Question

derive sin4A formulae

Answer 1

Sin4A = sin(2A +2A)
= sin2A *Cos 2A + sin2A* cos2A
= 2 Sin2A Cos2A
=2 [( 2sinA. CosA) * (1 - 2sin² A)]
= 4sinA CosA - 4Sin³A CosA

Answer 2

Therefore, $sin4A$ formula is $4(sinA.cos^{3}A- sin^{3}A.cosA)$

Step-by-step explanation:

Given,

Derive $sin4A$ formula.

∴ $sin4A$

$=sin2(2A)$

$=2sin2A.cos2A$  (∵$sin2A=2sinA.cosA$)

$=2(2sinA.cosA)(cos^{2} A-sin^{2} A)$

$=4sinA.cos^{3}A- 4sin^{3}A.cosA$

$=4(sinA.cos^{3}A- sin^{3}A.cosA)$

∴ $sin4A$ formula is $4(sinA.cos^{3}A- sin^{3}A.cosA)$