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,

            = 

            = (√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 (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)   {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    (proved)

Sum and Difference formulas 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.