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(Cos(x) + iSin(x))^9

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(Cos(x) + iSin(x))^9

Guest Feb 20, 2015

#2
+91436
+5

\$\$(Cos(x) + iSin(x))^9 = e^{ix}^9=e^{9xi}\$\$

Using Euler's formula

I think Geno's answer is correct too.  :)

Melody  Feb 21, 2015
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#1
+17711
+5

Using the expansion of  (x + y)9

=  x9 + 9x8y + 36x7y2 + 84x6y3 + 126x5y4 + 126x4y5 + 84x3y6 + 36x2y7 + 9xy8 + y9

[Cos(x) + iSin(x)]9  =

=  Cos(x)9 + 9 Cos(x)i Sin(x) + 36 Cos(x)7 i2 Sin(x)2 + 84 Cos(x)i3 Sin(x)3

+ 126 Cos(x)i4 Sin(x)4 + 126 Cos(x)i5 Sin(x)5 + 84 Cos(x)i6 Sin(x)6

+ 36 Cos(x)i7 Sin(x)7 + 9 Cos(x) i8 Sin(x)8 + i9 Sin(x)9

Since:     i2 = -1    i3 = -i     i4 = 1    i5 = i     i6 = -1     i7 = -i    i8 = 1     i9 = i

=  Cos(x)9 + 9 i Cos(x)Sin(x) - 36 Cos(x)7 Sin(x)2 - 84 i Cos(x)Sin(x)3

+ 126 Cos(x)Sin(x)4 + 126 i Cos(x)Sin(x)5 - 84 Cos(x)Sin(x)6

- 36 i Cos(x)Sin(x)7 + 9 Cos(x) Sin(x)8 + i Sin(x)9

geno3141  Feb 21, 2015
#2
+91436
+5

\$\$(Cos(x) + iSin(x))^9 = e^{ix}^9=e^{9xi}\$\$

Using Euler's formula

I think Geno's answer is correct too.  :)

Melody  Feb 21, 2015

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