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

Asked by aakashpachisia1488 | 7th Aug, 2011, 06:15: PM

Expert Answer:

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

Answered by  | 9th Aug, 2011, 03:33: AM

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