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Trigonometric Functions of Sum and Difference of Two Angles
  • Difficulty Level : Hard
  • Last Updated : 01 Apr, 2021

Trigonometry is a branch of mathematics, which deal with the angles, lengths, and heights of triangles and their relationships. It had played an important role to calculate complex functions or large distances which were not possible to calculate without trigonometry. While solving problems with trigonometry, we came across many situations where we have to calculate the trigonometric solutions for the sum of angles or differences of angles. E.g.

Here, \frac{BC}{AB} = \frac{Opposite \hspace{0.1cm}side} {Adjacent\hspace{0.1cm} side}

Which is a tangent trigonometric ratio, with an angle opposite to BC.

tan(θ+Φ) = \frac{BC}{AB}



If θ = 30° and Φ = 45°. We know the trigonometric angles of 45° and 30°, but we don’t know the trigonometric angle of (45° + 30° = 75°). So, to simplify these types of problems. We will get to learn trigonometric formulae or identities of sum and differences of two angles which will make things easier.

Before moving further first we will see the signs of the trigonometric functions in the four quadrants. These signs play an important role in trigonometry. 

Trigonometric identities

Now we are going to find the trigonometric identities. As we know that

sin(-x) = – sin x

cos(-x) = cos x

Because only cos and sec are positive in the fourth quadrant. So, now we prove some results regarding sum and difference of angles:

Let’s consider a unit circle (having radius as 1) with centre at the origin. Let x be the ∠DOA and y be the ∠AOB. Then (x + y) is the ∠DOB. Also let (– y) be the ∠DOC. 



Therefore, the coordinates of A, B, C and D are

A = (cos x, sin x)

B = [cos (x + y), sin (x + y)]

C = [cos (– y), sin (– y)] 

D = (1, 0).

As, ∠AOB = ∠COD

Adding, ∠BOC both side, we get

∠AOB + ∠BOC = ∠COD + ∠BOC

∠AOC = ∠BOD



In △ AOC and △ BOD

OA = OB (radius of circle)

∠AOC = ∠BOD (Proved earlier)

OC = OD (radius of circle)

△ AOC ≅ △ BOD by SAS congruency.

By using distance formula, for

AC2 = [cos x – cos (– y)]2 + [sin x – sin(–y]2

AC2 = 2 – 2 (cos x cos y – sin x sin y) …………….(i)

And, now

Similarly, using distance formula, we get

BD2 = [1 – cos (x + y)]2 + [0 – sin (x + y)]2

BD2 = 2 – 2 cos (x + y) …………….(ii)

As, △ AOC ≅ △ BOD

AC = BD, So AC2 = BD2

From eq(i) and eq(ii), we get

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

So, 

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

Take y = -y, we get

cos (x + (-y)) = cos x cos (-y) – sin x sin (-y)

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

Now, taking 

cos (\frac{\pi}{2}  -(x + y)) = cos ((\frac{\pi}{2}  -x) – y) (cos (\frac{\pi}{2}  -θ) = sin θ)

sin (x – y) = sin x cos y – cos x sin y

take y = -y, we get

sin (x – (-y)) = sin x cos (-y) – cos x sin (-y)

sin (x + y) = sin x cos y + cos x sin y

The derived formulae for trigonometric ratios of compound angles are as follows:

sin (A + B) = sin A cos B + cos A sin B     ………………..(1)

sin (A – B) = sin A cos B – cos A sin B       ………………..(2)

cos (A + B) = cos A cos B – sin A sin B     ..………………(3)

cos (A – B) = cos A cos B + sin A sin B     ………………..(4)

By using these formulae, we can obtain some important and mostly used form:

(1) Take, A = \mathbf{\frac{\pi}{2}}

In eq(1) and (3), we get

sin (\frac{\pi}{2}  +B) = cos B

cos (\frac{\pi}{2}  +B) = – sin A

(2) Take, A = π

In eq(1), (2), (3) and (4) we get

sin (π + B) = – sin B

sin (π – B) = sin B

cos (π ± B) = – cos B

(3) Take, A = 2π

In eq(2) and (4) we get

sin (2π – B) = – sin B

cos (2π – B) = cos B

Similarly for cot A, tan A, sec A, and cosec A

(4) \mathbf{tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}}

Here, A, B, and (A + B) is not an odd multiple of π/2, so, cosA, cosB and cos(A + B) are non-zero

