# Sum and Difference Formulae of Trigonometry

• Last Updated : 04 Jul, 2022

In trigonometry, using the sum and difference formulae, one can calculate the values of trigonometric functions at any angle where it is feasible to express the given angle as the sum or the difference of standard angles like 0°, 30°, 45°, 60°, 90°, and 180°. We can easily memorize the values of trigonometric ratios at standard angles; i.e., 0°, 30°, 45°, 60°, 90°, and 180°. For example, to evaluate the value of the cosine function at 15°, we can write 15° as the difference between 45° and 30°; i.e., cos 15° = cos (45°-15°). With the help of sum and difference formulae, we can solve various mathematics problems and also prove various trigonometric identities and problems.

We have mainly six trigonometric sum and difference formulae, they are;

Sum and Difference formulae

• sin (A + B) = sin A cos B + cos A sin B
• sin (A – B) = sin A cos B – cos A sin B
• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B
• tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
• tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

Derivation of Sum and Difference identities

To demonstrate, the trigonometric sum and difference formulas let us consider a unit circle, with coordinates given as (cos θ, sin θ). Consider points A and B, which form angles of α and β with the positive X-axis, respectively. The coordinates of A and B are (cos α, sin α) and (cos β, sin β), respectively. We can observe that the angle AOB is equal to (α – β). Now, consider another two points P and Q on the unit circle such that Q is a point on the X-axis with coordinates (1,0) and angle POQ is equal to (α – β), and thus the coordinates of the point P are (cos (α – β), sin (α – β)).

Now, OA = OP, and OB = OQ as they are the radii of the same unit circle, and also the measure of one of the included angles of both triangles is (α – β).

Hence, by the side-angle-side congruence, triangles AOB and triangle POQ are congruent.

We know that the corresponding parts of congruent triangles are congruent, hence AB = PQ.

So, AB = PQ.

Using the distance formula between two points we get,

dAB = √[(cos α – cos β)2 + (sin α – sin β)2]

= √[cos2 α – 2 cos α cos β + cos2 β + sin2 α – 2 sin α sin β + sin2 β]    {Since, (a – b)2 = a2 – 2ab + b2)}

= √[(cos2 α+ sin2 α) + (cos2 β+ sin2 β) – 2(cos α cos β + sin α sin β)]

= √[1 + 1 – 2(cos α  cos β + sin α  sin β)]         {Since, sin2 x + cos2 x = 1}

= √[2 – 2(cos α cos β+ sin α sin β)]          ———————— (1)

dPQ = √[(cos (α – β) – 1)2 + (sin (α – β) – 0)2

= √[cos2 (α – β) – 2 cos (α – β) + 1 + sin2 (α – β)]      {Since, (a – b)2 = a2 – 2ab + b2)}

= √[(cos2 (α – β) + sin2 (α – β)) + 1 – 2 cos (α – β)]

= √[1 + 1 – 2 cos (α – β)]            {Since, sin2 x + cos2 x = 1}

= √[2 – 2 cos (α – β)]                            ———————— (2)

Since AB = PQ, equate both equations (1) and (2).

√[2 – 2(cos α cos β+ sin α sin β)] = √[2 – 2 cos (α – β)]

By squaring on both sides, we get,

2 – 2(cos α cos β+ sin α sin β) = 2 – 2 cos (α – β) ________________(3)

### Cos (α – β) formula

from eq (3)

⇒ 2 (1 – cos α cos β – sin α sin β) = 2 (1 – cos (α – β))

⇒ 1 – cos α cos β – sin α sin β = 1 – cos (α – β)

⇒ cos (α – β) = cos α cos β + sin α sin β

cos (α – β) = cos α cos β + sin α sin β

### Cos (α + β) formula

To derive the sum formula of the cosine function substitute (-β) in the place of β in the difference of the cosine function.

Hence, we get cos (α + β) = cos (α – (β))

= cos α cos (-β) + sin α sin (-β)    {Since, cos (α – β) = cos α cos β + sin α sin β}

= cos α cos β – sin α sin β            {Since, cos (-θ) = cos θ, sin (-θ) = – sin θ}

⇒ cos (α + β) = cos α cos β – sin α sin β

cos (α + β) = cos α cos β – sin α sin β

### Sin (α – β) formula

We know that, sin (90° – θ) = cos θ and cos (90° – θ) = sin θ

So, sin (α – β) = cos (90° – (α – β))

= cos (90° – α + β)

= cos [(90° – α) + β]

= cos (90° – α) cos β – sin (90° – α) sin β      {Since,  cos (α + β) = cos α cos β – sin α sin β}

⇒ sin (α – β) = sin α cos β – cos α sin β

sin (α – β) = sin α cos β – cos α sin β

### Sin (α + β) formula

We know that, sin (90° – θ) = cos θ and cos (90° – θ) = sin θ.

So, sin (α + β) = cos (90° – (α + β))

= cos (90° – α – β)

= cos [(90° – α) – β]

= cos (90° – α) cos β + sin (90° – α) sin β    {Since, cos (α – β) = cos α cos β + sin α sin β}

⇒ sin (α + β) = sin α cos β + cos α sin β

sin (α + β) = sin α cos β + cos α sin β

### Tan (α – β) formula

We know that, tan θ = sin θ/cos θ

So, tan (α – β) = sin (α – β)/cos (α – β)

= (sin α cos β – cos α sin β)/(cos α cos β + sin α sin β)       {Since, sin (α – β) = sin α cos β – cos α sin β and cos (α – β) = cos α cos β + sin α sin β}

Now, divide the numerator and denominator with cos α cos β

= [(sin α cos β – cos α sin β)cos α cos β ]/[(cos α cos β + sin α sin β)/(cos α cos β)

= (sin α/cos α – sin β/cos β)/(1 + (sin α/cos α)×(sin β/cos β))

= (tan α – tan β)/(1 + tan α tan β)

⇒ tan (α – β) = (tan α – tan β)/(1 + tan α tan β)

tan (α – β) = (tan α – tan β)/(1 + tan α tan β)

### Tan (α + β) formula

To derive the tan (α + β) formula substitute (-β) in the place of β in the tan (α – β) formula.

