Friday, 30 May 2025
Wednesday, 28 May 2025
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
Sunday, 18 May 2025
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