Sample Problems on 2sinAsinB Formula

Problem 1: Solve the integral of 3 sin 5x sin (9x/2).

Solution:

Integral of 3 sin 5x sin (9x/2) = ∫3 sin 5x sin (9x/2) dx

From 2sinasinb formula we have,

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

3 sin 5x sin (9x/2) = 3/2 [2 sin 5x sin (9x/2)]

= 3/2 [cos (5x – (9x/2)) – cos (5x + (9x/2))]

= 3/2 [cos (x/2) – cos (19x/2)]

Now, ∫3 sin 5x sin (9x/2) dx = ∫3/2 [cos(x/2) – cos(19x/2)] dx

= 3/2 ∫cos(x/2) dx – ∫cos(19x/2) dx

= 3/2 [2 sin (x/2) – 2/19 sin (19x/2)]    {∫cos (ax) = 1/a sin (ax) + c}

= 3 sin (x/2) – 3/19 sin (19x/2) 

Hence, the  integral of 3 sin 5x sin (9x/2) = 3 [sin(x/2) – 1/19 sin (19x/2)]

Problem 2: Find the derivative of 4 sin 2x sin (5x/2).

Solution:

Derivative of 4 sin 2x sin (5x/2) = d(4 sin 2x sin (5x/2))/dx

From 2sinasinb formula we have,

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

Now, 4 sin 2x sin (5x/2) = 2 [2 sin 2x sin (5x/2)]

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

= 2 [cos (-x/2) – cos (9x/2)]

= 2[cos (x/2) – cos (9x/2)]      {Since, cos (-θ) = cos θ}

Now, d(4 sin 2x sin (5x/2))/dx = d{2[cos (x/2) – cos (9x/2)]}/dx

= 2{d(cos(x/2))/dx – d(cos(9x/2))/dx}

= 2{1/2 (-sin (x/2) – (- (9/2) sin (9x/2)}      {Since, d(cos ax)/dx = – a sin ax}

= 9 sin (9x/2) – sin (x/2)

Hence, the derivative of 4 sin 2x sin (5x/2) = 9 sin (9x/2) – sin (x/2).

Problem 3: Express 6 sin (11x/2) sin 7x in terms of the cosine function.

Solution: 

From 2sinasinb formula we have,

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

Now, 6 sin (11x/2) sin 7x = 3 [2 sin (11x/2) sin 7x]

= 3 [cos (11x/2 – 7x) – cos (11x/2 + 7)]

= 3 [cos (-3x/2) – cos (25x/2)]

= 3 [cos (3x/2) – cos (25x/2)]        {Since, cos (-θ) = cos θ}

Hence, 6 sin (11x/2) sin 7x = 3 [cos (3x/2) – cos (25x/2)].

Problem 4: Find the value of the expression 7 sin (17.5°) sin (72.5°) using the 2sinasinb formula.

Solution:

7 sin (17.5°) sin (72.5°) = 7/2 [2 sin (17.5°) sin (72.5°)]

From 2sinasinb formula we have,

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

Now, 7/2 [2 sin (17.5°) sin (72.5°)] = 7/2 [cos (17.7° – 72.5°) – cos (17.5° + 72.5°)]

= 7/2 [cos (-55°) – cos (90°)]

= 7/2 [cos 55° – cos 90°]      {Since, cos (-θ) = cos θ}

= 7/2 [0.5735 – 0]               {Since, cos 55° = 0.5735, cos 90° = 0}

= 2.00725‬            

Hence, 7 sin (17.5°) sin (72.5°) = 2.00725‬

Problem 5: Express 2 sin (12x) sin (17x/2) in terms of the cosine function.

Solution:

From 2sinasinb formula we have,

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

Now, 2 sin (12x) sin (17x/2) = [cos (12x – (17x/2) – cos(12x + (17x/2)]

= cos [(24x – 17x)/2] – cos [(24x + 17x)/2]

= cos (7x/2) – cos (41x/2)

Hence, 2 sin (12x) sin (17x/2) = cos (7x/2) – cos (41x/2).

Problem 6: Solve 4 sin (66.5°) sin (113.5°) using the 2sinasinb formula.

Solution:

From 2sinasinb formula we have,

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

Now, 4 sin (66.5°) sin (113.5°) = 2 [2 sin (66.5°) sin (113.5°)]

= 2 [cos (66.5° – 113.5°) – cos (66.5° + 113.5°)]

= 2 [cos (-47°) – cos (180°)]

= 2 [cos 47° – cos 180°]        {Since, cos (-θ) = cos θ}

= 2 [0.682 – 1]                       {Since, cos 47° = 0.682, cos 180° = -1}

= -0.636

Hence, 4 sin (66.5°) sin (113.5°) = -0.636

2sinasinb is one of the important trigonometric formulas which is equal to cos (a – b) – cos (a + b). 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