2SinASinB is one of the important trigonometry formulas that is used to simplify trigonometric expressions and solve various complex integration and differentiation problems. The formula for the 2SinASinB identity is given by the difference of the angle sum and angle difference formulas of the cosine function. Mathematically, we can write the 2SinASinB formula as 2SinASinB = cos(A - B) - cos(A + B), for angles A and B.

In this article, let us derive the formula and understand the proof of the 2SinASinB trigonometric identity. We will also explore its application with the help of solved examples for a better understanding of the usage of the 2SinASinB formula.

2SinASinB identity is a trigonometric formula and it is used for the simplification of trigonometric expressions. It is also used for evaluating integrals involving trigonometric functions for easy calculation. We can derive the formula for 2SinASinB using the angle sum and angle difference formulas of the cosine function. Using the 2SinASinB formula, we can derive another trigonometric formula for SinA SinB given by, sinA sinB = (1/2)[cos(A - B) - cos(A + B)]. We have mainly four trigonometric formulas of this kind as follows:

In this article, we will mainly focus on the 2SinASinB formula and derive its formula. Let us first go through its formula given below:

Formula for 2SinASinB is written as the difference of the cos(A - B) and cos(A + B). Therefore, the trigonometric formula for 2SinASinB is given by, 2SinASinB = cos(A - B) - cos(A + B), for compound angles A + B and A - B. Using this formula, we have the formula for sinAsinB given by, sinA sinB = (1/2)[cos(A - B) - cos(A + B)]. The image given below shows the 2SinASinB formula in trigonometry:

If the two angles A and B become equal, then we can write the formula as:

2SinASinA = cos(A - A) - cos(A + A)

⇒ 2Sin2A = cos 0 - cos2A

⇒ 2Sin2A = 1 - cos2A

⇒ cos2A = 1 - 2Sin2A

Hence, we have got the formula for the cos2A identity.

Now, to prove the formula for 2SinASinB, we will use the following trigonometric formulas of the cosine function:

Next, subtract the above two formulas.

cos(A + B) - cos(A - B) = (cosAcosB - sinAsinB) - (cosAcosB + sinAsinB)

⇒ cos(A + B) - cos(A - B) = cosAcosB - sinAsinB - cosAcosB - sinAsinB

⇒ cos(A + B) - cos(A - B) = - sinAsinB - sinAsinB

⇒ cos(A + B) - cos(A - B) = - 2sinAsinB

⇒ cos(A - B) - cos(A + B) = 2sinAsinB --- [Multiplying both sides by -1]

Hence, we have proved that the formula for 2SinASinB is cos(A - B) - cos(A + B) = 2sinAsinB.

In this section, we will understand the application of the 2SinASinB formula in simplifying trigonometric expressions and determining the integrals of combinations of trigonometric functions. Let us solve a few examples stepwise to understand how to apply the formula 2SinASinB = cos(A - B) - cos(A + B).

Example 1: Solve the integral ∫2 sinx sin4x dx.

Solution: We will use the formula 2SinASinB = cos(A - B) - cos(A + B) to solve the given integral. For the trigonometric expression, 2 sinx sin4x, A = x and B = 4x. Now, substituting these values into the 2SinASinB formula, we have 2 sinx sin4x = cos(x - 4x) - cos(x + 4x) = cos(-3x) - cos 5x = cos3x - cos5x (Because cos(-A) = cosA). Therefore, we have

∫2 sinx sin4x dx = ∫(cos3x - cos5x) dx

= ∫cos3x dx - ∫cos5x dx

= (1/3) sin3x - (1/5) sin5x + C

Answer: Hence, the integral of 2 sinx sin4x is equal to (1/3) sin3x - (1/5) sin5x + C using the 2SinASinB formula.

Example 2: Express the expression 2 sin6x sin3x in terms of the cosine function.

Solution: We know that 2SinASinB = cos(A - B) - cos(A + B). Now, substitute the values A = 6x and B = 3x into the formula.

2 Sin6x Sin3x = cos(6x - 3x) - cos(6x + 3x)

= cos3x - cos9x

Therefore the expression 2 sin6x sin3x in terms of the cosine function is written as cos3x - cos9x.

Important Notes on 2SinASinB

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Example 1: Find the integral of 2 sin5x sin2x.

Solution: To find the integral of 2 sin5x sin2x, we will use the 2sinAsinB formula given by 2SinASinB = cos(A - B) - cos(A + B). Substitute A = 5x and B = 2x into the formula.

∫2 sin5x sin2x dx = ∫[cos(5x - 2x) - cos(5x + 2x)] dx

= ∫(cos3x - cos7x) dx

= ∫cos3x dx - ∫cos7x dx

= (1/3) sin3x - (1/7) sin7x + C --- [Because the integral of cos(ax) is equal to (1/a) sin(ax) + C]

Answer: Integral of 2 sin5x sin2x is equal to (1/3) sin3x - (1/7) sin7x + C.

Example 2: Express 2 sin2x sin4x in terms of the cosine function.

Solution: The formula to be used is 2SinASinB = cos(A - B) - cos(A + B) to solve the problem. Consider A = 2x and B = 4x in the formula.

2 sin2x sin4x = cos(2x - 4x) - cos(2x + 4x)

= cos(-2x) - cos6x

= cos2x - cos6x --- [Because cos(-A) = cosA]

Answer: 2 sin2x sin4x is written as cos2x - cos6x in terms of the cosine function.

Example 3: Evaluate the derivative of 2 sin6x sin7x using the 2sinAsinB formula.

Solution: The formula for 2sinAsinB is 2SinASinB = cos(A - B) - cos(A + B). Substitute A = 6x and B = 7x in the formula.

2 sin6x sin7x = cos(6x - 7x) - cos(6x + 7x)

= cos(-x) - cos(13x)

= cosx - cos13x

Now, we know that the derivative of cos(ax) is equal to -a sin(ax). Therefore, we have

d(2 sin6x sin7x)/dx = d(cosx - cos13x)/dx

= d(cosx)/dx + d(cos13x)/dx

= -sinx + 13 sin13x

Answer: The derivative of 2 sin6x sin7x is -sinx + 13 sin13x

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2SinASinB identity is a trigonometric formula and it is used for the simplification of trigonometric expressions. It is also used for evaluating integrals involving trigonometric functions for easy calculation. Mathematically, we can write the 2SinASinB formula as 2SinASinB = cos(A - B) - cos(A + B), for angles A and B.

Formula for 2SinASinB is written as the difference of the cos(A - B) and cos(A + B). Therefore, the trigonometric formula for 2SinASinB is given by, 2SinASinB = cos(A - B) - cos(A + B), for compound angles A + B and A - B.

We can prove the 2sinAsinB formula using the angle sum and angle difference formulas of the cosine function, that is, the formulas of cos(A+B) and cos(A-B).

2SinASinB is one of the important trigonometry formulas that is used to simplify trigonometric expressions and solve various complex integration and differentiation problems.

We know that 2SinASinB = cos(A - B) - cos(A + B). Dividing both sides of the equation by 2, we have SinASinB = (1/2) [cos(A - B) - cos(A + B)] which is the formula for sinAsinB identity.