛑/4 – y) – Sin(𝛑/4 – x) Sin(𝛑/4 – y) is equals to Sin(x + y)

7. Prove that Tan(𝛑/4 + x)/Tan(𝛑/4 – x) = [(1 + Tanx)/(1 – Tanx)]2

8. Prove that Cos(𝛑 + x) Cos(- x)/Sin(𝛑 – x) Cos(𝛑/2 + x) = Cot2x

9. Prove that Cos(3𝛑/2 + x) Cos(2𝛑 + x) [Cot(3𝛑/2 – x) + Cot(2𝛑 + x)] = 1

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10. Prove that Sin(n + 1)x Sin(n + 2)x + Cos(n + 1)x Cos(n + 2)x = Cosx

Let’s simplify Left Hand Side of this equation
Sin(n + 1)x Sin(n + 2)x + Cos(n + 1)x Cos(n + 2)x

⇒ Sin(nx + x) Sin(nx + 2x) + Cos(nx + x) Cos(nx + 2x)

Multiplying and Dividing this whole equation by 2
⇒ [2 Sin(nx + x) Sin(nx + 2x) + 2 Cos(nx + x) Cos(nx + 2x)]/2 [Equation 1]

Using formulas
2 SinA SinB = Cos(A – B) – Cos(A + B)
2 CosA CosB = Cos(A + B) + Cos(A – B)

⇒ 2 Sin(nx + x) Sin(nx + 2x) = Cos(nx + x – nx – 2x) – Cos(nx + x + nx + 2x) = Cos(- x) – Cos(2nx + 3x)

⇒ 2 Cos(nx + x) Cos(nx + 2x) = Cos(nx + x + nx + 2x) + Cos(nx + x – nx – 2x) = Cos(2nx + 3x) + Cos(- x)

Replacing these values in [Equation 1]

⇒ [Cos(- x) – Cos(2nx + 3x) + Cos(2nx + 3x) + Cos(- x)]/2
⇒ [2 Cos(-x)]/2 = Cos(-x)
⇒ Cos(-x)

Using formula Cos(- A) = CosA
Cos(- x) = Cosx

Hence it’s proved that Sin(n + 1)x Sin(n + 2)x + Cos(n + 1)x Cos(n + 2)x is equals to Cosx

11. Prove that Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – √2 Sinx

Let’s simplify Left Hand Side of this equation
Using formula
CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2

Replacing
C = 3𝛑/4 + x
D = 3𝛑/4 – x
in above formula

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – 2 Sin(3𝛑/4 + x + 3𝛑/4 – x)/2 Sin(3𝛑/4 + x – 3𝛑/4 + x)/2

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – 2 Sin(3𝛑/4 + 3𝛑/4)/2 Sin(x + x)/2

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – 2 Sin(6𝛑/4)/2 Sin(2x)/2

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – 2 Sin(3𝛑/4) Sinx

Value of Sin(3𝛑/4) = 1/√2

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – 2 (1/√2) Sinx = – √2 Sinx

⇒ Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) = – √2 Sinx

Hence it’s proved that Cos(3𝛑/4 + x) – Cos(3𝛑/4 – x) is equals to – √2 Sinx

12. Prove that Sin26x – Sin24x = Sin2x Sin10x

Let’s simplify Left Hand Side of this equation

⇒ Sin26x – Sin24x

Using formula a2 – b2 = (a – b)(a + b)

Replacing
a = Sin6x
b = Sin4x

⇒ Sin26x – Sin24x = (Sin6x – Sin4x)(Sin6x + Sin4x) [Equation 1]

Using formulas
SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2
SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2

⇒ Sin6x – Sin4x = 2 Cos(6x + 4x)/2 Sin(6x – 4x)/2 = 2 Cos(10x/2) Sin(2x/2) = 2 Cos5x Sinx
⇒ Sin6x + Sin4x = 2 Sin(6x + 4x)/2 Cos(6x – 4x)/2 = 2 Sin(10x/2) Cos(2x/2) = 2 Sin5x Cosx

Replacing these values in [Equation 1]

⇒ Sin26x – Sin24x = (2 Cos5x Sinx) (2 Sin5x Cosx)

Rearranging values on Right Hand Side of above equation
⇒ Sin26x – Sin24x = (2 Sin5x Cos5x) (2 Sinx Cox) [Equation 2]

Using Formula 2 SinA CosA = Sin2A for any angle A
2 Sin5x Cos5x = Sin2(5x) = Sin10x
2 Sinx Cosx = Sin2(x) = Sin2x

Replacing these values in [Equation 2]
⇒ Sin26x – Sin24x = Sin2x Sin10x

Hence it’s proved that Sin26x – Sin24x is equals to Sin2x Sin10x.

13. Prove that Cos22x – Cos26x = Sin4x Sin8x

Let’s simplify Left Hand Side of this equation

⇒ Cos22x – Cos26x

Using formula a2 – b2 = (a – b)(a + b)

Replacing
a = Cos2x
b = Cos6x

⇒ Cos22x – Cos26x = (Cos2x – Cos6x)(Cos2x + Cos6x) [Equation 1]

Using formulas
CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2
CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

⇒ Cos2x – Cos6x = – 2 Sin(2x + 6x)/2 Sin(2x – 6x)/2 = – 2 Sin(8x/2) Sin(- 4x/2) = – 2 Sin4x Sin(- 2x)
⇒ Cos2x – Cos6x = 2 Cos(2x + 6x)/2 Cos(2x – 6x)/2 = 2 Cos(8x/2) Cos(- 4x/2) = 2 Cos(4x) Cos(- 2x)

Replacing these values in [Equation 1]
⇒ Cos22x – Cos26x = [-2 Sin4x Sin(-2x)][2 Cos4x Cos(-2x)]

Rearranging Right Hand Side of this equation
⇒ Cos22x – Cos26x = – [2 Sin4x Cos4x] [2 Sin(-2x) Cos(-2x)]

Using formula 2 SinA CosA = Sin2A for angle A

⇒ Cos22x – Cos26x = – [Sin2(4x)] [Sin2(-2x)] = – Sin8x Sin(-4x)

⇒ Cos22x – Cos26x = – Sin8x Sin(-4x)

Using formula Sin(- A) = – SinA for angle A

⇒ Cos22x – Cos26x = – Sin8x [- Sin4x] = Sin8x Sin4x

⇒ Cos22x – Cos26x = Sin4x Sin8x

Hence it’s proved that Cos22x – Cos26x is equals to Sin4x Sin8x

14. Prove that Sin2x + 2 Sin4x + Sin6x = 4 Cos2x Sin4x

Let’s simplify Left Hand Side of this equation

⇒ Sin2x + 2 Sin4x + Sin6x
⇒ (Sin2x + Sin6x) + 2 Sin4x

Using formula SinC + SinC = 2 Sin(C + D)/2 Cos(C – D)/2

⇒ 2 Sin(2x + 6x)/2 Cos(2x – 6x)/2 + 2 Sin4x

⇒ 2 Sin(8x/2) Cos(- 4x/2) + 2 Sin4x

⇒ 2 Sin4x Cos(- 2x) + 2 Sin4x

Using formula Cos(- A) = CosA for any angle A
⇒ 2 Sin4x Cos2x + 2 Sin4x

Taking 2 Sin4x common
⇒ 2 Sin4x(Cos2x + 1) [Equation 1]

Using formula Cos2A = 2 Cos2A – 1
Cos2x = 2 Cos2x – 1

Replacing value of Cos2x in [Equation 1]
⇒ 2 Sin4x(2 Cos2x – 1 + 1)

⇒ 2 Sin4x(2 Cos2x )

⇒ 4 Sin4x Cos2x which is equals to Right Hand Side of Trigonometric Equation given in the question.

