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Product to Sum Formulas

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  • Last Updated : 28 Nov, 2022
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As the name suggests, trigonometry means the study of triangles. Trigonometry is more precisely concerned with right triangles, where one of the internal angles is 90 degrees. It is a significant branch of mathematics that helps us to determine the missing or unknown angles or side lengths in a triangle and studies the relationships between angles and side lengths of a right triangle. Sine, cosine, tangent, cosecant, secant, and cotangent are the six trigonometric ratios or functions where a trigonometric ratio is defined as the ratio of side lengths of a right triangle.

 

The six trigonometric ratios/functions are:

  • sin θ = Opposite side/Hypotenuse
  • cos θ = Adjacent side/Hypotenuse
  • tan θ = Opposite side/Adjacent side
  • cosec θ = Hypotenuse/Opposite side
  • sec θ = Hypotenuse/Adjacent side
  • cot θ = Adjacent side/Opposite side

Product to Sum/Difference Formulae

The product-to-sum identities are used to express the product between sine and/or cosine functions as a sum or difference. The sum and difference formulas of sine and cosine functions are added or subtracted to derive these identities. The product-to-sum identities can be used to simplify the trigonometric expression. The integrals or derivatives of the trigonometric functions can be solved with ease by using these identities. For all possible sine and cosine product combinations, there are four products to sum or difference formulae in total.

Product to Sum/Difference identities

 Product of two cosine functions 

   cos A cos B =  (½) [cos (A + B) + cos (A – B)] 

 Product of cosine and sine functions 

 cos A sin B = (½) [sin (A + B) – sin (A – B)] 

 Product of sine and cosine functions 

 sin A cos B = (½) [sin (A + B) + sin (A – B)] 

 Product of two sine functions 

   sin A sin B = (½) [cos (A – B) – cos (A + B)] 

Derivation of Product to Sum/Difference identities

The product to sum/difference formulae can be derived using the trigonometric sum/difference formulae. 

The Sum/difference formulae are given below:

  • 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)

cos A cos B Formula

To derive the cos A cos B formula add equations (3) and (4)

⇒ cos (A + B) + cos (A – B) = [cos A cos B – sin A sin B] + [cos A cos B + sin A sin B]

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

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

Hence, 

  cos A cos B =  (½) [cos (A + B) + cos (A – B)]

cos A sin B Formula

To derive the cos A sin B formula subtract equations (2) from (1)

⇒ sin (A + B) – sin (A – B) = [sin A cos B + cos A sin B] – [sin A cos B – cos A sin B]

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

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

Hence,

cos A sin B = (½) [sin (A + B) – sin (A – B)]

sin A cos B Formula

To derive the cos A cos B formula add equations (1) and (2)

⇒ sin (A + B) + sin (A – B) = [sin A cos B + cos A sin B] + [sin A cos B – cos A sin B]

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

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

Hence,

sin A cos B = (½) [sin (A + B) + sin (A – B)]

sin A sin B Formula

To derive the cos A sin B formula subtract equations (4) from (3)

⇒ cos (A – B) – cos (A + B) = [cos A cos B + sin A sin B] – [cos A cos B – sin A sin B]

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

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

Hence,

sin A sin B = (½) [cos (A – B) – cos (A + B)]

Sample Problems

Problem 1: Express 6 cos 8x sin 5x as sum/difference.

Solution:

From one of the product to sum formulas, we have

cos A sin B = (½) [sin (A + B) – sin (A – B)]

So, by substituting A = 8x and B = 5x in the above formula, we get

cos 8x sin 5x = (½) [ sin (8x + 5x) – sin (8x – 5x) ]

cos 8x sin 5x = (½) [sin 13x – sin 3x]

Now, 6 cos 8x sin 5x = 6 × (½) [sin 13x – sin 3x]

Hence, 6 cos 8x sin 5x = 3 [sin 13x – sin 3x]

Problem 2: Determine the value of the integral of cos 4x cos 6x.

Solution:

From one of the product to sum formulas, we have

cos A cos B = (½) [cos (A + B) + cos (A – B)]

cos 4x cos 6x = ½ [cos (4x + 6x) + cos (4x – 6x)]

= (½) [cos 10x + cos (-2x)]

= (½) [cos 10x + cos 2x]       {Since, cos (-θ) = cos θ}

Now, integral of cos 4x cos 6x = ∫ cos 4x cos 6x dx

= ∫(½) [cos 10x + cos 2x] dx

= (½) [1/10 sin 10x + 1/2 sin 2x] + C   {Since, ∫cos(ax) dx = 1/a sin(ax) + c}

= 1/20 sin(10x) + 1/2 sin(2x) + C

Hence, integral of cos 4x cos 6x = 1/20 sin(10x) + 1/2 sin(2x) + C

Problem 3: Determine the value of sin 36° cos 54° without evaluating the sin 36° and cos 54° values. 

Solution:

From one of the product to sum formulas, we have

sin A cos B = (½) [sin (A + B) + sin (A – B)]

So, sin 36° cos 54° = (½) [sin (36° + 54°) + sin (36° – 54°)]

= (½) [sin (90°) + sin (-18°)]

= (½) [sin 90° – sin 18°]       {Since, sin (-θ) = – sin θ}

= (½) [1 – 0.3090]                {Since, sin 90° = 1, sin 18° = 0.3090}

= 0.3455

Hence, sin 36° cos 54° = 0.3455.

Problem 4: Determine the value of the derivative of 4 cos 3x sin 2x.

Solution:

From one of the product to sum formulas, we have

cos A sin B = (½) [sin (A + B) – sin (A – B)]

Now, 4 cos 3x sin 2x = 4 × (½) [sin (3x + 2x) – sin (3x – 2x)]

= 2 [sin 5x – sin x]

Now, derivative of 4 cos 3x sin 2x = d(4 cos 3x sin 2x)/dx

= d/(2 [sin 5x – sin x])/dx

= 2 [ d(sin 5x)/dx – d(sin x)/dx ]

= 2 [5 cos 5x – cos x]                 {Since, d(sin ax)/dx = a cos ax}

Hence, derivative of 4 cos 3x sin 2x = 2 [5 cos 5x – cos x] .

Problem 5: Determine the value of sin 15° sin 45° without evaluating the sin 15° and sin 45° values. 

Solution:

From one of the product to sum formulas, we have

sin A sin B = (½) [cos (A – B) – cos (A + B)]

Now, sin 15° sin 45° = (½)[cos (15° – 45°) – cos (15° + 45°)]

= (½) [cos (-30°) – cos (60° )]

= (½) [cos 30° – cos 60°]   {Since, cos (-θ) = cos θ}

= (½) [√3/2 – 1/2]            {Since, cos 30° = √3/2 and cos 60° = 1/2}

= (½) [(√3 -1)/2]

= (√3 -1)/4

Hence, sin 15° sin 45° = (√3 -1)/4.

Problem 6: Express 2 cos 9x cos 7x as sum/difference.

Solution:

From one of the product to sum formulas, we have

cos A cos B = (½) [cos (A + B) + cos (A – B)]

Now, 2 cos 9x cos 7x = 2 × (½) [cos (9x + 7x) + cos (9x – 7x)]

= [cos (16x) + cos (2x)]

Hence, 2 cos 9x cos 7x = [cos 16x + cos 2x]

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