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

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If find x. 1/(9!) + 1/(10!) = x/(11!).

  Here n! is read as n factorial. It can be written as n(n-1)! or n(n-1)(n-1)!