Hence it’s proved that Sin2x + 2 Sin4x + Sin6x is equals to 4 Sin4x Cos2x

15. Prove that Cot4x (Sin5x + Sin3x) = Cotx (Sin5x – Sin3x)

Let’s first simplify Left Hand Side of this equation
⇒ Cot4x (Sin5x + Sin3x) [Equation 1]

Using formula SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
Replacing
C = 5x
D = 3x

⇒ Sin5x + Sin3x = 2 Sin(5x + 3x)/2 Cos(5x – 3x)/2 = 2 Sin(8x/2) Cos(2x/2) = 2 Sin4x Cosx
⇒ Sin5x + Sin3x = 2 Sin4x Cosx

Replacing this value in [Equation 1]

⇒ Cot4x (2 Sin4x Cosx)

Cot4x can be rewritten as Cos4x/Sin4x
⇒ Cos4x/Sin4x (2 Sin4x Cosx)

⇒ Cos4x (2 Cosx) = 2 Cosx Cos4x

[Statement 1]
Thus Left Hand Side of given Trigonometric Equation Cot4x (Sin5x + Sin3x) = 2 Cosx Cos4x

Let’s now move onto simplifying Right Hand Side of given Trigonometric Equation
⇒ Cotx (Sin5x – Sin3x) [Equation 2]

Using formula SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2
Replacing
C = 5x
D = 3x
⇒ Sin5x – Sin3x = 2 Cos(5x + 3x)/2 Sin(5x – 3x)/2 = 2 Cos(8x/2) Sin(2x/2) = 2 Cos4x Sinx
⇒ Sin5x – Sin3x = 2 Cos4x Sinx

Replacing this value in [Equation 2]
⇒ Cotx (2 Cos4x Sinx)

Cotx can be written as Cosx/Sinx
⇒ Cosx/Sinx (2 Cos4x Sinx)
⇒ 2 Cosx Cos4x

[Statement 2]
Thus Right Hand Side of given Trigonometric Equation Cotx (Sin5x – Sin3x) = 2 Cosx Cos4x

Putting together [Statement 1] and [Statement 2]
Cot4x (Sin5x + Sin3x) = Cotx (Sin5x – Sin3x) = 2 Cosx Cos4x

Thus it’s proved that Cot4x (Sin5x + Sin3x) is equals to Cotx (Sin5x – Sin3x)

16. Prove that Cos9x – Cos5x/Sin17x – Sin3x = – Sin2x/Cos10x

Let’s first simplify Left Hand Side of this equation
⇒ Cos9x – Cos5x/Sin17x – Sin3x [Equation 1]

Using formulas
CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2
SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2

⇒ Cos9x – Cos5x = – 2 Sin(9x + 5x)/2 Sin(9x – 5x)/2 = – 2 Sin(14x/2) Sin(4x/2) = – 2 Sin7x Sin2x

⇒ Sin17x – Sin3x = 2 Cos(17x + 3x)/2 Sin(17x – 3x)/2 = 2 Cos(20x/2) Sin(14x/2) = 2 Cos10x Sin7x

Replacing these values in [Equation 1]
⇒ – 2 Sin7x Sin2x / 2 Cos10x Sin7x = – Sin2x/Cos10x which is equals to Right Hand Side of given Trigonometric Equation.

Hence it’s proved that Cos9x – Cos5x/Sin17x – Sin3x is equals to – Sin2x/Cos10x.

17. Prove that Sin5x + Sin3x/Cos5x + Cos3x = Tan4x

Let’s simplify Left Hand Side of this equation
⇒ Sin5x + Sin3x/Cos5x + Cos3x [Equation 1]

Using formulas
SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

⇒ Sin5x + Sin3x = 2 Sin(5x + 3x)/2 Cos(5x – 3x)/2 = 2 Sin(8x/2) Cos(2x/2) = 2 Sin4x Cosx
⇒ Cos5x + Cos3x = 2 Cos(5x + 3x)/2 Cos(5x – 3x)/2 = 2 Cos(8x/2) Cos(2x/2) = 2 Cos4x Cosx

Replacing these values in [Equation 1]
⇒ 2 Sin4x Cosx/2 Cos4x Cosx = Sin4x/Cos4x = Tan4x which is equals to Right Hand Side of Trigonometric Equation given in the question.

Hence it’s proved that Sin5x + Sin3x/Cos5x + Cos3x is equals to Tan4x

18. Prove that Sinx – Siny/Cosx + Cosy = Tan(x – y)/2

Let’s simplify Left Hand Side of the given Trigonometric Equation.
⇒ Sinx – Siny/Cosx + Cosy [Equation 1]

Using formulas
SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2
CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

Replacing
C = x
D = y

⇒ Sinx – Siny = 2 Cos(x + y)/2 Sin(x – y)/2
⇒ Cosx + Cosy = 2 Cos(x + y)/2 Cos(x – y)/2

Replacing these values in [Equation 1]
⇒ 2 Cos(x + y)/2 Sin(x – y)/2 / 2 Cos(x + y)/2 Cos(x – y)/2 = Sin(x – y)/2 / Cos(x – y)/2 = Tan(x – y)/2
Which is equal to Right Hand Side of the given Trigonometric Equation.

Hence it’s proved that Sinx – Siny/Cosx + Cosy is equals to Tan(x – y)/2.

19. Prove that Sinx + Sin3x/Cosx + Cos3x = Tan2x

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ Sinx + Sin3x/Cosx + Cos3x [Equation 1]

Using formulas
⇒ SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
⇒ CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

Replacing
C = x
D = 3x

⇒ Sinx + Sin3x = 2 Sin(x + 3x)/2 Cos(x – 3x)/2 = 2 Sin(4x/2) Cos(- 2x/2) = 2 Sin2x Cos(-x)
⇒ Sinx + Sin3x = 2 Sin2x Cos(-x)

⇒ Cosx + Cos3x = 2 Cos(x + 3x)/2 Cos(x – 3x)/2 = 2 Cos(4x/2) Cos(- 2x/2) = 2 Cos2x Cos(-x)
⇒ Cosx + Cos3x = 2 Cos2x Cos(-x)

Replacing these values in [Equation 1]
⇒ 2 Sin2x Cos(-x) / 2 Cos2x Cos(-x) = Sin2x/Cos2x = Tan2x
Which is equals to Right Hand Side of given Trigonometric Equation.

