Prove ${(\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x})}^{n} = cos 2 n x + i sin 2 n x$ $((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx$ ?

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Prove ${(\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x})}^{n} = cos 2 n x + i sin 2 n x$ $((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx$ ?

Prove ((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx ?

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2 Answers

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Answer 1 · 2017-12-17T17:48:39

Explanation is below

Explanation:

$\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x}$ $(1+cos2x+isin2x)/(1+cos2x-isin2x)$

= $\frac{2 {(cos x)}^{2} + 2 i \cdot sin x \cdot cos x}{2 {(cos x)}^{2} - 2 i \cdot sin x \cdot cos x}$ $[2(cosx)^2+2i*sinx*cosx]/[2(cosx)^2-2i*sinx*cosx]$

= $\frac{2 cos x \cdot (cos x + i sin x)}{2 cos x \cdot (cos x - i sin x)}$ $[2cosx*(cosx+isinx)]/[2cosx*(cosx-isinx)]$

= $\frac{cos x + i sin x}{cos x - i sin x}$ $(cosx+isinx)/(cosx-isinx)$

= $\frac{{(cos x + i sin x)}^{2}}{(cos x - i sin x) \cdot (cos x + i \cdot sin x)}$ $(cosx+isinx)^2/[(cosx-isinx)*(cosx+i*sinx)]$

= $\frac{{(cos x)}^{2} - {(sin x)}^{2} + 2 i \cdot sin x \cdot cos x}{{(cos x)}^{2} + {(sin x)}^{2}}$ $[(cosx)^2-(sinx)^2+2i*sinx*cosx]/[(cosx)^2+(sinx)^2]$

= $\frac{cos 2 x + i sin 2 x}{1}$ $(cos2x+isin2x)/1$

= $cos 2 x + i sin 2 x$ $cos2x+isin2x$

Thus,

${[\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x}]}^{n}$ $[(1+cos2x+isin2x)/(1+cos2x-isin2x)]^n$

= ${(cos 2 x + i sin 2 x)}^{n}$ $(cos2x+isin2x)^n$

= $cos (2 n x) + i sin (2 n x)$ $cos(2nx)+isin(2nx)$

Answer 2 · 2017-12-17T18:20:28

See below.

Explanation:

$1 + e^{i 2 x} = e^{i x} (e^{i x} + e^{- i x})$ $1+e^(i2x) = e^(ix)(e^(ix)+e^(-ix))$
$1 + e^{- i 2 x} = e^{- i x} (e^{i x} + e^{- i x})$ $1+e^(-i2x) = e^(-ix)(e^(ix)+e^(-ix))$ so

${(\frac{1 + e^{i 2 x}}{1 + e^{- i 2 x}})}^{n} = {(e^{i 2 x})}^{n} = e^{i 2 n x} = cos (2 n x) + i sin (2 n x)$ $((1+e^(i2x))/(1+e^(-i2x)))^n = (e^(i2x))^n = e^(i2nx) = cos(2nx)+isin(2nx)$

NOTE

We were using the de Moivre's identity

$e^{i ϕ} = cos ϕ + i sin ϕ$ $e^(i phi) = cos phi + i sin phi$

Science

Math

Humanities

... and beyond

Prove ${(\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x})}^{n} = cos 2 n x + i sin 2 n x$ $((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx$ ?

Prove ((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx ?

Thanks!

2 Answers

Explanation:

Explanation:

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Prove ((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx ? Thanks!

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Explanation:

Explanation:

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Prove ${(\frac{1 + cos 2 x + i sin 2 x}{1 + cos 2 x - i sin 2 x})}^{n} = cos 2 n x + i sin 2 n x$ $((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx$ ?

Prove ((1 + cos2 x + i sin2 x)/(1 + cos2 x - i sin2 x))^n=cos2nx+isin2nx ?

Thanks!