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. And the formula for the same is shown in the image added below:
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 of 2sinAcosB Formula
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 Using 2sinAcosB 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
Article Related to 2sinAcosB Formula:
Problems on 2sinAcosB Formula
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.