Hence it’s proved that Sinx + Sin3x/Cosx + Cos3x is equals to Tan2x

20. Prove that Sinx – Sin3x/Sin2x – Cos2x = 2 Sinx

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ (Sinx – Sin3x)/Sin2x – Cos2x [Equation 1]

Using formula
SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2

⇒ Sinx – Sin3x = 2 Cos(x + 3x)/2 Sin(x – 3x)/2 = 2 Cos(4x/2) Sin(- 2x/2) = 2 Cos2x Sin(-x)
⇒ Sinx – Sin3x = 2 Cos2x Sin(-x)

Replacing this value in [Equation 1]
⇒ 2 Cos2x Sin(-x) / Sin2x – Cos2x

Using formula Cos2A = Cos2A – Sin2A
Cos2x = Cos2x – Sin2x

⇒ 2 (Cos2x – Sin2x) Sin(-x) / Sin2x – Cos2x
⇒ – 2 (Sin2x – Cos2x) Sin(-x) / Sin2x – Cos2x
⇒ – 2 Sin(-x)

Using formula Sin(- A) = – SinA for any angle A
⇒ – 2 (- Sinx) = 2 Sinx which is equals to Right Hand Side of given Trigonometric Equation.

21. Prove that (Cos4x + Cos3x + Cos2x)/(Sin4x + Sin3x + Sin2x) = Cot3x

Let’s simplify Left Hand Side of this given Trigonometric Equation
⇒ (Cos4x + Cos3x + Cos2x)/(Sin4x + Sin3x + Sin2x)

⇒ [(Cos4x + Cos2x) + Cos3x] / [(Sin4x + Sin2x) + Sin3x] [Equation 1]

Using formulas
CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2
SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2

⇒ Cos4x + Cos2x = 2 Cos(4x + 2x)/2 Cos(4x – 2x)/2 = 2 Cos(6x/2) Cos(2x/2) = 2 Cos3x Cosx
⇒ Cos4x + Cos2x = 2 Cos3x Cosx

⇒ Sin4x + Sin2x = 2 Sin(4x + 2x)/2 Cos(4x – 2x)/2 = 2 Sin(6x/2) Cos(2x/2) = 2 Sin3x Cosx
⇒ Sin4x + Sin2x = 2 Sin3x Cosx

Replacing these values in [Equation 1]
⇒ [2 Cos3x Cosx + Cos3x] / [2 Sin3x Cosx + Sin3x]
⇒ [2 Cos3x (Cosx + 1)] / [2 Sin3x (Cosx + 1)]
⇒ 2 Cos3x / 2 Sin3x = Cos3x/Sin3x = Cot3x
Which is equal to Right Hand Side of given Trigonometric Equation.

22. Prove that Cotx Cot2x – Cot2x Cot3x – Cot3x Cotx = 1

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ Cotx Cot2x – Cot2x Cot3x – Cot3x Cotx
⇒ [Cotx Cot2x – Cot3x Cotx] – Cot2x Cot3x

⇒ Cotx[Cot2x – Cot3x] – Cot2x Cot3x
⇒ Cotx[Cos2x/Sin2x – Cos3x/Sin3x] – Cot2x Cot3x
⇒ Cotx[(Sin3x Cos2x – Cos3x Sin2x)/Sin2x Sin3x] – Cot2x Cot3x [Equation 1]

Using formula SinA CosB – CosA SinB = Sin(A – B)
⇒ Sin3x Cos2x – Cos3x Sin2x = Sin(3x – 2x) = Sinx
⇒ Sin3x Cos2x – Cos3x Sin2x = Sinx

Replacing this value in [Equation 1]
⇒ Cotx[Sinx/Sin2x Sin3x] – Cot2x Cot3x
⇒ Cosx/Sinx[Sinx/Sin2x Sin3x] – Cot2x Cot3x
⇒ Cosx/Sin2x Sin3x – Cot2x Cot3x

⇒ Cosx/Sin2x Sin3x – Cos2x Cos3x/Sin2x Sin3x
⇒ (Cosx – Cos2x Cos3x)/Sin2x Sin3x

Multiplying Numerator and Denominator of this equation by 2
⇒ [2 Cosx – 2 Cos2x Cos3x]/ 2Sin2x Sin3x [Equation 2]

Using formulas
2 CosA CosB = Cos(A + B) + Cos(A – B)
2 SinA SinB = Cos(A – B) – Cos(A + B)

⇒ 2 Cos2x Cos3x = Cos(2x + 3x) + Cos(2x – 3x) = Cos5x + Cos(-x)
⇒ 2 Cos2x Cos3x = Cos5x + Cos(-x)

⇒ 2 Sin2x Sin3x = Cos(2x – 3x) – Cos(2x + 3x) = Cos(-x) – Cos5x
⇒ 2 Sin2x Sin3x = Cos(-x) – Cos5x

Replacing these values in [Equation 2]
⇒ [2 Cosx – (Cos5x + Cos(-x))]/(Cos(-x) – Cos5x)
⇒ [2 Cosx – Cos5x – Cos(-x)]/(Cos(-x) – Cos5x) [Equation 3]

Using formula Cos(-A) = CosA for any angle A
⇒ Cos(-x) = Cosx

Replacing this value in [Equation 3]
⇒ [2 Cosx – Cos5x – Cosx]/(Cosx – Cos5x)
⇒ [Cosx – Cos5x]/(Cosx – Cos5x)
⇒ 1 which is equals to Right Hand Side of Trigonometric Equation in the question.

Hence it’s proved that Cotx Cot2x – Cot2x Cot3x – Cot3x Cotx is equals to 1

23. Prove that Tan4x = 4 Tanx (1 – Tan2x)/1 – 6 Tan2x + Tan4x

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24. Prove that Cos4x = 1 – 8 Sin2x Cos2x

Let’s simplify Left Hand Side of this given Trigonometric Equation
⇒ Cos4x

Using formula Cos2A = 2 Cos2A – 1
Replacing A = 2x
⇒ Cos2(2x) = 2 Cos2(2x) – 1
⇒ Cos4x = 2 Cos2(2x) – 1 [Equation 1]

Using same Cos2A = 2 Cos2A – 1 again
Replacing A = x
⇒ Cos2x = 2 Cos2x – 1

Replacing value of Cos2x in [Equation 1]
⇒ Cos4x = 2 (2 Cos2x – 1)2 – 1

Using formula (a – b)2 = a2 + b2 – 2ab to simplify (2 Cos2x – 1)2
⇒ Cos4x = 2 (4 Cos4x + 1 – 4 Cos2x) – 1

⇒ Cos4x = 8 Cos4x + 2 – 8 Cos2x – 1
⇒ Cos4x = 8 Cos4x – 8 Cos2x + 1

⇒ Cos4x = 8 Cos2x(Cos2x – 1) + 1
⇒ Cos4x = – 8 Cos2x(1 – Cos2x) + 1 [Equation 2]

Using formula Sin2A + Cos2A = 1 for any angle A
Replacing A = x
Sin2x + Cos2x = 1
⇒ Sin2x = 1 – Cos2x

Replacing 1 – Cos2x with Sin2x in [Equation 2]
⇒ Cos4x = – 8 Cos2x Sin2x + 1
⇒ Cos4x = 1 – 8 Cos2x Sin2x

