2sinAcosB Formula

2sinacosb is one of the important trigonometric formulas which is equal to sin (a + b) + sin (a – b). It is one of the product-to-sum formulae that is used to convert the product into a sum. The branch of mathematics that relates to the angles and the lengths of the sides of right-angled triangles is referred to as trigonometry. There are six trigonometric ratios or functions, i.e., sine, cosine, tangent, cosecant, secant, and cotangent functions, where a trigonometric ratio is defined as a ratio between the sides of a right-angled triangle. Cosecant, secant, and cotangent functions are the reciprocal functions of sine, cosine, and tangent functions, respectively.

• sin θ = opposite side/hypotenuse
• cos θ = adjacent side/hypotenuse 
• tan θ = opposite side/adjacent side
• cosec θ = 1/sin θ = hypotenuse/opposite side
• sec θ = 1/cos θ = hypotenuse/adjacent side
• cot θ = 1/tan θ = adjacent side/opposite side

2sinAcosB formula

2sinacosb is one of the product-to-sum formulae. Similarly, we have three other products to sum/difference formulas in trigonometry, namely, 2sinasinb, 2cosacosb, and 2cosasinb. By using the 2sinacosb formula, we can simplify trigonometric expressions and also solve integrals and derivatives involving expressions of the form 2sinacosb. This trigonometric identity is derived by adding the sin (a + b) and sin (a – b) identities.

The 2sinacosb formula is,

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

From the formula, we can observe that product of a sine function and a cosine function is converted into a sum of two other sine functions. For example,

The product-to-sum formulae for half angles are,

2 sin (x/2) cos (y/2) = sin [(x + y)/2] + sin [(x – y)/2]

Derivation

From the sum and difference formulae of trigonometry, we have,

sin (A + B) = sin A cos B + sin B cos A ⇢ (1)
sin (A – B) = sin A cos B – sin B cos A ⇢  (2)

Now, by adding equations (1) and (2) we get,

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

⇒ 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

Therefore, 2 sin A cos B = sin (A + B) + sin (A – B)

sin 2A formula

We have, 2 sin A cos B = sin (A + B) + sin (A – B)

Now, let us consider that A = B

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

⇒ 2 sin A cos A = sin 2A + sin 0°

⇒2 sin A cos A = sin 2A               {Since sin 0° = 0}

Hence, sin 2A = 2 sin A cos A

Solved Problems

Problem 1: Express 5 sin 2x cos 6x in terms of the sine function.

Solution:

5 sin 2x cos 6x

By multiplying and dividing the given equation by 2, we get

(2/2) 5 sin 2x cos 6x

= 5/2 [2 sin 2x cos 6x]

We have,

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

5/2 [2 sin 2x cos 6x] = 5/2 [sin (2x + 6x) + sin (2x – 6x)]

= 5/2 [sin (8x) + sin(-4x)]

= 5/2 [sin 8x – sin 4x] {since sin (-θ) = – sin θ}

Hence, 5 sin 2x cos 6x =  5/2 [sin 8x – sin 4x] 

Problem 2: Determine the derivative of 2 sin 3x cos (11x/2).

Solution: 

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

2 sin 3x cos (11x/2) = sin [3x+ (11x/2)] + sin [3x – (11x/2)]

= sin (17x/2) + sin (-5x/2)

= sin (17x/2) – sin (5x/2) {since sin (-θ) = – sin θ}

Now, derivative of 2 sin 3x cos (11x/2) = d [2 sin 3x cos (11x/2) ]/dx

= d [sin (17x/2) – sin (5x/2)]/dx

= 17/2 cos (17x/2) – 5/2 cos (5x/2) {Since, d[sin (ax)] = a cos (ax)}

= 1/2 [17 cos (17x/2) – 5 cos (5x/2)]

Hence, the derivative of 2 sin 3x cos (11x/2) = 1/2 [17 cos (17x/2) – 5 cos (5x/2)]

Problem 3: Write 8 sin 4y cos (7y/2) in terms of the sum function.

Solution:

8 sin 4y cos (7y/2) = 4 (2 sin 4y cos (7y/2))

We have,

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

4 (2 sin 4y cos (7y/2)) = 4 [sin (4y + (7y/2)) + sin (4y – (7y/2))]

= 4 [sin (15y/2) + sin (y/2)]

Hence, 8 sin 4y cos (7y/2) =  4 [sin (15y/2) + sin (y/2)]

Problem 4: Find the value of the expression 2 sin 38.5°cos 51.5‬° using the 2sinacosb formula.

Solution:

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

2 sin 38.5° cos 51.5‬° = sin (38.5° + 51.5°) + sin (38.5° – 51.5°)

= sin (90°) + sin (-13°)

= sin (90°) – sin (13°) {since sin (-θ) = – sin θ}                

= 1 – 0.22495 = 0.77505, sin 90° = 1 and sin 13° = -0.22495

Hence, 2 sin 38.5° cos 51.5‬° = 0.77505

Problem 5: What is the value of the integral of 3 sin (3x/2) cos (9x/2)?

Solution:

We have,

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

3 sin (3x/2) cos (9x/2) = 3/2 [sin (3x/2) cos (9x/2)]

= 3/2 [sin (3x/2 + 9x/2) + sin (3x/2 – 9x/2)]

= 3/2 [sin (12x/2) + (sin (-6x/2)]

= 3/2 [sin (6x) – sin (3x)] {since sin (-θ) = – sin θ}    

Now, integral of 3 sin (3x/2) cos (9x/2) =∫3 sin (3x/2) cos (9x/2) dx

= ∫3/2 [sin (6x) – sin (3x)] dx

= 3/2 [-1/6 (cos 6x) + 1/3 cos (3x) + C] {Since, the integral of sin(ax) is (-1/a) cos (ax) + C}

= 1/2 (cos 3x) – 1/4 (cos 6x) + C

Hence, ∫3 sin (3x/2) cos (9x/2) dx = 1/2 (cos 3x) – 1/4 (cos 6x) + C

Problem 6: Calculate the derivative of 9 sin (6y) cos (2y).

Solution:

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

9 sin (6y) cos (2y) = 9/2 [2 sin (6y) cos (2y)]

= 9/2 [sin [6y + 2y] + sin [6y – 2y]

= 9/2 [sin 8y + sin 4y]

Now, derivative of 9 sin (6y) cos (2y) = d [9 sin (6y) cos (2y) ]/dy

= d [9/2 (sin 8y + sin 4y)]/dy

= 9/2 [8 cos 8y + 4 cos 4y] {Since, d[sin (ax)] = a cos (ax)}

= 36 cos 8y + 18 cos 4y

Hence, the derivative of 9 sin (6y) cos (2y) = 36 cos 8y + 18 cos 4y.