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
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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