Hence its proved that Cos4x is equals to 1 – 8 Sin2x Cos2x

25. Cos6x = 32 Cos6x – 48 Cos4x + 18 Cos2x – 1

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ Cos6x

Using formula Cos3A = 4 Cos3A – 3 CosA
Replacing A = 2x
⇒ Cos3(2x) = 4 Cos3(2x) – 3 Cos(2x)
⇒ Cos6x = 4 Cos3(2x) – 3 Cos(2x)
⇒ Cos6x = Cos(2x)[4 Cos2(2x) – 3] [Equation 1]

Using formula Cos2A = 2 Cos2A – 1
Replacing A = x
⇒ Cos2x = 2 Cos2x – 1

Replacing Cos2x with 2 Cos2x – 1 in [Equation 1]
⇒ Cos6x = [2 Cos2x – 1][4 (2 Cos2x – 1)2 – 3]

Using formula (a – b)2 = a2 + b2 – 2ab
To simplify (2 Cos2x – 1)2

⇒ Cos6x = [2 Cos2x – 1][4 (4 Cos4x + 1 – 4 Cos2x) – 3]
⇒ Cos6x = [2 Cos2x – 1][16 Cos4x + 4 – 16 Cos2x – 3]

⇒ Cos6x = [2 Cos2x – 1][16 Cos4x – 16 Cos2x + 1]
⇒ Cos6x = 2 Cos2x[16 Cos4x – 16 Cos2x + 1] – 1[16 Cos4x – 16 Cos2x + 1]

⇒ Cos6x = 32 Cos6x – 32 Cos4x + 2 Cos2x – 16 Cos4x + 16 Cos2x – 1
⇒ Cos6x = 32 Cos6x – 48 Cos4x + 18 Cos2x – 1

Hence it’s proved that Cos6x is equals to 32 Cos6x – 48 Cos4x + 18 Cos2x – 1

## Exercise 3.4

1. Find the principal and general solutions of the following equations.

1. Tanx = √3
2. Secx = 2
3. Cotx = – √3
4. Cosecx = – 2

Summary of this whole question

Let’s solve each of these one by one.

1. Tanx = √3
Principal Solution of Tanx = √3
Value of Tan(𝛑/3) = √3 and Tan(4𝛑/3) = √3

⇒ Tanx = √3 = Tan(𝛑/3) = Tan(4𝛑/3)

As both 𝛑/3 and 4𝛑/3 lies in interval [0, 2𝛑) thus both of these are Principal Solutions of Tanx = √3.

Hence Principal Solution of Tanx = √3 is 𝛑/3 and 4𝛑/3.

General Solution of Tanx = √3
Using formula General Solution of Tanx = Tanθ is x = n𝛑 + θ where θ ∈ (- 𝛑/2, 𝛑/2) and n ∈ I
Tanx = √3 = Tan(𝛑/3)

Tanx = √3 = Tan(𝛑/3) as θ = 𝛑/3 ∈ (- 𝛑/2, 𝛑/2]
⇒ General Solution of Tanx = √3 is x = n𝛑 + 𝛑/3 = (3n + 1)𝛑/2

Thus General Solution of Tanx = √3 is x = (3n + 1)𝛑/2 where θ ∈ (- 𝛑/2, 𝛑/2)

2. Secx = 2

Principal Solution of Secx = 2
Secx = 2 can be rewritten as 1/Cosx = 2
Cosx = 1/2

Principal Solution of Secx = 2 will be same as Principal Solution of Cosx = 1/2
Value of Cos(𝛑/3) = Cos(5𝛑/3) = 1/2

As both 𝛑/3 and 5𝛑/3 lies in interval [0, 2𝛑) thus both of these are Principal Solutions of Secx = 2

Hence Principal Solution of Secx = 2 is 𝛑/3 and 5𝛑/3.

General Solution of Secx = 2
Secx = 2 can be rewritten as 1/Cosx = 2
Cosx = 1/2

Using formula General Solution of Cosx = Cosθ is x = 2n𝛑 ± θ where θ ∈ (0, 𝛑] and n ∈ I
⇒ General Solution of Cosx = 1/2 = Cos(𝛑/3) is x = 2n𝛑 ± 𝛑/3 = (6n ± 1)𝛑/3

Hence General Solution of Secx = 2 is x = (6n ± 1)𝛑/3 where θ ∈ (0, 𝛑] and n ∈ I

3. Cotx = – √3

Principal and General Solution both of Cotx = – √3 will be same as
Principal and General Solution of Tanx = – 1/√3

⇒ Value of Tan(5𝛑/6) = Tan(11𝛑/6) = – 1/√3

As both 5𝛑/6 and 11𝛑/6 lies in interval [0, 2𝛑) thus both of these are Principal Solutions of Cotx = – √3

Hence Principal Solution of Cotx = – √3 is 5𝛑/6 and 11𝛑/6

General Solution of Cotx = – √3
Cotx = – √3 can be rewritten as 1/Tanx = – √3
⇒ Tanx = – 1/√3

Using formula General Solution of Tanx = Tanθ is x = n𝛑 + θ where θ ∈ (- 𝛑/2, 𝛑/2) and n ∈ I
⇒ Tanx = – 1/√3 = Tan(- 𝛑/6)

As – 𝛑/6 lies in interval (- 𝛑/2, 𝛑/2)
Thus
General Solution of Tanx = – 1/√3 = Tan(- 𝛑/6) is x = n𝛑 + (- 𝛑/6) = (6n – 1)𝛑/6

Hence General Solution of Cotx = – √3 is x = (6n – 1)𝛑/6 where n ∈ I

4. Cosecx = – 2
Both Principal and General Solutions of Cosecx = – 2 will be
same as Principal and General Solutions of Sinx = – 1/2

⇒ Sinx = Sin(7𝛑/6) = Sin(11𝛑/6) = – 1/2

As both 7𝛑/6 and 11𝛑/6 lies in interval [0, 2𝛑) thus both of these are Principal Solutions of Cosecx = – 2

Hence Principal Solutions of Cosecx = – 2 are 7𝛑/6 and 11𝛑/6

General Solution of Cosecx = – 2 will be
same as General Solution of Sinx = – 1/2

⇒ Sinx = – 1/2 = Sin(- 𝛑/6)

Using formula General Solution of Sinx = Sinθ is x = n𝛑 + (-1)nθ where θ ∈ [- 𝛑/2, 𝛑/2] and n ∈ I

⇒ General Solution of Sinx = Sin(- 𝛑/6) is x = n𝛑 + (-1)n(- 𝛑/6) = [6n + (-1)n]𝛑/6

Hence General Solution of Cosecx = – 2 is x = [6n + (-1)n]𝛑/6 where n ∈ I

Summary of this whole question

5. Find General Solution of Cos4x = Cos2x

For find out General Solution of Trigonometric Equation Cos4x = Cos2x
We first need to simplify it
⇒ Cos4x = Cos2x

⇒ Cos4x – Cos2x = 0 [Equation 1]

Using formula CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2
Replacing
C = 4x
D = 2x

⇒ Cos4x – Cos2x = – 2 Sin(4x + 2x)/2 Sin(4x – 2x)/2 = – 2 Sin(6x/2) Sin(2x/2) = – 2 Sin3x Sinx

⇒ Cos4x – Cos2x = – 2 Sin3x Sinx

Replacing value of Cos4x – Cos2x with – 2 Sin3x Sinx in [Equation 1]
⇒ – 2 Sin3x Sinx = 0

⇒ Either Sin3x = 0 OR Sinx = 0
Let’s find out General Solution for each of these.

