2 Sin a Sin b formula

Formula of 2 sin a sin b

Trigonometry identity 2 Sin A Sin B Formula is very important which is regularly used to solve many trigonometric functions. By using this 2 Sin A Sin B = Cos(A-B) – Cos(A+B) we can solve various differentiation and integration questions. This 2SinASinB trigonometry formula is simply derived by the difference of the Cosine angle difference and cosine function angle addition.

In this article we will drive the formula of 2 sin a sin b. For better understanding of this formula we will solve some examples as well. Using the 2 Sin A Sin B trigonometric identity formula, we can derive Sin A Sin B trigonometric formula which is given by, Sin A Sin B = (1/2) [cos (A – B) – cos (A + B)]. Same kind of trigonometric formulas are there, please go through below list –

• 2SinASinB
• 2CosACosB
• 2SinACosB
• 2CosASinB

Cos 2A Formula

Lets go through another fact – we know that 2 Sin A Sin B = cos(A – B) – cos(A + B). If these two angles A and B become equal then this formula will change to cos 2A trigonometric identity.

• If A =B then
• 2 Sin A Sin A = cos(A – A) – cos(A + A)
• 2 Sin² A = cos (0) – cos (2A)
• So we can rewrite it as 2 Sin² A = 1 – cos 2A
• ⇒ Cos 2A = 1 – 2Sin²A

Proof of 2 Sin a Sin b formula

To prove this 2 Sin a Sin b formula we need two trigonometric identities cos (A – B) = cos A cos B + sin A sin B and cos (A + B) = cos A cos B – sin A sin B.

• Now, to prove the formula for 2SinASinB, we will use the following trigonometric formulas of the cosine function:
• cos(A – B) = cosAcosB + sinAsinB — (1)
• cos(A + B) = cosAcosB – sinAsinB — (2)
• 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
• Rewrite 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.

Solved Examples

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

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.

Example 3: 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
• So ∫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]

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