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

From eq(1) and (3), we get

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

Now divide the numerator and denominator by cos A cos B we get

tan(A + B) = \frac{\frac{sin A cos B}{cosAcosB} + \frac{cos A sin B}{cosAcosB}}{\frac{cos A cos B}{cosAcosB} - \frac{sin A sin B}{cosAcosB}}

tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}

(5) \mathbf{tan(A - B) = \frac{tan A - tan B}{1+tan A tan B}}

As we know that 

tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}

So, on putting B = -B, we get

tan(A - B) = \frac{tan A - tan B}{1+tan A tan B}

(6) \mathbf{cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}}

Here, A, B, and (A + B) is not a multiple of π, so, sinA, sinB and sin(A + B) are non-zero

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

From eq(1) and (3), we get

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

Now divide the numerator and denominator by sin A sin B we get

cot(A + B) = \frac{\frac{cos A cos B}{sinAsinB} - \frac{sin A sin B}{sinAsinB}}{\frac{sin A cos B}{sinAsinB} + \frac{cos A sin B}{sinAsinB}}

cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}

(7) \mathbf{cot(A - B) = \frac{cotAcot B+1}{cot A - cot B}}

As we know that 

cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}

So, on putting B = -B, we get

cot(A - B) = \frac{cotAcot B+1}{cot A - cot B}

Here, we will establish two sets of transformation formulae: Factorization and Defactorization formulae. 

Defactorization Formulae

In trigonometry, defactorisation means converting a product into a sum or difference. The defactorization formulae are:

(1) 2 sin A cos B = sin (A + B) + sin (A – B)

Proof:

As we know that 

sin (A + B) = sin A cos B + cos A sin B     ………………………(1)

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

By adding eq(1) and (2), we get

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

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

Proof:

As we know that 

sin (A + B) = sin A cos B + cos A sin B     ………………………(1)

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

By subtracting eq(2) from (1), we get

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

(3) 2 cos A cos B = cos (A + B) + cos (A – B)

Proof:

As we know that 

cos (A + B) = cos A cos B – sin A sin B     ………………………(1)

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

By adding eq(1) and (2), we get

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

(4) 2 sin A sin B = cos (A – B) – cos (A + B) 

Proof:

cos (A + B) = cos A cos B – sin A sin B     ………………………(1)

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

By subtracting eq(3) from (4), we get

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

Example 1. Convert each of the following products into the sum or difference.

(i) 2 sin 40° cos 30°

(ii) 2 sin 75° sin 15°

(iii) cos 75° cos 15°

Solution:

(i) Given: A = 40° and B = 30°

Now put all these values in the formula, 

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

We get

2 sin 40° cos 30° = sin (40 + 30) + sin (40 – 30) 

= sin (70°) + sin (10°) 

(ii) Given: A = 75° and B = 15°

Now put all these values in the formula, 

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

We get

2 sin 75° sin 15° = cos (75-15) – cos (75+15)

= cos (60°) – cos (90°)

(iii) Given: A = 75° and B = 15°

Now put all these values in the formula, 

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

We get

cos 75° cos 15° = 1/2(cos (75+15) + cos (75-15))

= 1/2 (cos (90°) + cos (60°))

Example 2. Solve for 2cos \frac{\pi}{13}cos \frac{9\pi}{13}+cos \frac{3\pi}{13}+cos \frac{5\pi}{13}

Solution:

Using the formula

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

2cos \frac{\pi}{13}cos \frac{9\pi}{13} =cos (\frac{9\pi}{13}+\frac{\pi}{13})+cos (\frac{9\pi}{13}-\frac{\pi}{13}) + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}     

cos \frac{10\pi}{13}+cos \frac{8\pi}{13} + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}

cos (\pi-\frac{3\pi}{13})+cos (\pi-\frac{5\pi}{13}) + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}

-cos \frac{3\pi}{13}-cos \frac{5\pi}{13} + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}

Hence, 

2cos \frac{\pi}{13}cos \frac{9\pi}{13}+cos \frac{3\pi}{13}+cos \frac{5\pi}{13}  = 0

