2cosacosb Formula

2cosacosb is an important trigonometric formula and is equal to cos (A + B) + cos (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

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

2cosacosb is one of the product-to-sum formulae. Similarly, we have three more product-to-sum formulae in trigonometry, i.e., 2sinasinb, 2sinacosb, and 2cosasinb. We can use the 2cosacosb identity for simplifying trigonometric expressions and also for calculating integrals and derivatives involving expressions of the form 2cosacosb.

The 2cosacosb formula is,

From the formula, we can observe that product of two cosine functions is converted into a sum of two other cosine functions.

For example,

The product-to-sum formulae for half angles is,

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

From the sum and difference formulae of trigonometry, we have,

• 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, by adding equations (1) and (2), we get

⇒ 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 + Cos A cos B 

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

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

Article Related to 2cosacosb Formula:

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

Problem 1: Express 3 cos 5x cos 7x in terms of the sum function.

Solution:

3 cos 5x cos 7x

Now multiply and divide the given equation by 2.

(2/2) 3 cos 5x cos 7x

= 3/2 [2 cos 5x cos 7x]

We have, 

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

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

= 3/2 [cos (12x) + cos (-2x)]

= 3/2 [cos 12x + cos 2x]    {since cos (-θ) = cos θ}

Hence, 3 cos 5x cos 7x = 3/2 [cos 12x + cos 2x]

Problem 2: Prove that, cos 2x cos (3x/2) – cos 3x cos (5x/2) = sin x sin (9x/2).

Solution:

Let us consider the equation on left hand side,

L.H.S = cos 2x cos (3x/2) – cos 3x cos (5x/2)

= 1/2 [2 cos 2x cos (3x/2) – 2 cos 3x cos (5x/2)}

We have, 2 Cos A cos B = cos (A + B) + cos (A – B)

= 1/2 [cos (2x + 3x/2) + cos (2x – 3x/2) – cos (3x + 5x/2) – 2 cos (3x – 5x/2)]

= 1/2 [cos (7x/2) + cos (x/2) – cos (11x/2) – cos (x/2)]

= 1/2 [cos (7x/2) – cos (11x/2)]

By using cos A – cos B = – 2 sin [(A + B)/2] sin [(A – B)/2] we get,

= 1/2 {-2 sin [(7x/2 + 11x/2)/2] sin [(7x/2 – 11/2)/2]}

= – sin (18x/4) sin(-4x/4)

= – sin (9x/2) sin (-x)

=  sin x sin (9x/2) {Since sin (-θ) = -sin θ}

= R.H.S

Hence, it is proved that cos 2x cos (3x/2) – cos 3x cos (5x/2) = sin x  sin (9x/2)

Problem 3: What is the value of the integral of 2 cos 4x cos (5x/2) dx?

Solution:

By,

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

2 cos 4x cos (5x/2) = cos [4x + (5x/2)] + cos [4x – (5x/2)]

= cos (13x/2) + cos (3x/2)

Now, integral of 2 cos 4x cos (5x/2) dx =∫2 cos 4x cos (5x/2) dx 

= ∫[cos (13x/2) + cos (3x/2)] dx

= 2/13 sin (13x/2) + 2/3 cos (3x/2) + C   {Since, the integral of cos(ax) is (1/a) sin (ax) + C}

Hence, ∫ 2 cos 4x cos (5x/2) dx = (2/3) sin (3x/2) + (2/13) cos (13x/2) + C

Problem 4: Determine the derivative of 2 cos (x/2) cos (3x/2).

Solution:

By,

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

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

= cos (4x/2) + cos (-2x/2)

= cos (2x) + cos (-x) 

= cos x + cos 2x     {since cos (-θ) = cos θ}

Now, derivative of 2 cos (x/2) cos (3x/2) = d [2 cos (x/2) cos (3x/2) ]/dx 

= d [cos x + cos 2x]/dx

= – sin x – 2 sin 2x     {Since, d[cos (ax)] = -a sin (ax)}

= – (sin x + sin 2x)

Hence, the derivative of 2 cos (x/2) cos (3x/2) = – (sin x + sin 2x).

Problem 5: Find the value of the expression 3 cos 37.5° cos 52.5° using the 2coscosb formula.

Solution:

3 cos 37.5° cos 52.5° = 3/2 [2 cos 37.5° cos 52.5°]

By,

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

3/2 [2 cos 37.5° cos 52.5°] = 3/2 [cos (37.5° + 52.5°) + cos (37.5° – 52.5°)]

= 3/2 [cos (90°) + cos (-12°)]

= 3/2 [cos 90° + cos 12°]  {since cos (-θ) = cos θ}

cos 90° = 0 and cos 12° = 0.9781

= 3/2 [0 + 0.9781]

= 1.46715

Hence, 3 cos 37.5° cos 52.5° = 1.46715

Problem 6: Write 4 cos 2y cos 5y in terms of the sum function.

Solution:

4 cos 2y cos 5y = 2 ( 2 cos 2y cos 5y)

We have,

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

2 ( 2 cos 2y cos 5y) = 2 [cos (2y + 5y) + cos (2y – 5y)]

= 2 [cos 7y + cos (-3y)]

= 2 [cos 7y + cos 3y]   {since cos (-θ) = cos θ}

Hence, 4 cos 2y cos 5y =  2 [cos 7y + cos 3y] 

Problem 7: Find the value of the expression 2 cos 44.5° cos 135.5° using the 2coscosb formula.

Solution:

By,

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

2 cos 44.5° cos 135.5° = cos (44.5° + 135.5°) + cos (44.5° – 135.5°)

= cos (180°) + cos (-91°)

= cos (180°) + cos (91°) {since cos (-θ) = cos θ}                     

= -1 + (-0.01745) = -1.01745 {cos 180° = -1 and cos 91° = -0.01745}

Hence, 2 cos 44.5° cos 135.5° = -1.01745

What is the formula for 2cos a sin b?

The formula for 2cos a sin b is, 2cosAsinB = sin(A + B) – sin(A – B).

What is the formula for 2 sin a cos b?

The formula for 2 sin a cos b is, 2sinAcosB = sin(A + B) + sin(A – B).

What is the formula for 2 sin a sin b?

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

What is the formula for 2 cos a cos b?

The formula for 2 cos a cos b is, 2 Cos A Cos B = Cos (A + B) + Cos (A – B).

What are Applications of 2cosAcosB Formula?

2cosAcosB formula is used in solving various trigonometric, differential and integral questions.