The 2cosacosb formula belongs to the category of product-to-sum formulas, which is utilized to convert a product into a sum. Trigonometry is a specialized field of study that concerns itself with the interrelationships between angles, heights, and lengths of right triangles. The trigonometric ratios are defined as the ratios of the sides of a right triangle. There are six main ratios in trigonometry, including sin, cos, tan, cot, sec, and cosec, each with its own set of formulas. These formulas utilize the three sides and angles of a right-angled triangle. In this context, we will examine the 2cosacosb formula in greater detail.
Trigonometric identities play an important role in solving problems in mathematics, physics, engineering, and other fields. One such important identity is the 2cosacosb formula, which is used to express the product of two cosine functions in terms of their sum and difference. In this article, we will discuss the 2cosacosb formula, its derivation, and some examples to help you understand it better.
What is 2 Cosacosb Formula?
The 2cosacosb formula represents the equation 2 cos A cos B = cos (A + B) + cos (A – B). This formula is used to convert the product of two cos functions into the sum of two other cos functions. To illustrate, consider the following example:
Derivation of 2cosacosb formula:
The 2cosacosb formula can be derived using the following trigonometric identity:
cos(A + B) = cos A cos B – sin A sin B
If we rearrange this formula, we can get:
cos A cos B = cos(A + B) + sin A sin B
Similarly, we can also write:
cos A cos (-B) = cos(A – B) + sin A sin (-B)
Using the fact that sin(-x) = -sin(x), we get:
cos A cos B = cos(A + B) + sin A sin B
cos A cos B = cos(A – B) – sin A sin B
Adding these two equations, we get:
2cos A cos B = cos(A + B) + cos(A – B)
Dividing both sides by 2, we get:
cos A cos B = (cos(A + B) + cos(A – B))/2
This is the 2cosacosb formula, which expresses the product of two cosine functions in terms of their sum and difference.
Example 1
Find the value of cos 60° cos 30° using the 2cosacosb formula.
Solution:
Using the 2cosacosb formula, we have:
cos 60° cos 30° = (cos(60° + 30°) + cos(60° – 30°))/2
= (cos 90° + cos 30°)/2 = (0 + √3/2)/2
= √3/4
Therefore, cos 60° cos 30° = √3/4.
Example 2
Express 6 cos x cos 3x in terms of sum function.
Solution:
Using the 2cosacosb formula, we have:
2 cos A cos B = cos (A + B) + cos (A – B)
Here, A = x and B = 3x, so we can write:
6 cos x cos 3x = 3(2cosx cos3x)
= 3 [cos (x + 3x) + cos (x – 3x)]
= 3 [cos 4x + cos (-2x)]
= 3 cos 4x + 3 cos 2x
Therefore, 6 cos x cos 3x in terms of sum function is equal to 3 cos 4x + 3 cos 2x.
Example 3
Write 10 cos x cos 3x as sum.
Solution:
To express 10 cos x cos 3x as a sum of cosine functions, we can use the 2cosacosb formula:
2 cos A cos B = cos (A + B) + cos (A – B)
Here, A = x and B = 3x, so we can write:
10 cos x cos 3x = 5 (2 cos x cos 3x)
= 5 [cos (x + 3x) + cos (x – 3x)]
= 5 [cos 4x + cos (-2x)]
= 5 cos 4x + 5 cos 2x
Therefore, 10 cos x cos 3x can be expressed as a sum of cosine functions as 5 cos 4x + 5 cos 2x.
Example 4
Evaluate the integral ∫ 3 cos x cos 4x dx.
Solution:
To solve this integral, we can use the product-to-sum formula to rewrite the integrand in terms of cosine functions:
cos A cos B = 1/2 [cos (A + B) + cos (A – B)]
Here, A = x and B = 4x, so we have:
cos x cos 4x = 1/2 [cos (x + 4x) + cos (x – 4x)]
= 1/2 [cos 5x + cos (-3x)]
= 1/2 cos 5x + 1/2 cos 3x
Substituting this into the original integral, we get:
∫ 3 cos x cos 4x dx = 3 ∫ [1/2 cos 5x + 1/2 cos 3x] dx
= 3/2 ∫ cos 5x dx + 3/2 ∫ cos 3x dx
= 3/10 sin 5x + 3/6 sin 3x + C
where C is the constant of integration.
Therefore, the value of the integral ∫ 3 cos x cos 4x dx is (3/10) sin 5x + (1/2) sin 3x + C.
