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.
1.  What is 2SinACosB in Trigonometry? 
2.  2SinACosB Formula 
3.  Proof of 2SinACosB Formula 
4.  How to Apply 2sinAcosB Formula? 
5.  FAQs on 2SinACosB 
What is 2SinACosB in Trigonometry?
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.
2SinACosB Formula
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(AB). 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)].
Proof of 2SinACosB Formula
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.
How to Apply 2sinAcosB Formula?
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:
2SinACosB Examples

Example 1: Find the integral of 2 sin7x cos4x using the 2sinAcosB formula.
Solution: We know that 2sinAcosB = sin(A + B) + sin(A  B). Substitute A = 7x and B = 4x in the formula.
2 sin7x cos4x = sin(7x + 4x) + sin(7x  4x)
= sin11x + sin3x
To find the intergal of 2 sin7x cos4x, we have
∫2 sin7x cos4x dx = ∫(sin11x + sin3x) dx
= ∫sin11x dx + ∫sin3x dx
= (1/11) cos11x  (1/3) cos3x + C
Answer: ∫2 sin7x cos4x dx = (1/11) cos11x  (1/3) cos3x + C

Example 2: Express 2 sin2x cos3x in terms of sine function.
Solution: To express 2 sin2x cos3x in terms of sine function, substitute A = 2x and B = 3x in 2sinAcosB = sin(A + B) + sin(A  B).
2 sin2x cos3x = sin(2x + 3x) + sin(2x  3x)
= sin5x + sin(x)
= sin5x  sinx  [Because sin(A) = sinA]
Answer: 2 sin2x cos3x = sin5x  sinx
FAQs on 2SinACosB
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(AB). 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.
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