Factorisation Formulae

In trigonometry, factorisation means converting sum or difference into the product. The factorisation formulae are:

(1) sin (C) + sin (D) = 2 sin (\mathbf{\frac{C+D}{2}}) cos (\mathbf{\frac{C-D}{2}})

Proof:

We have 

2 sin A cos B = sin (A + B) + sin (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = \frac{C+D}{2}  and B = \frac{C-D}{2}

Now put all these values in eq(1), we get

2 sin (\frac{C+D}{2} ) cos (\frac{C-D}{2} ) = sin (C) + sin (D)

Or 

sin (C) + sin (D) = 2 sin (\frac{C+D}{2} ) cos (\frac{C-D}{2} )

(2) sin (C) – sin (D) = 2 cos (\mathbf{\frac{C+D}{2}}) sin (\mathbf{\frac{C-D}{2}})     

Proof:

We have 

2 cos A sin B = sin (A + B) – sin (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = \frac{C+D}{2}  and B = \frac{C-D}{2}

Now put all these values in eq(1), we get

2 cos (\frac{C+D}{2} ) sin (\frac{C-D}{2} ) = sin (C) – sin (D)

Or 

sin (C) – sin (D) = 2 cos (\frac{C+D}{2} ) sin (\frac{C-D}{2} )

(3) cos (C) + cos (D) = 2 cos (\mathbf{\frac{C+D}{2}}) cos (\mathbf{\frac{C-D}{2}})     

Proof:

We have 

2 cos A cos B = cos (A + B) + cos (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = \frac{C+D}{2}  and B = \frac{C-D}{2}

Now put all these values in eq(1), we get

2 cos (\frac{C+D}{2} ) cos (\frac{C-D}{2} ) = cos (C) + cos (D)

Or 

cos (C) + cos (D) = 2 cos (\frac{C+D}{2} ) cos (\frac{C-D}{2} )

(4) cos (C) – cos (D) = 2 sin (\mathbf{\frac{C+D}{2}}) sin (\mathbf{\frac{D-C}{2}})      

Proof:

We have 

2 sin A sin B = cos (A – B) – cos (A + B)  ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = \frac{C+D}{2}  and B = \frac{C-D}{2}

Now put all these values in eq(1), we get

2 sin (\frac{C+D}{2} ) sin (\frac{C-D}{2} ) = cos (C) – cos (D)

Or 

cos (C) – cos (D) = 2 sin (\frac{C+D}{2} ) sin (\frac{C-D}{2} )

Explain 1. Express each of the following as a product

(i) sin 40° + sin 20°

(ii) sin 60° – sin 20°

(iii) cos 40° + cos 80°

Solution:

(i) Given: C = 40° and D = 20°

Now put all these values in the formula, 

sin (C) + sin (D) = 2 sin (\frac{C+D}{2}) cos (\frac{C-D}{2})     

We get

sin 40° + sin 20° = 2 sin (\frac{40+20}{2}) cos (\frac{40-20}{2})

= 2 sin (\frac{60}{2})  cos (\frac{20}{2})

= 2 sin 30° cos 10°

(ii) Given: C = 60° and D = 20°

Now put all these values in the formula, 

sin (C) – sin (D) = 2 cos (\frac{C+D}{2}) sin (\frac{C-D}{2})

We get

sin 60° – sin 20° = 2 cos (\frac{60+20}{2}) sin (\frac{60-20}{2})

= 2 cos (\frac{80}{2})  sin (\frac{40}{2})

= 2 cos 40° sin 20°

(iii) Given: C = 80° and D = 40°

 Now put all these values in the formula, 

cos (C) + cos (D) = 2 cos (\frac{C+D}{2}) cos (\frac{C-D}{2})

We get

cos 40° + cos 80° = 2 cos (\frac{40+80}{2})  cos (\frac{80-40}{2})

= 2 cos (\frac{120}{2})  cos (\frac{40}{2})

= 2 cos 60° cos 20°

Example 2. Prove that: 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x

Solution:

Lets take LHS

1 + cos 2x + cos 4x + cos 6x

Here, cos 0x = 1

So,

(cos 0x + cos 2x) + (cos 4x + cos 6x)

