`If (costheta +isintheta)(cos2theta +isin2theta).....(cosntheta + isinntheta) = 1 ` then the value of `theta` is:

Options

A. `theta = (4mpi)/(n(n+1))`
B. `theta = (2mpi)/(n(n+1))`
C. `theta = (4m)/(n(n+1))`
D. `theta = (2m)/(n(n+1))`

Answer

A

Explanation

`(costheta+isintheta)(cos2theta+isin2theta)....(cosntheta+isinntheta) = 1`

we know, `costheta+isintheta = e^itheta`

So, our equation becomes,

`=>(e^(itheta))(e^(i2theta)).....(e^(i ntheta)) = 1`

`=>e^(itheta(1+2+3...n))= 1`

`=>e^(itheta((n(n+1))/2)) = 1`

`=>cos((n(n+1))/2)theta+isin((n(n+1))/2)theta = 1`

Comparing real and imaginary parts,

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