Question

Solve for θ : (cos θ + i sin θ) (cos 2θ + i sin 2θ) …. (cos nθ + i sin nθ) = i .

Answer 1

Given :   (cos θ + i sin θ) (cos 2θ + i sin 2θ) …. (cos nθ + i sin nθ) = i

To Find :   θ

Solution:

(cos θ + i sin θ)  =  $e^{\iota \theta}$

(cos 2θ + i sin 2θ)  =  $e^{\iota 2\theta}$

(cos nθ + i sin nθ) = $e^{\iota n\theta}$

(cos θ + i sin θ) (cos 2θ + i sin 2θ) …. (cos nθ + i sin nθ)

=  $e^{\iota \theta}$ . $e^{\iota 2\theta}$ .. .. .. $e^{\iota n\theta}$

=  $e^{\iota (1+2+3 + .. + n )\theta}$

1 + 2 + 3 +  ... + n  = n(n + 1)/2

= $e^{\iota (\frac{n(n+1)}{2} )\theta}$

= cos {n(n+1)/2}θ  + i sin {n(n+1)/2}θ

cos {n(n+1)/2}θ  + i sin {n(n+1)/2}θ = i

=> cos {n(n+1)/2}θ  = 0

    sin {n(n+1)/2}θ   = 1

{n(n+1)/2}θ   =  π/2  

=> θ   =  π/n(n+1)

or {n(n+1)/2}θ = 2k π +  π/2    generalized  k  being integer

=> θ  =  (4k π +  π)/n(n+1)

=> θ  = (4k + 1)π/n(n+1)

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Answer 2

Answer:

LHS=2(sin 6 θ+cos 6 θ)−3(sin 4 θ+cos 4θ)+1

=2[(sin 2 θ) 3 +(cos 2 θ) 3 ]−3[(sin 2 θ) 2+(cos 2 θ) 2 ]+1

=2[(sin 2θ+cos 2 θ) 3 −3sin 2θcos 2θ(sin 2 θ+cos 2 θ)]−3[(sin 2 θ+cos 2 θ) 2 −2sin 2 θcos2θ]+1

=2(1−3sin 2 θcos 2 θ)−3(1−2sin θ cos 2 θ)+1

=2−6sin 2 θcos 2 θ−3+6sin 2θcos 2 θ+1=0=RHS