General Solution of Trigonometric Equation Sinθ = 0 is θ = n𝛑 where θ ∈ I
⇒ General Solution of Sin3x = 0 is 3x = n𝛑
⇒ General Solution of Sin3x = 0 is x = n𝛑/3 where n ∈ I

⇒ General Solution of Sinx = 0 is x = n𝛑 where n ∈ I

Hence General Solution of Trigonometric Equation Cos4x = Cos2x is x = n𝛑/3 or n𝛑 where n ∈ I

6. Find General Solution of Trigonometric Equation Cos3x + Cosx – Cos2x = 0

Let’s first simplify this Trigonometric Equation Cos3x + Cosx – Cos2x = 0
⇒ Cos3x + Cosx – Cos2x = 0
⇒ [Cos3x + Cosx] – Cos2x = 0 [Equation 1]

Using formula CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2
Replacing
C = 3x
D = x

⇒ Cos3x + Cosx = 2 Cos(3x + x)/2 Cos(3x – x)/2 = 2 Cos(4x/2) Cos(2x)/2 = 2 Cos2x Cosx
⇒ Cos3x + Cosx = 2 Cos2x Cosx

Replacing Cos3x + Cosx in [Equation 1] with 2 Cos2x Cosx
⇒ 2 Cos2x Cosx – Cos2x = 0
⇒ Cos2x(2 Cosx – 1) = 0

⇒ Either Cos2x = 0 OR 2 Cosx – 1 = 0
Let’s find out General Solutions of each of these Trigonometric Equations

Hence General Solution of Cos3x + Cosx – Cos2x = 0 is x = (2n + 1)𝛑/4 or x = (2n ± 1)𝛑/3 where n ∈ I

7. Find General Solution of Trigonometric Equation Sin2x + Cosx = 0

Let’s first simplify this Trigonometric Equation and then find out General Solution.
⇒ Sin2x + Cosx = 0 [Equation 1]

Using formula Sin2A = 2 SinA CosA
Replacing A = x
⇒ Sin2x = 2 Sinx Cosx

Replacing Sin2x in [Equation 1] with 2 Sinx Cosx
⇒ 2 Sinx Cosx + Cosx = 0
⇒ Cosx(2 Sinx + 1) = 0

⇒ Either Cosx = 0 OR 2 Sinx + 1 = 0
Let’s not find out General Solutions of each of these equations.

Hence General Solution of Sin2x + Cosx = 0 is x = (2n + 1)𝛑/2 or [6n + 7(-1)n]𝛑/6 where n ∈ I

8. Find General Solution of Trigonometric Equation Sec2(2x) = 1 – Tan(2x)

Let’s start solving this question by simplifying
Sec2(2x) = 1 – Tan(2x) [Equation 1]

Using formula Sec2A = 1 + Tan2A
Replacing
A = 2x
⇒ Sec2(2x) = 1 + Tan2(2x)

Replacing Sec2(2x) with 1 + Tan2(2x) in [Equation 1]
⇒ 1 + Tan2(2x) = 1 – Tan(2x)

⇒ 1 + Tan2(2x) – 1 + Tan(2x) = 0
⇒ Tan2(2x) + Tan(2x) = 0
⇒ Tan(2x)[Tan(2x) + 1] = 0

⇒ Either Tan(2x) = 0 OR Tan(2x) + 1 = 0

Let’s find out General Solutions of each of these Trigonometric Equations.

Hence General Solution of Sec2(2x) = 1 – Tan(2x) is x = n𝛑/2 or (4n – 1)𝛑/8 where n ∈ I

9. Find General Solution of Trigonometric Equation Sinx + Sin3x + Sin5x = 0

Let’s first simplify this Trigonometric Equation and then find out General Solutions.
⇒ Sinx + Sin3x + Sin5x = 0
⇒ [Sinx + Sin5x] + Sin3x = 0 [Equation 1]

Using formula SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
Replacing
C = x
D = 5x

⇒ Sinx + Sin5x = 2 Sin(x + 5x)/2 Cos(x – 5x)/2 = 2 Sin(6x/2) Cos(- 4x/2) = 2 Sin3x Cos(- 2x)
⇒ Sinx + Sin5x = 2 Sin3x Cos(- 2x)

Replacing Sinx + Sin5x in [Equation 1] with 2 Sin3x Cos(- 2x)
⇒ 2 Sin3x Cos(- 2x) + Sin3x = 0

Using formula Cos(- A) = CosA for any value of angle A
⇒ 2 Sin3x Cosx + Sin3x = 0
⇒ Sin3x (2 Cosx + 1) = 0

⇒ Either Sin3x = 0 OR 2 Cosx + 1 = 0
Let’s now find out General Solutions of each of these equations

Hence General Solution of Sinx + Sin3x + Sin5x = 0 is x = n𝛑/3 or (3n ± 1)2𝛑/3 where n ∈ I

## Miscellaneous Exercise Chapter 3

1. Prove that 2Cos(𝛑/13) Cos(9𝛑/13) + Cos(3𝛑/13) + Cos(5𝛑/13) = 0

Let’s simplify Left Hand Side of this trigonometric equation
⇒ 2Cos(𝛑/13) Cos(9𝛑/13) + Cos(3𝛑/13) + Cos(5𝛑/13) [Equation 1]

Using formula 2 CosA CosB = Cos(A + B) + Cos(A – B)
Replacing
A = 𝛑/13
B = 9𝛑/13
⇒ 2Cos(𝛑/13) Cos(9𝛑/13) = Cos(𝛑/13 + 9𝛑/13) + Cos(𝛑/13 – 9𝛑/13) = Cos(10𝛑/13) + Cos(- 8𝛑/13)
⇒ 2Cos(𝛑/13) Cos(9𝛑/13) = Cos(10𝛑/13) + Cos(- 8𝛑/13)

Replacing 2Cos(𝛑/13) Cos(9𝛑/13) in [Equation 1] with Cos(10𝛑/13) + Cos(- 8𝛑/13)
⇒ Cos(10𝛑/13) + Cos(- 8𝛑/13) + Cos(3𝛑/13) + Cos(5𝛑/13)

Above Trigonometric Equation can be rewritten as
⇒ [Cos(10𝛑/13) + Cos(5𝛑/13)] + [Cos(- 8𝛑/13) + Cos(3𝛑/13)] [Equation 2]

Using formula CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2
⇒ Cos(10𝛑/13) + Cos(5𝛑/13) = 2 Cos(10𝛑/13 + 5𝛑/13)/2 Cos(10𝛑/13 – 5𝛑/13)/2

⇒ Cos(- 8𝛑/13) + Cos(3𝛑/13) = 2 Cos(- 8𝛑/13 + 3𝛑/13)/2 Cos(- 8𝛑/13 – 3𝛑/13)/2

Replacing these values in [Equation 2]
⇒ 2 Cos(15𝛑/26) Cos(5𝛑/26) + 2 Cos(- 5𝛑/26) Cos(- 11𝛑/26) [Equation 3]

