2cos a cos b Formula with Solved Examples

2cos a cos b Formula

2 Cos A Cos B is the product of trigonometric sum mathematical statements utilised to rewrite the product of cosines into difference or sum.

The 2 cos a cos b expression can help solve integration mathematical statements referring to the product of trigonometric ratios, such as cosine. 

Furthermore, the expression of 2 Cos a Cos b can assist in altering the trigonometric demonstration by reckoning the product term, such as Cos a Cos b, and changing it into a sum. 

The Formula for 2 Cos a Cos b

The formula for 2 Cos a Cos b can be gained by observing the difference and sum for the cosine formula.

We know that,

Cos (a + b) = Cos a Cos b – Sin a Sin b…… (i)

Cos (a – b) = Cos a Cos b + Sin a Sin b……(ii)

Adding the above two equations, 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) = 2 Cos a Cos b 

Hence, the formula for the 2 Cos a Cos b is supplied as:

2 Cos a Cos b = Cos (a + b) + Cos (a – b)  

In mathematics, there are a total of six trigonometric functions of angles, which are cited below:

• Sine
• Cosine
• Tangent
• Cotangent
• Secant
• Cosecant

What do you mean by 2 Cos a Cos b?

The formula of 2 Cos a Cos b is – Cos (a + b) + Cos (a – b) = 2 Cos a Cos b.

This mathematical formula converts the product of two functions of the Cos function as the sum.

Let’s take an example;

2 cos (2x) cos (2y) = cos (2x + 2y) + cos (2x – 2y)

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

Solved Examples

1. Express 8 cos y cos 2y in terms of sum functions.

8 cos y cos 2y = 4 [2 cos y cos 2y]

Using the 2cos a cos b Formula,

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

= 4[cos (y + 2y) + cos (y – 2y)]

= 4[cos 3y + cos (-y)]

= 4 [cos 3y + cos y]

Hence, 8 cos y cos 2y in terms of functions of the sum is 4 [cos 3y + cos y]

1. Show 6 cos x cos 2x in terms of the sum functions.

Consider,

6 cos x cos 2x = 3 [2 cos x cos 2x]

Using the formula 2 cos a cos b = cos (a + b) + cos (a – b),

= 3[cos (x + 2x) + cos (x – 2x)]

= 3[cos 3x + cos (-x)]

= 3 [cos 3x + cos x]