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.

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.

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:

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.

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:

• Cos3x
• Derivative of xsinx
• Derivative of sin3x

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(A-B). We can subtract the formula of sin(A-B) 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)].