2sinAcosB

2sinAcosB is a trigonometric formula that can be derived using the compound angle formulas of the sine function. The formula for 2sinAcosB is given by, 2sinAcosB = sin(A + B) + sin(A - B). We can use this formula to solve various mathematical problems including simplification of trigonometric expressions and calculation of integrals and derivatives. We have four such trigonometric formulas which are 2sinAsinB, 2cosAcosB, 2sinAcosB, and 2cosAsinB.

In this article, we will explore the concept of 2sinAcosB and derive its formula using trigonometric formulas of the sine function. We will also find out how to apply the 2sinAcosB formula and solve a few examples for a better understanding of its application.

2sinAcosB is one of the important trigonometric formulas in trigonometry. Its formula can be used to solve various trigonometric problems. It is used to simplify trigonometric expressions and solve complex integrals and derivatives. The formula of 2sinAcosB is derived by taking the sum of the compound angle formulas (angle sum and angle difference) of the sine function, that is, sin(A - B) and sin(A + B). We can apply the formula of 2sinAcosB when the sum and difference of two angles A and B are known.

The formula for the 2sinAcosB identity in trigonometry is 2sinAcosB = sin(A + B) + sin(A - B). We can derive this formula by adding the sine function formulas sin(A+B) and sin(A-B). We can use the formula of 2sinAcosB when pair values of the angles A and B or their sum and difference A + B and A - B are known. If the two angles A and B become equal, then we get the formula for the sin2A identity in trigonometry. The image given below shows the formula for 2sinAcosB:

If we divide both sides of the formula 2sinAcosB = sin(A + B) + sin(A - B) by 2, we get the formula for sinAcosB as sinAcosB = (1/2) [sin(A + B) + sin(A - B)].

Now that we know that the formula for 2sinAcosB is equal to sin(A + B) + sin(A - B), we will derive this using the compound angle formulas of the sine function. We will use the following formulas to derive the formula of 2sinAcosB:

• sin(A + B) = sinAcosB + sinBcosA --- (1)
• sin(A - B) = sinAcosB - sinBcosA --- (2)

Adding the above two formulas (1) and (2), we have

sin(A + B) + sin(A - B) = (sinAcosB + sinBcosA) + (sinAcosB - sinBcosA)

⇒ sin(A + B) + sin(A - B) = sinAcosB + sinBcosA + sinAcosB - sinBcosA

⇒ sin(A + B) + sin(A - B) = sinAcosB + sinAcosB --- [Cancelling out sinBcosA and -sinBcosA]

⇒ sin(A + B) + sin(A - B) = 2sinAcosB

Hence, we have derived the formula of 2sinAcosB using the angle sum and angle difference formulas of the sine function.

In this section, we will understand the application of the 2sinAcosb formula in simplifying trigonometric expressions and calculating complex integration and differentiation problems. Let us solve a few examples below stepwise to understand how to apply the formula of 2sinAcosB.

Example 1: Find the derivative of 2 sinx cos2x using the 2sinAcosB formula.

Solution: To find the derivative of 2 sinx cos2x, substitute A = x and B = 2x into the formula 2sinAcosB = sin(A + B) + sin(A - B) to simplify and express it in terms of sine function. Therefore, we have

2 sinx cos2x = sin(x - 2x) + sin(x + 2x)

= sin(-x) + sin3x

= -sinx + sin3x --- [Because sin(-A) = -sinA]

Now, the derivative of 2 sinx cos2x is given by,

d(2 sinx cos2x)/dx = d(-sinx + sin3x)/dx

= d(-sinx)/dx + d(sin3x)/dx

= -d(sinx)/dx + 3cos3x

= -cosx + 3cosx

Answer: The derivative of 2 sinx cos2x is -cosx + 3cosx.

Example 2: Find the value of 2 sin135° cos45°.

Solution: We know values of trigonometric functions at specific angles including 0°, 30°, 45°, 60°, and 90°. So, we will use the 2sinAcosB formula to find the value of the expression 2 sin135° cos45°.

2 sin135° cos45° = sin(135° + 45°) + sin(135° - 45°)

= sin180° + sin90°

= 0 + 1

= 1

Answer: 2 sin135° cos45° = 1

Important Notes on 2sinAcosB

• The formula of 2sinAcosB is 2sinAcosB = sin(A + B) + sin(A - B).
• We can derive the formula using sin(A + B) and sin(A - B).
• The formula for 2sinAcosB is used to simplify and determine values of trigonometric expressions, integrals and derivatives.

☛ Related Topics:

• Cot3x
• Cot2x
• Antiderivative Rules

What is 2SinACosB in Trigonometry?

2sinAcosB is one of the important trigonometric formulas in trigonometry. The value of 2sinAcosB is equal to sin(A + B) + sin(A - B), for angles A and B. This formula can be derived using the compound angle formulas of the sine function.

What is the Formula of 2sinAcosB?

The formula for the 2sinAcosB identity in trigonometry is 2sinAcosB = sin(A + B) + sin(A - B). We can use the formula of 2sinAcosB when pair values of the angles A and B or their sum and difference A + B and A - B are known.

How to Prove 2sinAcosB Formula?

We can derive the formula of 2sinAcosB by adding the sine function formulas sin(A+B) and sin(A-B). We have sin(A + B) + sin(A - B) = (sinAcosB + sinBcosA) + (sinAcosB - sinBcosA) which implies 2sinAcosB = sin(A + B) + sin(A - B).

What is 2SinACosB Equal to?

2sinAcosB is equal to the sum of sin(A + B) and sin(A - B), that is, 2sinAcosB is equal to sin(A + B) + sin(A - B).

What are the Applications of 2sinAcosB?

Some of the common applications of 2sinAcosB are simplifying and determining values of trigonometric expressions, integrals, and derivatives.