Using formula Cos(- A) = CosA for any angle A
⇒ Cos(- 5𝛑/26) = Cos(5𝛑/26)
⇒ Cos(- 11𝛑/26) = Cos(11𝛑/26)

Replacing these values in [Equation 3]
⇒ 2 Cos(15𝛑/26) Cos(5𝛑/26) + 2 Cos(5𝛑/26) Cos(- 11𝛑/26)
⇒ 2 Cos(5𝛑/26)[Cos(15𝛑/26) + Cos(11𝛑/26)] [Equation 4]

Using formula CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2
⇒ Cos(15𝛑/26) + Cos(11𝛑/26) = 2 Cos(15𝛑/26 + 11𝛑/26)/2 Cos(15𝛑/26 – 11𝛑/26)/2

Replacing Cos(15𝛑/26) + Cos(11𝛑/26) in [Equation 4] with 2 Cos(15𝛑/26 + 11𝛑/26)/2 Cos(15𝛑/26 – 11𝛑/26)/2

⇒ 2 Cos(5𝛑/26)[2 Cos(26𝛑/26)/2 Cos(4𝛑/26)/2]

⇒ 2 Cos(5𝛑/26)[2 Cos(𝛑/2) Cos(4𝛑/52)]

Value of Cos(𝛑/2) = 0

⇒ 2 Cos(5𝛑/26)[2 Cos(𝛑/2) Cos(4𝛑/52)] = 0

This means that 2Cos(𝛑/13) Cos(9𝛑/13) + Cos(3𝛑/13) + Cos(5𝛑/13) is equals to 0.

2. Prove that (Sin3x + Sinx) Sinx + (Cos3x – Cosx) Cosx = 0

Let’s simplify Left Hand Side of this Trigonometric Equations
⇒ (Sin3x + Sinx) Sinx + (Cos3x – Cosx) Cosx [Equation 1]

Using formula SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
⇒ Sin3x + Sinx = 2 Sin(3x + x)/2 Cos(3x – x)/2 = 2 Sin(4x/2) Cos(2x/2) = 2 Sin2x Cosx
⇒ Sin3x + Sinx = 2 Sin2x Cosx

Using formula CosC – CosD = – 2 Sin(C + D)/2 Sin(C – D)/2
⇒ Cos3x – Cosx = – 2 Sin(3x + x)/2 Sin(3x – x)/2 = – 2 Sin(4x/2) Sin(2x/2) = – Sin2x Sinx
⇒ Cos3x – Cosx = – Sin2x Sinx

Replacing these values in [Equation 1]
⇒ 2 Sin2x Cosx Sinx – 2 Sin2x Sinx Cosx = 0

Hence (Sin3x + Sinx) Sinx + (Cos3x – Cosx) Cosx is equals to 0.

3. Prove that (Cosx + Cosy)2 + (Sinx – Siny)2 = 4 Cos2(x + y)/2

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ (Cosx + Cosy)2 + (Sinx – Siny)2 [Equation 1]

Using formulas
(a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab

⇒ (Cosx + Cosy)2 = Cos2x + Cos2y + 2 Cosx Cosy
⇒ (Sinx – Siny)2 = Sin2x + Sin2y – 2 Sinx Cosy

Replacing these values in [Equation 1]
⇒ Cos2x + Cos2y + 2 Cosx Cosy + Sin2x + Sin2y – 2 Sinx Cosy

⇒ (Cos2x + Sin2x) + (Cos2y + Sin2y) + 2 Cosx Cosy – 2 Sinx Siny

Using formula for Cos2A + Sin2A = 1 for any angle A
⇒ 1 + 1 + 2 Cosx Cosy – 2 Sinx Cosy
⇒ 2 + 2 [Cosx Cosy – Sinx Siny]

Using formula CosA CosB – SinA SinB = Cos(A + B)
⇒ 2 + 2 Cos(x + y)
⇒ 2[1 + Cos(x + y)] [Equation 2]

Using formula Cos2θ = 2 Cos2θ – 1
Replacing
θ = (x + y)/2
⇒ Cos2(x + y)/2 = 2 Cos2(x + y)/2 – 1
⇒ Cos(x + y) = 2 Cos2(x + y)/2 – 1

Replacing this value in [Equation 2]
⇒ 2[1 + 2 Cos2(x + y)/2 – 1]
⇒ 2[2 Cos2( x + y)/2]
⇒ 4 Cos2(x + y)/2

Hence it’s proved that (Cosx + Cosy)2 + (Sinx – Siny)2 is equals to 4 Cos2(x + y)/2

4. Prove that (Cosx – Cosy)2 + (Sinx – Siny)2 = 4 Sin2(x – y)/2

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ (Cosx – Cosy)2 + (Sinx – Siny)2 [Equation 1]

Using formulas
(a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab

⇒ (Cosx – Cosy)2 = Cos2x + Cos2y – 2 Cosx Cosy
⇒ (Sinx – Siny)2 = Sin2x + Sin2y – 2 Sinx Siny

Replacing these values in [Equation 1]
⇒ Cos2x + Cos2y – 2 Cosx Cosy + Sin2x + Sin2y – 2 Sinx Siny

⇒ [Cos2x + Sin2x] + [Cos2y + Sin2y] – 2 Cosx Cosy – 2 Sinx Siny

Using formula for Cos2A + Sin2A = 1 for any angle A
⇒ 1 + 1 – 2 Cosx Cosy – 2 Sinx Cosy

⇒ 2 – 2[Cosx Cosy + Sinx Siny] [Equation 2]

Using formula CosA CosB + SinA SinB = Cos(A – B)
Replacing
A = x
B = y

⇒ Cosx Cosy + Sinx Siny = Cos(x – y)

Replacing this value in [Equation 2]
⇒ 2 – 2 Cos(x – y)
⇒ 2 [1 – Cos(x – y)] [Equation 3]

Using formula Cos2A = 1 – 2 Sin2A
Replacing
A = (x – y)/2
⇒ Cos2(x – y)/2 = 1 – 2 Sin2(x – y)/2
⇒ Cos(x – y) = 1 – 2 Sin2(x – y)/2

Replacing this value in [Equation 3]
⇒ 2 [1 – (1 – 2 Sin2(x – y)/2)]

⇒ 2 [1 – 1 + 2 Sin2(x – y)/2]

⇒ 2[2 Sin2(x – y)/2]

⇒ 4 Sin2(x – y)/2
which is equals to Right Hand Side of the given Trigonometric Equation.

