2cosAsinB Formula

2cosasinb is one of the important trigonometric formulas which is equal to sin (a + b) – sin (a-b). 2cosAsinB Formula is used in trigonometry to solve various equations and problems more efficiently. This formula is found using the sum and difference of the sine function.

In this article, we have covered, in brief, trigonometric ratios, 2Cos(a)Sin(b) Formula, its derivation and others in detail.

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:

Right Triangle ABC

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

2Cos(a)Sin(b) Formula

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

2Cos(a)Sin(b) Formula

2cosasinb formula is,

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

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

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

Derivation of 2CosaSinb Formula

We can derive the 2cosasinb formula with the help of the sum and difference of formulae of the sine function.

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

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

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

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

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

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

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

Articles Related to 2cosAsinB Formula:

• Trigonometric Identites
• Real-life Application of trigonometry
• Sin and Cos Formulas

Examples on 2CosaSinb Formula

Example 1: Solve the integral of 2cos 3x sin (5x/2).

Solution:

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

From 2cosasinb formula we have,

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

2 cos 3x sin (5x/2) = sin (3x + (5x/2)) – sin (3x – (5x/2))

= sin (11x/2) – sin (x/2)

Now, ∫2 cos 3x sin (5x/2)) dx = ∫[sin (11x/2) – sin (x/2)] dx

= ∫sin (11x/2) dx – ∫sin (x/2) dx

= -2/11 cos (11x/2) – (-2 cos (x/2))   {∫sin (ax) = -1/a cos (ax) + c}

= 2[cos (x/2) – 1/11 cos (11x/2)]

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

Example 2: Express 5cos (7x/2) sin 3x in terms of the sine function.

Solution:

From 2cosasinb  formula we have,

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

Now, 5 cos (7x/2) sin 3x = 5/2 [2 cos (7x/2) sin 3x]

= 5/2 [sin (7x/2 + 3x) – sin (7x/2 – 3x)]

= 5/2 [sin (13x/2) – sin (x/2)]

Hence, 5 cos (7x/2) sin 3x = 5/2 [sin (13x/2) – sin (x/2)].

Example 3: Find the value of the expression 4 cos (27.5°) sin (62.5°) using the 2cosasinb formula.

Solution: 

4 cos (27.5°) sin (62.5°) = 2 [2 cos (27.5°) sin (62.5°)]

From 2cosasinb formula we have,

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

Now, 2 [2 cos (27.5°) sin (62.5°)] = 2 [sin (27.5° + 62.5°) – sin (27.5° – 62.5°)]

=2 [sin (90°) – sin (-35°)]

= 2 [sin 90°+ sin 35°]      {Since, sin (-θ) = – sin θ}

= 2 [1 + 0.5735]               {Since, sin 35° = 0.5735, sin 90° = 1}

= 3.147          

Hence, 4 cos (27.5°) sin (62.5°) = 3.147

Example 4: Find the derivative of 7 cos 4x sin 11x.

Solution:

Derivative of 7 cos 4x sin 11x = d(7 cos 4x sin 11x)/dx

From 2cosasinb formula we have,

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

Now, 7 cos 4x sin 11x  = 7/2 [2 cos 4x sin 11x ]

= 7/2 [sin (4x + 11x) – sin (4x – 11x)]

= 5/2 [sin (15x) – sin (-7x)]

=  5/2 [sin (15x) + sin (7x)]     {Since, sin (-θ) = – sin θ}

Now, d(7 cos 4x sin 11x)/dx = d{5/2 [sin 15x + sin 7x]}/dx

= 5/2{d(sin 15x)/dx + d( sin 7x)/dx}

= 5/2 [15 cos 15x + 7 cos 7x]    {Since, d(sin ax)/dx = a cos ax}

= 37.5 cos 15x + 17.5 cos 7x

Hence, the derivative of 7 cos 4x sin 11x = [37.5 cos 15x + 17.5 cos 7x] .

Example 5: Express 2 cos 14x sin (3x/2) in terms of the sine function.

Solution:

From 2cosasinb  formula we have,

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

Now, 2 cos 14x sin (3x/2) =  sin (14x + 3x/2) – sin (14x – 3x/2)

= sin [(28x + 3x)/2] – sin [(28x – 3x)/2]

= sin (31x/2) – sin (25x/2)

Hence, 2 cos 14x sin (3x/2) = [sin (31x/2) – sin (25x/2)].

Example 6: Solve 6 sin (52.5 °) sin (127.5‬°) using the 2cosasinb formula.

Solution:

6 sin (52.5 °) sin (127.5‬°) = 3 [2 sin (52.5 °) sin (127.5‬°)]

From 2cosasinb formula we have,

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

Now, 3 [2 sin (52.5 °) sin (127.5‬°)] = 3 [sin (52.5° + 127.5°) – sin (52.5° – 127.5°)]

=2 [sin (180°) – sin (-75°)]

= 3 [sin 180°+ sin 75°]      {Since, sin (-θ) = – sin θ}

= 3 [1 + 0.9659]               {Since, sin 35° = 0.5735, sin 180° = 0}

= 5.8977       

Hence, 6 sin (52.5 °) sin (127.5‬°) = 5.8977

FAQs on 2cosAsinB

What is the Formula for 2 sin a sin b?

The 2sinasinb formula is, 2 sin A sin B = cos (A-B) – cos (A + B).

What is 2cosAsinB?

The 2cosasinb formula is, 2 cos A sin B = sin (A + B) – sin (A – B).

What is the Formula for 2 sinacosb?

The 2sinacosb formula is, 2sinAcosB = sin(A + B) + sin(A – B).

What is the Formula for 2cosAcosB?

The formula for 2cosAcosB 2cosAcosB formula is cos (A + B) + cos (A – B).