Question

sin4A= who's the formula

Answer 1

Hello,
we have:
sin 4A =sin2(2A);
=2sin2Acos2A;

since, sin2A=2sinAcosA:
=2(2sinAcosA)cos2A;
=4sinAcosAcos2A;

since, cos2A=cos²A-sin²A:
=4sinAcosA(cos²A-sin²A);
=4sinAcos³A-4sin³AcosA;
=4sinAcos³A-4cosAsin³A

bye :-)

Answer 2

Given : $Sin4A=?$

$Sin4A=Sin2(2A)$

$=2\times Sin2A \times Cos2A$

$=2\times 2\times SinA\times CosA \times Cos2A$

$=4\times SinA\times CosA \times (Cos^{2}A -Sin^{2}A )$

$=4\times SinA\times Cos^{3}A -4\times CosA\times Sin^{3}A$

$Sin4A=4\times SinA\times Cos^{3}A -4\times CosA\times Sin^{3}A$.