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.

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.

This formula is derived using the angle sum and angle difference formulas. Before learning more about the 2sinAsinB Formula, let’s first learn in brief about, Trigonometric Ratios