Product to Sum Formulas

Product-to-sum formulas are trigonometric identities that convert the product of two trigonometric functions into a sum or difference of trigonometric functions.

Formula Derivation

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

The Sum/difference formulae are given below:

• sin(A+B) = sinA cosB + cosA sinB ———— (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 (Product of Cosines)

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 (Product of Sine and Cosine)

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 (Product of Sine and Cosine)

To derive the sin 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 (Product of Sines)

To derive the sin 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

= 1/20 ​sin(10x) + 41​sin(2x) + C

Answer : ∫cos4x cos6xdx = 1/20 ​sin(10x) + 1/4​sin(2x) + C

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

Solution:

sin36 cos54 = 1/2 ​[sin90 + sin(−18)]

=1/2​ [1−sin18]

If an exact numerical answer is required, use:

\frac{1}{2}\left[1-\frac{\sqrt{5}-1}{4}\right]

Final Answer: \frac{5-\sqrt{5}}{8}

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/dx​(2[sin5x−sinx])

= 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]