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De Moivre's formula using prove that,

If a=cosθ+isinθ,b=cosϕ+isinϕ prove that

(i) cos(θ+ϕ)=12[ab+1ab]

(ii) sin(θϕ)=12i[abba]

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Solution:

Given, a=cosθ+isinθ

&  b=cosϕ+isinϕ

(i) cos(θ+ϕ)=12[ab+1ab] prove that,

Now, ab=(cosθ+isinθ)(cosϕ+isinϕ)

ab=cos(θ+ϕ)+isin(θ+ϕ)

also,1ab=(ab)1=[cos(θ+ϕ)+isin(θ+ϕ)] 1ab=cos(θ+ϕ)isin(θ+ϕ)..(2) (1)+(2)

ab+1ab=cos(θ+ϕ)+isin(θ+ϕ)+cos(θ+ϕ)isin(θ+ϕ)ab+1ab=2cos(θ+ϕ)cos(θ+ϕ)=12[ab+1ab]

(ii) sin(θϕ)=12i[abba] prove that,

=(cosθ+isinθ)(cosϕ+isinϕ)1=(cosθ+isinθ)(cosϕisinϕ)

ab=cos(θϕ)+isin(θϕ)also,ba=(ab)1=[cos(θϕ)+isin(θϕ)]1ba=cos(θϕ)isin(θϕ)(3)(4)abba=cos(θϕ)+isin(θϕ)cos(θϕ)+isin(θϕ)abba=2isin(θϕ)

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