Question

sin(2A+3B).cos(2A-3B)+cos(2A+3B).sin(2A-3B)= sin 4A prove that
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Answer 1

Given :

• sin(2A + 3B).cos(2A - 3B) + cos(2A + 3B).sin(2A - 3B)

To prove :

• sin(2A + 3B).cos(2A - 3B) + cos(2A + 3B).sin(2A - 3B) = sin4A

Proof :

Let's assume sin(2A + 3B) = R and cos(2A - 3B) = T

• As we know that

✒ sin(A + B) = sinA.cosB + cosA.sinB

From LHS

→ sin(2A + 3B).cos(2A - 3B) + cos(2A + 3B).sin(2A - 3B)

→ sinR.cosT + cosR.sinT (By assuming)

• According to the formula of sin(A+B)

→ sin(R + T)

• Put the value of "R" & "T"

→ sin[(2A + 3B) + (2A - 3B)]

→ sin(2A + 3B + 2A - 3B)

→ sin(2A + 2A + 3B - 3B)

→ sin4A = RHS

•°• LHS = RHS proved

Answer 2

$\bf SOLUTION$

sin(2A+3B).cos(2A-3B)+cos(2A+3B).sin(2A-3B)

sin(2A + 3B + 2A - 3B)

sin4A