Using formula

cos (C) + cos (D) = 2 cos (\frac{C+D}{2}) cos (\frac{C-D}{2})     

We get

(2 cos (\frac{0x+2x}{2}) cos (\frac{2x-0x}{2}) ) + (2 cos (\frac{4x+6x}{2})  cos (\frac{6x-4x}{2}) )

(2 cos x cos x) + (2 cos 5x cos x)

Taking 2 cos x common, we have

2 cos x (cos x + cos 5x)

Again using the forumla 

cos (C) + cos (D) = 2 cos (\frac{C+D}{2}) cos (\frac{C-D}{2})

We get

2 cos x (2 cos (\frac{x+5x}{2}) cos (\frac{5x-x}{2}) )

2 cos x (2 cos 3x cos 2x)

4 cos x cos 2x cos 3x

LHS = RHS

Hence proved

Trigonometric ratios of multiple angles (2A) in terms of angle A

The trigonometric ratios of an angle in a right triangle define the relationship between the angle and the length of its sides. sin 2x or cos 2x, etc. are also, one such trigonometrical formula, also known as double angle formula, as it has a double angle in it.

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

Proof:

As we know that 

sin (A + B) = sin A cos B + cos A sin B ………………..(1)

Now taking B = A, in eq(1), we get

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

sin 2A = 2 sin A cos A  

(2) cos 2A = cos2 A – sin2 A

Proof:

As we know that 

cos (A + B) = cos A cos B – sin A sin B ………………..(1)

Now taking B = A, in eq(1), we get

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

cos 2A = cos2 A – sin2 A

(3) cos 2A = 2cos2 A – 1  

Proof:

As we know that 

cos 2A = cos2 A – sin2 A ………………..(1)

We also know that 

sin2 A + cos2 A = 1

So, sin2 A = 1 – cos2 A

Now put the value of sin2 A in eq(1), we get

cos 2A = cos2 A – (1 – cos2 A)

cos 2A = cos2 A – 1 + cos2 A

cos 2A = 2cos2 A – 1

(4) cos 2A = 1 – 2sin2 A

Proof:

As we know that 

cos 2A = 2cos2 A – 1 ………………..(1)

We also know that 

sin2 A + cos2 A = 1

So, cos2 A = 1 – sin2 A

Now put the value of sin2 A in eq(1), we get

cos 2A = 2(1 – sin2 A) – 1

cos 2A = 2 – 2sin2 A) – 1

cos 2A = 1 – 2sin2

(5) cos 2A = \mathbf{\frac{1-tan^2A}{1+tan^2A}}     

Proof:

As we know that 

cos 2A = cos2 A – sin2

So, now dividing, by sin2 A + cos2 A = 1, we get

cos 2A = \frac{cos^2 A - sin^2 A}{sin^2 A + cos^2 A}

Again dividing the numerator and denominator by cos2 A, we get

cos 2A = \frac{\frac{cos^2 A - sin^2 A}{cos^2A}}{\frac{sin^2 A + cos^2 A}{cos^2A}}

cos 2A = \frac{1-tan^2A}{1+tan^2A}

(6) sin 2A = \mathbf{\frac{2tanA}{1+tan^2A}}

Proof:

As we know that 

sin (A + B) = sin A cos B + cos A sin B ………………..(1)

Now taking B = A, in eq(1), we get

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

sin 2A = 2 sin A cos A  

As we also know that sin2 A + cos2 A = 1

So, now dividing, by sin2 A + cos2 A = 1, we get

sin 2A = \frac{2 sin A cos A}{cos^2A + sin^2A}

Now, on dividing the numerator and denominator by cos2 A, we get

sin 2A = \frac{2 tanA}{1+tan^2A}

(7) tan 2A = \mathbf{\frac{2 tanA}{1-tan^2A}}

Proof:

As we know that 

tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}  ………………..(1)

Now taking B = A, in eq(1), we get

tan(A + A) =  \frac{tan A + tan A}{1-tan A tan A}

tan 2A =  \frac{2tan A}{1-tan^2 A}

Example: Prove that

(i) \mathbf{\frac{sin 2\theta}{1+cos 2\theta}}  = tan θ

(ii) \mathbf{\frac{sin 2\theta}{1-cos 2\theta}}  = cot θ

(iii) cos 4x = 1 – 8 sin2x cos2x

Solution:

