Let `z= cos theta +i sin theta`  .Then the value of `sum_(m=1)^15` `lm(Z^(2m-1))` at `theta=2` deg is

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You need to evaluate the sum of imaginaries of the given powers of complex number z, such that:

`Sigma_(m=1)^15 Im(z^(2m-1)) = Im(z^(2*1-1)) + Im(z^(2*2-1)) + Im(z^(2*3-1)) + ... + Im(z^(2*15-1))`

`Sigma_(m=1)^15 Im(z^(2m-1)) = Im(z^1) + Im(z^3) + Im(z^5) + ... + Im(z^29)`

All the powers of the complex number z, may be evaluated with De Moivre formula, such that:

`z^n = (cos theta + i*sin theta)^n => z^n = (cos (n*theta) + i*sin (n*theta))`

`z^3 = cos (3theta) + i*sin(3 theta)`

........

`z^29 = cos (29theta) + i*sin(29 theta)`

The imaginary part of the sum is `Im(z^1+z^3+...z^29) = cos theta + cos (3theta) + ... + cos(29 theta)`

You may group the terms such that:

`cos theta + cos (29 theta) = 2cos ((theta + 29theta)/2)*cos((theta - 29theta)/2)`

`cos theta + cos (29 theta) = 2cos (15theta)cos(-14theta)`

Since `cos(-theta)=cos theta`

`cos theta + cos (29 theta) = 2cos (15theta)cos(14theta)`

You may calculate the next sum `cos (3 theta) + cos (27theta),` such that:

`cos (3 theta) + cos (27theta) = 2cos (15theta)cos(12theta)`

`cos (5 theta) + cos (25 theta) = 2cos (15theta)cos(10theta)`

`cos (7 theta) + cos (23 theta) = 2cos (15theta)cos(8theta)`

`cos (9 theta) + cos (21 theta) = 2cos (15theta)cos(6theta)`

`cos (11 theta) + cos (19 theta) = 2cos (15theta)cos(4theta)`

`cos (13 theta) + cos (17 theta) = 2cos (15theta)cos(2theta)`

Notice that the term `cos (15 theta)` remains. You may take out the common factor `cos (15theta)` such that:

`Im(z^1+z^3+...z^29) = cos (15 theta)(2cos(12theta) + 2cos(10theta) + 2cos(8theta) + 2cos(6theta) + 2cos(4theta) + 2cos(2theta) + 1)`

You may put now` theta = 2^o` , such that:

`cos (15*2^o) = cos 30^o = sqrt3/2`

You may group again the terms, such that:

`2cos(12theta) + 2cos(2theta) = 4 cos (7 theta)*cos (5 theta)`

`2cos(10theta) + 2cos(4theta) = 4 cos (7 theta)*cos (3 theta)`

`2cos(8theta) + 2cos(6theta) = 4 cos (7 theta)*cos ( theta)`

Factoring out 4` cos (7 theta)` yields:

`2cos(12theta) + 2cos(10theta) + 2cos(8theta) + 2cos(6theta) + 2cos(4theta) + 2cos(2theta) = 4 cos (7 theta)*(cos (5 theta) + cos (3 theta) + cos ( theta))`

You may group again the terms, such that:

`cos (5 theta) + cos ( theta) = 2cos (3theta)*cos (2theta)`

`2cos (3theta)*cos (2theta) + cos (3theta) = cos(3theta)(2cos(2theta) + 1))`

Using the formula of double angle yields:

`cos (2theta) = 2cos^2 theta - 1`

`cos(3theta)(2cos(2theta) + 1)) = cos(3theta)(4cos^2 theta -2 + 1))`

`cos(3theta)(2cos(2theta) + 1)) = cos(3theta)(4cos^2 theta -1))`

`2cos(12theta) + 2cos(10theta) + 2cos(8theta) + 2cos(6theta) + 2cos(4theta) + 2cos(2theta) = 4 cos (7 theta)*cos(3 theta)*(4cos^2 theta -1))`

`Im(z^1+z^3+...z^29) = cos (15 theta)(4 cos (7 theta)*cos(3 theta)*(4cos^2 theta -1)) + 1)`

Hence, evaluating `Sigma_(m=1)^15 Im(z^(2m-1)) = (sqrt3/2)*)(4 cos (14^o)*cos(6^o)*(4cos^2(2^o) -1)) + 1).`