5. Prove that Sinx + Sin3x + Sin5x + Sin7x = 4 Cosx Cos2x Cos4x

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ Sinx + Sin3x + Sin5x + Sin7x [Equation 1]

Using formula SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
⇒ Sinx + Sin3x = 2 Sin(x + 3x)/2 Cos(x – 3x)/2 = 2 Sin(4x/2) Cos(- 2x/2) = 2 Sin2x Cos(-x)
⇒ Sinx + Sin3x = 2 Sin2x Cos(-x)

⇒ Sin5x + Sin7x = 2 Sin(5x + 7x)/2 Cos(5x – 7x)/2 = 2 Sin(12x/2) Cos(-2x/2) = 2 Sin6x Cos(-x)
⇒ Sin5x + Sin7x = 2 Sin6x Cos(-x)

Replacing these values in [Equation 1]
⇒ 2 Sin2x Cos(-x) + 2 Sin6x Cos(-x)

Using formula Cos(- A) = CosA for any angle A
⇒ 2 Sin2x Cosx + 2 Sin6x Cosx

⇒ 2 Cosx[Sin2x + Sin6x] [Equation 2]

Again using SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2 formula for simplifying Sin2x + Sin6x
⇒ Sin2x + Sin6x = 2 Sin(2x + 6x)/2 Cos(2x – 6x)/2 = 2 Sin(8x/2) Cos(- 4x/2) = 2 Sin4x Cos(- 2x)
⇒ Sin2x + Sin8x = 2 Sin4x Cos(- 2x)

Replacing this value in [Equation 2]
⇒ 2 Cosx[2 Sin4x Cos(- 2x)]

Using formula Cos(- A) = CosA for any angle A
⇒ 2 Cosx [2 Sin4x Cos2x]
⇒ 4 Cosx Cos2x Sin4x
which is equals to Right Hand Side of the given Trigonometric Equation.

6. Prove that (Sin7x + Sin5x) + (Sin9x + Sin3x)/(Cos7x + Cos5x) + (Cos9x + 3x) = Tan6x

Let’s simplify each part of Left Hand Side of this Trigonometric Equation
⇒ (Sin7x + Sin5x) + (Sin9x + Sin3x)/(Cos7x + Cos5x) + (Cos9x + Cos3x) [Equation 1]

Using formula SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2
and CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2

⇒ Sin7x + Sin5x = 2 Sin(7x + 5x)/2 Cos(7x – 5x)/2 = 2 Sin(12x/2) Cos(2x/2) = 2 Sin6x Cosx
⇒ Sin7x + Sin5x = 2 Sin6x Cosx

⇒ Sin9x + Sin3x = 2 Sin(9x + 3x)/2 Cos(9x – 3x)/2 = 2 Sin(12x/2) Cos(6x/2) = 2 Sin6x Cos3x
⇒ Sin9x + Sin3x = 2 Sin6x Cos3x

⇒ Cos7x + Cos5x = 2 Cos(7x + 5x)/2 Cos(7x – 5x)/2 = 2 Cos(12x/2) Cos(2x/2) = 2 Cos6x Cosx
⇒ Cos7x + Cos5x = 2 Cos6x Cosx

⇒ Cos9x + Cos3x = 2 Cos(9x + 3x)/2 Cos(9x – 3x)/2 = 2 Cos(12x/2) Cos(6x/2) = 2 Cos6x Cos3x
⇒ Cos9x + Cos3x = 2 Cos6x Cos3x

Replacing all these values in [Equation 1]
⇒ 2 Sin6x Cosx + 2 Sin6x Cos3x / 2 Cos6x Cosx + 2 Cos6x Cos3x
⇒ 2(Sin6x Cosx + Sin6x Cos3x) / 2(Cos6x Cosx + Cos6x Cos3x)

⇒ Sin6x Cosx + Sin6x Cos3x / Cos6x Cosx + Cos6x Cos3x
⇒ Sin6x(Cosx + Cos3x) / Cos6x(Cosx + Cos3x)

⇒ Sin6x / Cos6x = Tan6x

Hence (Sin7x + Sin5x) + (Sin9x + Sin3x)/(Cos7x + Cos5x) + (Cos9x + Cos3x) is equals to Tan6x

7. Prove that Sin3x + Sin2x – Sinx = 4 Sinx Cos(x/2) Cos(3x/2)

Let’s simplify Left Hand Side of this Trigonometric Equation
⇒ Sin3x + Sin2x – Sinx
⇒ [Sin3x – Sinx] + Sin2x

Using formula SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2
⇒ 2 Cos(3x + x)/2 Sin(3x – x)/2 + Sin2x
⇒ 2 Cos(4x/2) Sin(2x/2) + Sin2x
⇒ 2 Cos2x Sinx + Sin2x [Equation 1]

Using Sin2A = 2 SinA CosA for any angle A
⇒ Sin2x = 2 Sinx Cosx

Replacing this value in [Equation 1]
⇒ 2 Cos2x Sinx + 2 Sinx Cosx
⇒ 2 Sinx(Cos2x + Cosx)

Using formula CosC + CosD = 2 Cos(C + D)/2 Cos(C – D)/2
⇒ 2 Sinx[2 Cos(2x + x)/2 Cos(2x – x)/2]

⇒ 2 Sinx[2 Cos(3x/2) Cos(x/2)]

⇒ 4 Sinx Cos(x/2) Cos(3x/2)
Which is equals to Right Hand Side of the given Trigonometric Equation.

Hence it’s proved that Sin3x + Sin2x – Sinx is equals to 4 Sinx Cos(x/2) Cos(3x/2)

8. Find value of Sin(x/2), Cos(x/2) and Tan(x/2) if Tanx = – 4/3 and x is in quadrant II

As per question it’s given that Tanx = – 4/3 and x is in quadrant II

Using formula Sec2A = 1 + Tan2A
Replacing A = x
Sec2x = 1 + Tan2x

Sec2x = 1 + (-4/3)2 = 1 + 16/9 = 9 + 16/9 = 25/9
Sec2x = 25/9
Secx = ± √25/9 = ± 5/3
Secx = ± 5/3

⇒ Using formula Cosx = 1/Secx = ± 3/5
⇒ Cosx = ± 3/5

As per question it’s given that x is in quadrant II this means that Cosx must be negative
Cosx = – 3/5

Using formula Cos2A = 2 Cos2A – 1
Replacing A = x/2
⇒ Cos2(x/2) = 2 Cos2(x/2) – 1
⇒ Cosx = 2 Cos2(x/2) – 1

Putting value of Cosx = – 3/5 in this formula
⇒ – 3/5 = 2 Cos2(x/2) – 1
– 3/5 + 1 = 2 Cos2(x/2)
2/5 = 2 Cos2(x/2)
1/5 = Cos2(x/2)
Cos(x/2) = ± √1/5 = ± 1/√5
Cos(x/2) = ± 1/√5

As per question x is in quadrant II so x/2 will be in quadrant I ⇒ Cos(x/2) must be positive
Cos(x/2) = 1/√5

Using formula Sin2A + Cos2A = 1
Replacing A = x/2
⇒ Sin2(x/2) + Cos2(x/2) = 1
⇒ Sin2(x/2) + (1/√5)2 = 1
Sin2(x/2) + 1/5 = 1
Sin2(x/2) = 1 – 1/5 = 4/5
Sin2(x/2) = 4/5
Sin(x/2) = ± √4/5 = ± 2/√5
Sin(x/2) = ± 2/√5
As per question x is in quadrant II so x/2 will be in quadrant I ⇒ Sin(x/2) must be positive
Sin(x/2) = 2/√5

Using formula TanA = SinA/CosA
Replacing A = x/2
⇒ Tan(x/2) = Sin(x/2) / Cos(x/2) = 1/√5 / 2/√5 = 1/√5 × √5/2 = 1/2
Tan(x/2) = 1/2