Hence, we get, tan (α + β) = tan(α – (-β))

= (tan α – tan (-β))/(1 + tan α tan (-β))            {Since, tan (α – β) = (tan α – tan β)/(1 + tan α tan β)}

= (tan α + tan β)/(1 – tan α tan β)                   {Since, tan (-θ) = – tan θ}

⇒ tan (α + β) = (tan α + tan β)/(1 – tan α tan β)

tan (α + β) = (tan α + tan β)/(1 – tan α tan β)

### Sample Problems

Problem 1: Prove the triple angle formulae of sine and cosine functions using the sum and difference formulae.

Solution:

To Prove: sin 3A = 3 sin A – 4 sin3A

Proof:

We can write sin 3A as sin (2A + A)

⇒ sin 3A = sin (2A + A)

We have, sin (A + B) = sin A cos B + cos A sin B

So, sin (2A + A) = sin 2A cos A + cos 2A sin A

We know that,

sin 2A = 2 sin A cos A and cos 2A = 1 – 2sin2

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

= 2 sin A cos2 A + sin A – 2 sin3 A

We have, cos2 A = 1 – sin2

= 2 sin A (1 – sin2 A) + sin A – 2 sin3 A

= 2 sin A – 2sin3 A + sin A – 2 sin3 A

= 3 sin A – 4 sin3 A

Thus, sin 3A = 3 sin A – 4 sin3 A

Hence proved

To Prove: cos 3A = 4 cos3 A – 3 cos A

Proof:

We can write cos 3A as cos (2A + A)

⇒ cos 3A = cos (2A + A)

We have, cos (A + B) = cos A cos B – sin A sin B

So, cos (2A + A) = cos 2A cos A – sin 2A sin A

We know that,

sin 2A = 2sin A cos A and cos 2A = 2cos2 A – 1

⇒ cos (2A + A) = (2 cos2 A – 1) cos A – (2 sin A cos A) sin A

= 2 cos3 A – cos A – 2 sin2 A cos A

We have, sin2 A = 1- cos2 A

= 2 cos3 A – cos A – 2 (1- cos2 A) cos A

= 2 cos3 A – cos A – 2 cos A + 2 cos3 A = 4 cos3 A – 3 cos A

Thus, cos 3A = 4 cos3 A – 3 cos A

Hence proved

Problem 2: Find the value of cos 75° using the sum and difference formulae.

Solution:

We can write 75° as the sum of 45° and 30°.

By using the sum formula of the cosine function we get,

cos 75° = cos (45° + 30°)

= cos 45° cos 30° – sin 45° sin 30° {Since, cos (A + B) = cos A cos B – sin A sin B}

= (1/√2) (√3/2) – (1/√2)(1/2)        {Since, cos 45° = sin 45° = (1√2) , cos 30° = √3/2, sin 30° = 1/2}

= (√3 -1)/2√2

Hence, cos 75° = (√3 – 1)/2√2

Problem 3: Find the value of tan 105° using the sum and difference formulae.

Solution:

We can write 105° as the sum of 60° and 45°.

By using the sum formula of the tangent function we get,

tan 105° = tan (60° + 45°)

= (tan 60° + tan 45°)/(1 – tan 60° tan 45°)   {Since, tan (A + B) = (tan A + tan B)

= (√3 + 1)/(1 – (√3 × 1))                                {Since, tan 60° = √3, tan 45° = 1}

= (√3 + 1)/(1 – √3)

Rationalize the above expression with the conjugate of the denominator.

= (√3 + 1)2/(1 – (√3)2)

= (3 + 2√3 + 1)/(1 – 3)

= (4 + 2√3)/(-2)

= -2 – √3

Hence, tan 105° = -2 – √3.

Problem 4: Evaluate the value of sin 15° using the sum and difference formulae.

Solution:

We can write 15° as the difference between 45° and 30°.

By using the difference formula of sine function we get,

sin 15° = sin (45° – 30°)

= sin 45° cos 30° – cos 45° sin 30°  {Since, sin (A – B) = sin A cos B – cos A sin B}

= (1/√2) (√3/2) – (1/√2)(1/2)          {Since, cos 45° = sin 45° = (1√2) , cos 30° = √3/2, sin 30° = 1/2}

= (√3 – 1)/2√2

Hence, sin 15° = (√3 – 1)/2√2

Problem 5: Prove that sin (π/4 – a) cos (π/4 – b) + cos (π/4 – a) sin (π/4 – b)  = cos (a + b).

Solution:

L.H.S = sin (π/4 – a) cos (π/4 – b) + cos (π/4 – a) sin (π/4 – b)

If we observe it in the form of sin A cos B + cos A sin B

We know that, sin (A + B) = sin A cos B + cos A sin B

= sin [(π/4 – a) + (π/4 – b)]

= sin [(π/2) – (a + b)]

= cos (a + b)             {Since, sin (90° – θ) = cos θ}

= R. H. S

Hence, proved.

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