2sinAsinB Formula

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

Trigonometric ratios are ratios of sides in a triangle and there are six trigonometric ratios. In a right-angle triangle, the six trigonometric ratios are defined as:

1. sin θ = (Opposite Side/Hypotenuse = AB/AC
2. cos θ = Adjacent Side/Hypotenuse = BC/AC
3. tan θ = Opposite side/adjacent side = AB/BC
4. cosec θ = 1/sin θ = Hypotenuse/Opposite Side = AC/AB
5. sec θ = 1/cos θ = Hypotenuse/Adjacent Side = AC/BC
6. cot θ = 1/tan θ = Adjacent Side/Opposite Side = BC/AB

The 2sinasinb formula is a trigonometric formula that is used to simplify trigonometric expressions and also solve complex integrals and derivatives of trigonometric expressions. The 2sinasinb formula is equal to the difference between the angle sum and the angle difference of the cosine functions, i.e., for two angles A and B,

2sinasinb formula is,

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

From the formula, we can observe that twice the product of two sine functions is converted into the difference between the angle sum and the angle difference of the cosine functions. With the help of the 2 sin A sin B formula, we can extract the formula of sin A sin B. 

sin A sin B = ½ [cos (A – B) – cos (A + B)]

We can derive the 2sinasinb formula with the help of the sum and difference of formulae of the cosine function.

• cos (A + B) = cos A cos B – sin A sin B    ———— (1)
• cos (A – B) = cos A cos B + sin A sin B    ———— (2)

Now subtract the equation(1) from the equation (2)

⇒ 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) = sin A sin B + sin A sin B

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

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

Article Related to 2sinAsinB Formula:

• Trigonometric Identites
• Trigonometric Function
• Trigonometric Table
• Sin Cos Tan values

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

What is 2 sin a sin b equal to?

2sinasinb is equal to, 2sin A sin B = cos (A-B) – cos (A + B).

What is the formula for 2 sin into cos b?

The formula for the 2sinAcosB is 2sinAcosB = sin(A + B) + sin(A – B).

What is the formula of 2 cos a cos b?

The formula for 2cosacosb is, 2 cos A cos B = cos (A + B) + cos (A – B).

What is the formula for 2 sin a cosa?

The formula for 2 sin a cosa is sin 2a, i.e. 2sin(a)cos(a) = sin 2a.