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2cosacosb Formula

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2cosa2cosb is one of the important trigonometric formulas which 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. Trigonometry is an important branch of mathematics that deals with the relationship between angles and lengths of sides of right-angled triangles. There are six trigonometric ratios, where a trigonometric ratio is a ratio between the sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant, and cotangent are the six trigonometric ratios, where cosecant, secant, and cotangent functions are the reciprocal functions of sine, cosine, and tangent functions, respectively.

  • sin θ = opposite side/hypotenuse
  • cos θ = adjacent side/hypotenuse 
  • tan θ = opposite side/adjacent side
  • cosec θ = 1/sin θ = hypotenuse/opposite side
  • sec θ = 1/cos θ = hypotenuse/adjacent side
  • cot θ = 1/tan θ = adjacent side/opposite side

2cosacosb Formula

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

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

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]

Derivation

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)

Sample Problems

Problem1: 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



Last Updated : 10 Jan, 2024
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