Prove that Sin5A= 5Cos^5 A. SinA -10cos^2 A. Sin^3 A+ Sin^5 A

By de moivres:

cos5A + i sin5A = (cosA + i sinA)⁵

Expanding, we get:

cos5a + i sin5a = cos⁵a + 5 cos⁴a i sina + 10 cos³a i² sin²a + 10 cos²a i³ sin³a + 5 cosa i⁴ sin⁴a + i⁵ sin⁵a

Using i²=-1, i³= -i, i⁴=1, i⁵=i, we get

cos5a + i sin5a = cos⁵a + 5 cos⁴a i sina - 10 cos³a sin²a - 10 cos²a i sin³a + 5 cosa sin⁴a + i sin⁵a

Equating imaginary parts,

sin5a = 5 cos⁴a sina - 10 cos²a sin³a + sin⁵a