With θ=π/10, the numerator 1+sinθ+icosθ and denominator 1+sinθ−icosθ are conjugates; factoring each using 1+sinθ=(cos2θ+sin2θ)2 and cosθ=cos22θ−sin22θ turns the quotient into a single cis expression that de Moivre's theorem raises to the 10th power.
Step 1. Let θ=10π and factor the numerator using half-angle identities. Since 1=cos22θ+sin22θ and sinθ=2sin2θcos2θ,
1+sinθ=cos22θ+sin22θ+2sin2θcos2θ=(cos2θ+sin2θ)2.
Also, cosθ=cos22θ−sin22θ=(cos2θ−sin2θ)(cos2θ+sin2θ).
Step 2. Factor out (cos2θ+sin2θ) from the numerator.
1+sinθ+icosθ=(cos2θ+sin2θ)[(cos2θ+sin2θ)+i(cos2θ−sin2θ)].
Regroup the bracket by real/imaginary contributions of cos2θ and sin2θ:
(cos2θ+sin2θ)+i(cos2θ−sin2θ)=cos2θ(1+i)+sin2θ(1−i).
Since 1−i=−i(1+i), this is (1+i)[cos2θ−isin2θ]. So
1+sinθ+icosθ=(cos2θ+sin2θ)(1+i)(cos2θ−isin2θ).
Step 3. Do the same for the denominator (conjugate throughout).
1+sinθ−icosθ=(cos2θ+sin2θ)(1−i)(cos2θ+isin2θ).
Step 4. Divide and simplify. The common real factor (cos2θ+sin2θ) cancels:
1+sinθ−icosθ1+sinθ+icosθ=1−i1+i⋅cos2θ+isin2θcos2θ−isin2θ.
Now 1−i1+i=(1−i)(1+i)(1+i)2=22i=i=cis2π, and cos2θ+isin2θcos2θ−isin2θ=cis(−θ) (dividing cis(−θ/2) by cis(θ/2)). So the whole quotient is
cis2π⋅cis(−θ)=cis(2π−θ).
Step 5. Raise to the 10th power using de Moivre's theorem, with θ=10π.
[cis(2π−θ)]10=cis(10(2π−10π))=cis(5π−π)=cis(4π)=cos4π+isin4π=1.