(i) sin 2θ = 2 sin θ cos θ ………..(from identity 1)

and, 1 + cos 2θ = 2cos2θ  ………..(from identity 3)

\frac{sin 2\theta}{1+cos 2\theta} = \frac{2 sin \theta cos \theta}{2cos^2\theta}

\frac{sin \theta }{cos\theta}

= tan θ 

Hence Proved

(ii) sin 2θ = 2 sin θ cos θ ………..(from identity 1)

and, 1 – cos 2θ = 2sin2θ  ………..(from identity 4)

\frac{sin 2\theta}{1-cos 2\theta} = \frac{2 sin \theta cos \theta}{2sin^2\theta}

\frac{cos \theta }{sin\theta}

= cot θ

Hence Proved

(iii) cos 4x = cos 2(2x)

= 1 – 2sin2(2x) (using 16)

= 1 – 2(sin(2x))2

= 1 – 2(2 sin x cos x)   (using identity 1)

= 1 – 2(4 sin2 x cos2 x)

cos 4x = 1 – 8 sin2 x cos2 x

Hence Proved

Trigonometric ratios of multiple angles (3A) in terms of angle A

The trigonometric ratios of an angle in a right triangle define the relationship between the angle and the length of its sides. sin 3x or cos 3x, etc. are also, one such trigonometrical formula, also known as triple angle formula, as it has a triple angle in it.

(1) sin 3A = 3sin A – 4 sin3A

Proof:

Let’s take LHS

sin 3A = sin(2A + A)

Using identity 

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

We get

sin 3A = sin 2A cos A + cos 2A sin A

= 2sin A cos A cos A + (1 – 2 sin2A)sin A

= 2sin A(1 – sin2A) + sin A – 2 sin3A

= 2sin A – 2sin3A + sin A – 2 sin3A

sin 3A = 3sin A – 4 sin3A

(2) cos 3A = 4 cos3A – 3cos A 

Proof:

Let’s take LHS

sin 3A = sin(2A + A)

Using identity 

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

We get

cos 3A = cos 2A cos A – sin 2A sin A

= (2cos2A – 1)cos A – 2sin A cos A sin A

= (2cos2A – 1)cos A – 2cos A(1 – cos2A)

= 2cos3A – cos A – 2cos A + 2cos3A)

cos 3A = 4 cos3A – 3cos A 

(3) tan 3A = \mathbf{\frac{3tan A-tan^3A}{1-3tan^2A}}

Proof:

Let’s take LHS

tan 3A = tan(2A + A)

Using identity 

tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}

We get

tan 3A = \frac{tan 2A + tan A}{1-tan 2A tan A}

\frac{\frac{tan 2A}{1-tan^2A} + tan A}{1-\frac{tan 2A tan A}{1-tan^2A}}

\frac{2tan A + tan A-tan^3A}{1-tan^2A-2 tan^2 A}

\frac{3tan A-tan^3A}{1-3tan^2A}

Example 1. Solve 2sin3xsinx.

Solution:

We have 2sin3xsinx

We also write as y = y1 . y2  ….(1)

Here, y1 = 2sin3x

y2 = sinx

So let’s solve y1 = 2sin3x

Using identity 

sin 3A = 3sin A – 4 sin3A

We get

y1 = 2(sin x – 4 sin3x)

= 2sin x – 8 sin3x

Now put these values in eq(1), we get

y = (2sin x – 8 sin3x)(sinx)

= 2sin2 x – 8 sin4x

Example 2. Solve 2tan3xtanx.

Solution:

We have 2tan3xtanx

We also write as y = y1 . y2  ….(1)

Here, y1 = 2tan3x

y2 = tanx

So let’s solve y1 = 2tan3x

Using identity 

tan 3A = \frac{3tan A-tan^3A}{1-3tan^2A}

We get

y1 = 2(\frac{3tan x-tan^3x}{1-3tan^2x} )

\frac{6tan x-2tan^3x}{1-3tan^2x}

Now put these values in eq(1), we get

y = (\frac{6tan x-2tan^3x}{1-3tan^2x} )(tanx)

\frac{6tan^2 x-2tan^4x}{1-3tan^2x}

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