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Sum and Difference Formulas

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Sum and Difference formulae of trigonometry are used to 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°. 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°). Now with the help of sum and difference formulae, we can easily solve the above problem. In this article, we will learn about various Sum and Difference formulae used in trigonometry in detail.

Trigonometry Functions

Various functions used in trigonometry are called trigonometry functions they define the relationships between angles and sides of the triangle. The six basic trigonometric formulas are,

  • sin
  • cosine
  • tan
  • cosec
  • sec
  • cot

What are Sum and Difference Formulas?

Sum and Difference formulas are used to calculate the trigonometric function for those angles where standard angle cant be used. We have six main sum and difference formulas which we mainly used in trigonometry.

The six main trigonometric sum and difference formulae 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)

Proof 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)). 

Sum and Difference Identities

 

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)

Sum and Difference Formulas for Cosine

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) 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,

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

Sum and Difference Formulas for Sine

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) 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

Sum and Difference Formulas for Tangent

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)

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) 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)

Sum and Difference Formulae Table

In the previous section, we derived the formulas of all the sum and difference identities of the trigonometric functions sine, cosine, and tangent. Now, let us summarize these formulas in the table below for a quick revision.

 

Sum Formulae

Difference Formulae

Sine function

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

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

Cosine function

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

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

Tangent function

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

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

How to Apply Sum and Difference Formulas

Sum and Difference Formulas of trigonometry are used to solve various trigonometry problems and find the values of trigonometric functions without standard values. To Apply Sum and Difference Formulas study the following example,

Example: Find the value of sin 15°

Solution:

Step 1: Write the given function in the sum and difference of the standard function,

sin 15° = sin (45 -30)°

Step 2: Use the required Sum and Difference Formulas, here we use, sin (α – β) = sin α cos β – cos α sin β

sin (45 -30)° = sin 45° cos 30° – cos 45° sin 30°

Step 3: Substitute the value of these standard trigonometric functions using the trigonometric table.

sin (45 -30)° = 1/√2 × √3/2 – 1/√2 × 1/2

Step 4: Simplify the value obtained in the above step.

sin (45 -30)° = 1/√2 × √3/2 – 1/√2 × 1/2

                    = (√3 -1)/ 2√2 

sin 15° = (√3 -1)√2 / 4

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Solved Examples on Sum and Difference Formulas

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

  •  sin 3A = 3 sin A – 4 sin3A
  • cos 3A = 4 cos3 A – 3 cos A

Solution:

To Prove: sin 3A = 3 sin A – 4 sin3A

sin 3A = sin (2A + A)     [sin (A + B) = sin A cos B + cos A sin B]

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 A, and cos2 A = 1 – sin2 A

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 

                    = 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  (proved)

To Prove: cos 3A = 4 cos3 A – 3 cos A

cos 3A = cos (2A + A)   [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, and sin2 A = 1- cos2 A

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

                    = 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  (proved)

Example 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°

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

Example 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°.

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,

            = \left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right]\times\left[\frac{1+\sqrt{3}}{1+\sqrt{3}}\right]

            = (√3 + 1)2/(1 – (√3)2)

            = (3 + 2√3 + 1)/(1 – 3)

            = (4 + 2√3)/(-2)

            = -2 – √3

Hence, tan 105° = -2 – √3.

Example 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°

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

Example 5: Prove that sin (Ï€/4 – a) cos (Ï€/4 – b) + cos (Ï€/4 – a) sin (Ï€/4 – b)  = cos (a + b).

Solution:

L.H.S = sin (Ï€/4 – a) cos (Ï€/4 – b) + cos (Ï€/4 – a) sin (Ï€/4 – b)   {sin (A + B) = sin A cos B + cos A sin B}

         = sin [(Ï€/4 – a) + (Ï€/4 – b)]

         = sin [(Ï€/2) – (a + b)]

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

           = R. H. S    (proved)

FAQs on Sum and Difference Formulas

Question 1: What are Trigonometric Identities?

Answer:

Trigonometric Identities are the various identities which are used to solve trigonometric problems. Some of the common trigonometric identities are,

  • sin2 θ + cos2 θ = 1
  • 1+tan2 θ = sec2 θ
  • cosec2 θ = 1 + cot2 θ

For more, click on Trigonometric Identities

Question 2: What are the Six Sum and Difference Formulas in Trigonometry?

Answer:

Sum and Difference Formulas are widely used in trigonometry and the six sum and difference formulas of trigonometry are,

  • sin(A + B) = sinA cosB + cosA sinB
  • sin(A – B) = sinA cosB – cosA sinB
  • cos(A + B) = cosA cosB – sinA sinB
  • cos(A – B) = cosA cosB + sinA sinB
  • tan(A + B) = (tanA + tanB) / (1 – tanA tanB)
  • tan(A – B) = (tanA – tanB) / (1 + tanA tanB)

Question 3: How to Prove Sum and Difference Formulas?

Answer:

Sum and Difference formulas of trigonometry are found using a unit circle, standard trigonometric formulas, and distance formulas.

Question 4: Why do we Use Sum and Difference Identities?

Answer:

Sum and Difference Identities are used to evaluate the value of the trigonometric functions at special angles. These angles can easily be expressed as the sum or difference of standard angles 0°, 30°, 45°, 60°, 90°, and 180°. Such as for finding cos 15° we can write it as cos (45 -30)° then using Sum and Difference Identities we can evaluate its value.



Last Updated : 10 Jan, 2024
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