Tuesday, 20 May 2025

Prove that sin3A=3sinA-4sin^3A.

 sin3A = sin(2A+A)

= sin 2A cos A+cos 2AsinA

= 2sinAcosA.cos A+(1-2sin²A)sinA

[Here sin2A=2sinAcosA and cos2A=2cos²-1=1-2sin²A]

=2sinAcos²A+(1-2sin²A)sinA

=2sinA(1-sin²A)+(1-2sin²A)sinA

[Here sin²A+cos²A=1 => cos²A=1-sin²A]

=2sinA-2sin^3A+sinA-2sin^3A

=3sinA-4sin^3A

No comments:

Post a Comment

Check whether 7+3x is a factor of 3x^3+7x ? [ncert]

                               Now take 7+3x=0 =>3x= -7 => x= -7/3 => If p(x)=3x^3+7x , then, => p(-7/3) = 3(-7/3)^3+7(-7/3) =...