Question

sin 3A=3 sin A-4 sin³A please answer fast........

Answer 1

Step-by-step explanation:

sin 3A = sin ( 2A + A)

=> sin 2A cos A + cos 2A sin A

=> 2 sin A cos A cos A + (1 - 2sin²A)sin A

=> 2 sin A cos²A + sin A - 2 sin³ A

=> 2 sin A ( 1 - sin²A) + sinA - 2 sin³ A

=> 2 sin A - 2 sin³ A + sin A - 2 sin³ A

=> 3 sin A - 4 sin³A hence proved

[ sin (A+B) = sin A cos B + cos A sin B ]

[ sin 2A = 2 sin A cos A and

cos 2A = 1-2sin³A]

these are the identities used in deriving the above formula