Summary of this question

9. Find value of Sin(x/2), Cos(x/2) and Tan(x/2) if Cosx = – 1/3 and x is in quadrant III

As per question it’s given that Cosx = – 1/3 and x is in quadrant III

Using formula Cos2A = 2 Cos2A – 1
Replacing A = x/2
⇒ Cos2(x/2) = 2 Cos2(x/2) – 1
Cosx = 2 Cos2(x/2) – 1

– 1/3 = 2 Cos2(x/2) – 1
– 1/3 + 1 = 2 Cos2(x/2)
– 1 + 3/3 = 2 Cos2(x/2)
2/3 = 2 Cos2(x/2)
1/3 = Cos2(x/2)

Cos(x/2) = ± √1/3 = ± 1/√3
Cos(x/2) = ± 1/√3

As per question it’s given that x is in quadrant III this means that x/2 will be in quadrant II
⇒ Cos(x/2) must be negative

Cos(x/2) = – 1/√3

Using formula Sin2A + Cos2A = 1
Replacing A = x/2
⇒ Sin2(x/2) + Cos2(x/2) = 1
Sin2(x/2) + (- 1/√3)2 = 1
Sin2(x/2) + 1/3 = 1
Sin2(x/2) = 1 – 1/3 = 2/3
Sin2(x/2) = 2/3
Sin(x/2) = ± √2/3 = ± √2/√3
Sin(x/2) = ± √2/√3

As per question it’s given that x is in quadrant III this means that x/2 will be in quadrant II
⇒ Sin(x/2) must be positive

Sin(x/2) = √2/√3

Using formula TanA = SinA / CosA
Replacing A = x/2
Tan(x/2) = Sin(x/2) / Cos(x/2) = √2/√3 / – 1/√3 = – √2

Tan(x/2) = – √2

Summary of this question

10. Find value of Sin(x/2), Cos(x/2) and Tan(x/2) if Sinx = 1/4 and x is in quadrant II

As per question it’s given that Sinx = 1/4 and x is in quadrant II

Using formula Sin2A + Cos2A = 1
Replacing A = x
⇒ Sin2x + Cos2x = 1
(1/4)2 + Cos2x = 1
1/16 + Cos2x = 1
Cos2x = 1 – 1/16 = 15/16
Cos2x = 15/16
Cosx = ± √15/16 = ± √15/√16 = ± √15/4
Cosx = ± √15/4

As per question it’s given that x is in quadrant II this means
Cosx must be negative

Cosx = – √15/4

Using formula Cos2A = 1 – 2 Sin2A
Replacing A = x/2
⇒ Cos2(x/2) = 1 – 2 Sin2(x/2)
Cosx = 1 – 2 Sin2(x/2)

\begin{equation} Cosx = 1 – 2 Sin^{2}\left(\frac{x}{2}\right) \\ – \frac{\sqrt{15}}{4} = 1 – 2 Sin^{2}\left(\frac{x}{2}\right) \\ – \frac{\sqrt{15}}{4} – 1 = – 2 Sin^{2}\left(\frac{x}{2}\right) \\ \frac{- \sqrt{15} – 4}{4} = – 2 Sin^{2}\left(\frac{x}{2}\right) \\ (-1)\frac{\sqrt{15} + 4}{4} = – 2 Sin^{2}\left(\frac{x}{2}\right) \\ \frac{\sqrt{15} + 4}{4} = 2 Sin^{2}\left(\frac{x}{2}\right) \\ \frac{\sqrt{15} + 4}{4 × 2} = Sin^{2}\left(\frac{x}{2}\right) \\ \frac{\sqrt{15} + 4}{8} = Sin^{2}\left(\frac{x}{2}\right) \\ \pm \sqrt{\frac{\sqrt{15} + 4}{8}} = Sin\left(\frac{x}{2}\right) \\ \end{equation}

As per question it’s given that x is in quadrant II this means that x/2 must lie in quadrant I
⇒ Sin(x/2) must be positive only

\begin{equation} \sqrt{\frac{\sqrt{15} + 4}{8}} = Sin\left(\frac{x}{2}\right) \\ \end{equation}

Using formula Sin2A + Cos2A = 1
Replacing A = x/2
⇒ Sin2(x/2) + Cos2(x/2) = 1

\begin{equation} Sin^{2}\left(\frac{x}{2}\right) + Cos^{2}\left(\frac{x}{2}\right) = 1 \\ \left(\sqrt{\frac{\sqrt{15} + 4}{8}}\right)^{2} + Cos^{2}\left(\frac{x}{2}\right) = 1 \\ \frac{\sqrt{15} + 4}{8} + Cos^{2}\left(\frac{x}{2}\right) = 1 \\ Cos^{2}\left(\frac{x}{2}\right) = 1 – \frac{\sqrt{15} + 4}{8} \\ Cos^{2}\left(\frac{x}{2}\right) = \frac{8 – \sqrt{15} – 4}{8} = \frac{4 – \sqrt{15}}{8} \\ Cos^{2}\left(\frac{x}{2}\right) = \frac{4 – \sqrt{15}}{8} \\ Cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{4 – \sqrt{15}}{8}} \\ \end{equation}

As per question it’s given that x is in quadrant II this means that x/2 must lie in quadrant I
⇒ Cos(x/2) must be positive only

\begin{equation} Cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 – \sqrt{15}}{8}} \\ \end{equation}

Using formula TanA = SinA / CosA
Replacing A = x/2
Tan(x/2) = Sin(x/2) / Cos(x/2)

\begin{equation} Tan\left(\frac{x}{2}\right) = \frac{Sin\left(\frac{x}{2}\right)}{Cos\left(\frac{x}{2}\right)} \\ Tan\left(\frac{x}{2}\right) = \frac{\sqrt{\frac{\sqrt{15} + 4}{8}}}{\sqrt{\frac{4 – \sqrt{15}}{8}}} \\ Tan\left(\frac{x}{2}\right) = \frac{\sqrt{\sqrt{15} + 4}}{\sqrt{4 – \sqrt{15}}} \\ Tan\left(\frac{x}{2}\right) = \frac{\sqrt{4 + \sqrt{15}}}{\sqrt{4 – \sqrt{15}}} \end{equation}

In order to simplify let’s multiply and divide numerator/denominator of this equation by 4 – √15

\begin{equation} Tan\left(\frac{x}{2}\right) = \frac{\sqrt{(4 + \sqrt{15})(4 – \sqrt{15})}}{\sqrt{(4 – \sqrt{15})(4 – \sqrt{15})}} \\ \end{equation}

Using formula (a + b)(a – b) = a2 – b2
⇒ (4 + √15)(4 – √15) = 42 – (√15)2 = 16 – 15 = 1
⇒ (4 + √15)(4 – √15) = 1

Using formula (a – b)(a – b) = (a – b)2 = a2 + b2 -2ab
⇒ (4 – √15)(4 – √15) = (4 – √15)2 = 42 + (√15)2 – 2(4)(√15) = 16 + 15 – 8√15 = 31 – 8√15
⇒ (4 – √15)(4 – √15) = 31 – 8√15

Replacing these values in Tan(x/2) equation

\begin{equation} Tan\left(\frac{x}{2}\right) = \frac{1}{31 – 8\sqrt{15}} \end{equation}