if sin2a=2sina*cosa
So, sin4a=?
Answers
Answered by
105
Sin2a= 2sinAcosA
sin4A=sin 2*(2A)
= 2sin2Acos2A
sin4A=sin 2*(2A)
= 2sin2Acos2A
Anonymous:
If u want I can simplify this further
Answered by
72
(4 sin A cos^2 A - sin A) cos A + sin A ( (2 cos^2 A
-1) cos A - (2 sinAcosA) sin A)
4 sin A cos^3 A - sin A cos A + 2 sin A cos^3 A -sin A cos A - 2sin^3A cosA
6 sin A cos^3 A - 2 sin A cos A - 2sin^3A cosA
2 sin A cos A ( 3 cos^2 A - 1 - sin^2 A)
2 sin A cos A ( 3 cos^2 A - 1 - (1 - cos^2 A))
2 sin A cos A ( 3 cos^2 A - 1 - 1 + cos^2 A)
2 sin A cos A ( 4 cos^2 A - 2)
8 sin A cos^3 A - 4 sin A cos A
sin 4A = 8 sin A cos^3 A - 4 sin A cos A
sin A = sin A sin 2A = 2 sinAcos A
sin 3A = 4 sin A cos^2 A - sin A
sin 4A = 8 sin A cos^3 A - 4 sin A cos A
4 sin A cos^3 A - sin A cos A + 2 sin A cos^3 A -sin A cos A - 2sin^3A cosA
6 sin A cos^3 A - 2 sin A cos A - 2sin^3A cosA
2 sin A cos A ( 3 cos^2 A - 1 - sin^2 A)
2 sin A cos A ( 3 cos^2 A - 1 - (1 - cos^2 A))
2 sin A cos A ( 3 cos^2 A - 1 - 1 + cos^2 A)
2 sin A cos A ( 4 cos^2 A - 2)
8 sin A cos^3 A - 4 sin A cos A
sin 4A = 8 sin A cos^3 A - 4 sin A cos A
sin A = sin A sin 2A = 2 sinAcos A
sin 3A = 4 sin A cos^2 A - sin A
sin 4A = 8 sin A cos^3 A - 4 sin A cos A
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