Question

derive sin4A in terms of SinA​

Answer 1

Answer:

[√(1 - SIn²A)](4SinA - 8Sin³A)

Step-by-step explanation:

sin 4A =sin2(2A)   

=2sin2Acos2A    (∵sin2A=2sinAcosA)

=2(2sinAcosA)cos2A

= 4sinAcosAcos2A

= 4SinA√(1 - SIn²A)(1-2Sin²A) (∵Cos2A = 1 - 2Sin²A; CosA = √(1 - SIn²A))

= [√(1 - SIn²A)](4SinA - 8Sin³A)