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2cosAsinB Formula

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  • Last Updated : 03 Jul, 2022
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2cosasinb is one of the important trigonometric formulas which is equal to sin (a + b) – sin (a-b). In mathematics, trigonometry is an important branch that studies the relationship between angles and sides of a right-angled triangle, which has a wide range of applications in numerous fields like astronomy, architecture, marine biology, aviation, etc. There are six trigonometric ratios, of which three ratios are the reciprocals of the other three trigonometric ratios. A trigonometric ratio is a ratio between the lengths of the sides of a right triangle. 

 

Trigonometric Ratios

  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

2cosasinb formula

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

The 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

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)

Solved Examples

Problem 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)]

Problem 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)].

Problem 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

Problem 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] .

Problem 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)].

Problem 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.

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