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Q.If nn is an integer then show that (1+cos⁡θ+isin⁡θ)n+(1+cos⁡θ−isin⁡θ)n=2n+1cos⁡n(θ2)cos⁡(nθ2)(1 + \cos\theta + i\sin\theta)^n + (1 + \cos\theta - i\sin\theta)^n = 2^{n+1}\cos^n\left(\dfrac{\theta}{2}\right)\cos\left(\dfrac{n\theta}{2}\right).

Telangana TsbieTelangana Board of Intermediate Education 2026Subjective· 7mImportance★★★★★
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Each bracket factors as 2cos⁡(θ/2) cis⁡(±θ/2)2\cos(\theta/2)\,\operatorname{cis}(\pm\theta/2); De Moivre's theorem plus cis⁡ϕ+cis⁡(−ϕ)=2cos⁡ϕ\operatorname{cis}\phi+\operatorname{cis}(-\phi)=2\cos\phi gives the result.

Use the half-angle identities: 1+cos⁡θ=2cos⁡2(θ/2)1 + \cos\theta = 2\cos^2(\theta/2) and sin⁡θ=2sin⁡(θ/2)cos⁡(θ/2)\sin\theta = 2\sin(\theta/2)\cos(\theta/2).

Then

1+cos⁡θ+isin⁡θ=2cos⁡2(θ/2)+2isin⁡(θ/2)cos⁡(θ/2)=2cos⁡(θ/2)[cos⁡(θ/2)+isin⁡(θ/2)]1 + \cos\theta + i\sin\theta = 2\cos^2(\theta/2) + 2i\sin(\theta/2)\cos(\theta/2) = 2\cos(\theta/2)\big[\cos(\theta/2) + i\sin(\theta/2)\big].

So 1+cos⁡θ+isin⁡θ=2cos⁡(θ/2) cis⁡(θ/2)1 + \cos\theta + i\sin\theta = 2\cos(\theta/2)\,\operatorname{cis}(\theta/2).

Similarly 1+cos⁡θ−isin⁡θ=2cos⁡(θ/2) cis⁡(−θ/2)1 + \cos\theta - i\sin\theta = 2\cos(\theta/2)\,\operatorname{cis}(-\theta/2).

Raising to the power nn (De Moivre's theorem):

(1+cos⁡θ+isin⁡θ)n=2ncos⁡n(θ/2) cis⁡(nθ/2)(1 + \cos\theta + i\sin\theta)^n = 2^n\cos^n(\theta/2)\,\operatorname{cis}(n\theta/2),

(1+cos⁡θ−isin⁡θ)n=2ncos⁡n(θ/2) cis⁡(−nθ/2)(1 + \cos\theta - i\sin\theta)^n = 2^n\cos^n(\theta/2)\,\operatorname{cis}(-n\theta/2).

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