2cosAsinB
2cosAsinB is equal to sin(A + B)  sin(A  B) which is one of the important formulas in trigonometry. We use the 2cosAsinB formula to solve different mathematical problems such as expressing trigonometric functions in terms of the sine function and evaluating integrals and derivatives involving trigonometric functions. We can derive the formula for 2cosAsinB using the compound angle formulas of the sine function given by sin(A + B) and sin(A  B).
Further, in this article, we will explore the trigonometric formula of 2cosAsinB and derive its formula. We will also solve different examples based on the concept to understand the application of the 2cosAsinB formula in simplifying trigonometric expressions.
1.  What is 2cosAsinB in Trigonometry? 
2.  2cosAsinB Formula 
3.  2cosAsinB Identity Proof 
4.  2cosAsinB Formula Application 
5.  FAQs on 2cosAsinB 
What is 2cosAsinB in Trigonometry?
2cosAsinB is an important trigonometric formula. We have four formulas in trigonometry of this kind given by, 2sinAsinB, 2sinAcosB, 2cosAcosB, and 2cosAsinB. In this article, we will focus on the 2cosAsinB formula in detail. We can derive its formula by subtracting the angle difference formula of the sine function, that is, sin(A  B) from the angle sum formula of the sine function, that is, sin(A + B). We can use the 2cosAsinB identity to simplify trigonometric expressions and calculate integrals and derivatives involving expressions of the form 2cosAsinB.
2cosAsinB Formula
The formula for 2cosAsinB is given by, 2cosAsinB = sin(A + B)  sin(A  B). Assume A = B in this formula, then we have 2cosAsinA = sin(A + A)  sin(A  A) = sin2A  sin0 = sin2A which gives us the formula sin2A = 2cosAsinA for the sin2A identity in trigonometry. The image given below shows the formula for the trigonometric identity 2cosAsinB:
2cosAsinB Identity Proof
In this section, we will prove that the formula for 2cosAsinB is equal to sin(A + B)  sin(A  B) using trigonometric formulas and identities. We will use the following formulas of the sine function:
 sin(A + B) = sinAcosB + cosAsinB  (1)
 sin(A  B) = sinAcosB  cosAsinB  (2)
Subtracting the formula (2) from (1), we have
sin(A + B)  sin(A  B) = (sinAcosB + cosAsinB)  (sinAcosB  cosAsinB)
⇒ sin(A + B)  sin(A  B) = sinAcosB + cosAsinB  sinAcosB + cosAsinB
⇒ sin(A + B)  sin(A  B) = cosAsinB + cosAsinB
⇒ sin(A + B)  sin(A  B) = 2cosAsinB
Hence, we have derived the formula for 2cosAsinB.
2cosAsinB Formula Application
Now that we know the formula of 2cosAsinB, let us now understand its application. Now, we will solve a few examples using the 2cosAsinB Formula to know how to apply the formula and how it simplifies the problem.
Example 1: Express 2 cos3x sin5x in terms of the sine function.
Solution: To express 2 cos3x sin5x in terms of the sine function, we will use the 2cosAsinB formula. We know that 2cosAsinB is equal to sin(A + B)  sin(A  B). Substitute A = 3x and B = 5x in the formula.
2 cos3x sin5x = sin(3x + 5x)  sin(3x  5x)
= sin8x  sin(2x)
= sin8x + sin2x  [Because sin(A) = sinA]
Answer: 2 cos3x sin5x in terms of the sine function is written as sin8x + sin2x.
Example 2: What is the integral of 2 cos7x sin4x using the 2cosAsinB formula?
Solution: To find the integral of 2 cos7x sin4x, we will use the 2cosAsinB formula. Substitute A = 7x and B = 4x in the formula.
2 cos7x sin4x = sin(7x + 4x)  sin(7x  4x)
= sin11x  sin3x
So, the integral of 2 cos7x sin4x is given by,
∫2 cos7x sin4x dx = ∫(sin11x  sin3x) dx
= ∫sin11x dx  ∫sin3x dx
= (1/11) cos11x + (1/3) cos3x + C [Because the intergal of sin(ax) is (1/a) cos(ax) + C]
Answer: ∫2 cos7x sin4x dx = (1/11) cos11x + (1/3) cos3x + C
Important Notes on 2cosAsinB
 The formula for 2cosAsinB is 2cosAsinB = sin(A + B)  sin(A  B)
 We can derive the formula for 2cosAsinB using the compound angle formulas of the sine function given by sin(A + B) and sin(A  B).
 It is used to simplify trigonometric expressions and solve integrals and derivatives.
☛ Related Topics:
2cosAsinB Examples

Example 1: Find the value of the expression 2 cos112.5° sin67.5° using the 2cosAsinB formula.
Solution: To find the value of 2 cos112.5° sin67.5°, we know that 2cosAsinB = sin(A + B)  sin(A  B). Substitute A = 112.5° and B = 67.5° in the formula.
2 cos112.5° sin67.5° = sin(112.5° + 67.5°)  sin(112.5°  67.5°)
= sin(180°)  sin(45°)
= 0  1/√2  [Because sin180° = 0 and sin45° = 1/√2]
= 1/√2
Answer: 2 cos112.5° sin67.5° = 1/√2

Example 2: What is the derivative of 2 cos4x sinx in terms of cosine function?
Solution: To find the derivative of 2 cos4x sinx, we will use the formula 2cosAsinB = sin(A + B)  sin(A  B). First, we will express 2 cos4x sinx in terms of sine by substituting A = 4x and B = x in the formula.
2 cos4x sinx = sin(4x + x)  sin(4x  x)
= sin5x  sinx
Now, the derivative of 2 cos4x sinx is given by,
d(2 cos4x sinx)/dx = d(sin5x  sinx)/dx
= d(sin5x)/dx  d(sinx)/dx
= 5cos5x  cosx
Answer: The derivative of 2 cos4x sinx in terms of cosine is 5cos5x  cosx.
FAQs on 2cosAsinB
What is 2cosAsinB in Trigonometry?
2cosAsinB is equal to sin(A + B)  sin(A  B). It is an important trigonometric formula that is used to solve various mathematical problems involving trigonometric functions.
How to Find the Value of 2cosAsinB?
We can derive the formula of 2cosAsinB using the formulas of sin(A+B) and sin(AB). We can subtract the formula of sin(AB) from sin(A+B) to find the formula of 2cosAsinB.
What is the Formula for 2cosAsinB?
The formula for 2cosAsinB is given by, 2cosAsinB = sin(A + B)  sin(A  B). This formula is used to express trigonometric expressions in terms of the sine function, and solve differentiation and integration problems.
How to Apply the 2cosAsinB Formula?
We can apply the 2cosAsinB formula to express trigonometric expressions in terms of sine function. We can also use the formula of 2cosAsinB to find the integrals and derivatives of expressions of the form 2cosAsinB.
How to Derive Sin2A Formula Using 2cosAsinB?
We can derive the sin2A formula using 2cosAsinB by substituting A = B in the formula 2cosAsinB = sin(A + B)  sin(A  B). We have 2cosAsinA = sin(A + A)  sin(A  A) = sin2A  sin0 = sin2A.
What is the Relationship Between 2cosAsinB and cosAsinB Formula?
We know that the formula of 2cosAsinB is 2cosAsinB = sin(A + B)  sin(A  B). Dividing both sides of the equation by 2, we have cosAsinB = (1/2) [sin(A + B)  sin(A  B)].
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