Question

hey give me appropriate answer otherwise stay away with this question​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\left\{\dfrac{1+cos\left(\dfrac{\pi}{8}\right)+i\,sin\left(\dfrac{\pi}{8}\right)}{1+cos\left(\dfrac{\pi}{8}\right)-i\,sin\left(\dfrac{\pi}{8}\right)}\right\}^8}$

$\sf{=\left\{\dfrac{2\,cos^2\left(\dfrac{\pi}{16}\right)+2\,i\,sin\left(\dfrac{\pi}{16}\right)\,cos\left(\dfrac{\pi}{16}\right)}{2\,cos^2\left(\dfrac{\pi}{16}\right)-2\,i\,sin\left(\dfrac{\pi}{16}\right)\,cos\left(\dfrac{\pi}{16}\right)}\right\}^8}$

$\sf{=\left[\dfrac{2\,cos\left(\dfrac{\pi}{16}\right)\left\{cos\left(\dfrac{\pi}{16}\right)+i\,sin\left(\dfrac{\pi}{16}\right)\right\}}{2\,cos\left(\dfrac{\pi}{16}\right)\left\{cos\left(\dfrac{\pi}{16}\right)-i\,sin\left(\dfrac{\pi}{16}\right)\right\}}\right]^8}$

$\sf{=\left[\dfrac{cos\left(\dfrac{\pi}{16}\right)+i\,sin\left(\dfrac{\pi}{16}\right)}{cos\left(\dfrac{\pi}{16}\right)-i\,sin\left(\dfrac{\pi}{16}\right)}\right]^8}$

$\sf{=\left[\dfrac{cos\left(\dfrac{\pi}{16}\right)+i\,sin\left(\dfrac{\pi}{16}\right)}{cos\left(-\dfrac{\pi}{16}\right)+i\,sin\left(-\dfrac{\pi}{16}\right)}\right]^8}$

$\sf{=\left[\dfrac{e^{\frac{i\,\pi}{16}}}{e^{\frac{-i\,\pi}{16}}}\right]^8}$

$\sf{=\left[e^{\frac{i\,\pi}{16}+\frac{i\,\pi}{16}}\right]^8}$

$\sf{=\left[e^{\frac{i\,2\pi}{16}}\right]^8}$

$\sf{=\left[e^{\frac{i\,\pi}{8}}\right]^8}$

$\sf{=e^{i\,\pi}}$

$\sf{=-1}$

Answer 2

Given Question :-

$\sf \: {\bigg[\dfrac{1 + cos\bigg(\dfrac{\pi}{8} \bigg) + isin\bigg(\dfrac{\pi}{8} \bigg) \: }{1 + cos\bigg(\dfrac{\pi}{8} \bigg) - i \: sin\bigg(\dfrac{\pi}{8} \bigg)} \bigg]}^{8} = - - - -$

$\green{\large\underline{\sf{Solution-}}}$

Given complex number is

$\rm :\longmapsto\: \sf \: {\bigg[\dfrac{1 + cos\bigg(\dfrac{\pi}{8} \bigg) + isin\bigg(\dfrac{\pi}{8} \bigg) \: }{1 + cos\bigg(\dfrac{\pi}{8} \bigg) - i \: sin\bigg(\dfrac{\pi}{8} \bigg)} \bigg]}^{8}$

We know,

$\\ \boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }} \\ \\ \boxed{ \tt{ \: sin2x = 2sinx \: cosx \: }}$

So, using this, we get

$\rm \: = \:{ \bigg\{\dfrac{2 {cos}^{2} \bigg(\dfrac{\pi}{16} \bigg)+ i \: 2sin\bigg(\dfrac{\pi}{16} \bigg)cos\bigg(\dfrac{\pi}{16} \bigg)}{2 {cos}^{2}\bigg(\dfrac{\pi}{16} \bigg) - i \: 2sin\bigg(\dfrac{\pi}{16} \bigg)cos\bigg(\dfrac{\pi}{16} \bigg)} \bigg\}}^{8}$

$\rm \: = \:{ \bigg\{\dfrac{2 {cos}^{} \bigg(\dfrac{\pi}{16} \bigg)\bigg[cos\bigg(\dfrac{\pi}{16} \bigg)+ i \: sin\bigg(\dfrac{\pi}{16} \bigg)\bigg]}{2 {cos}^{}\bigg(\dfrac{\pi}{16} \bigg)\bigg[cos\bigg(\dfrac{\pi}{16} \bigg) - i \: sin\bigg(\dfrac{\pi}{16} \bigg)\bigg]} \bigg\}}^{8}$

$\rm \: = \:{ \bigg\{\dfrac{\bigg[cos\bigg(\dfrac{\pi}{16} \bigg)+ i \: sin\bigg(\dfrac{\pi}{16} \bigg)\bigg]}{\bigg[cos\bigg(\dfrac{\pi}{16} \bigg) - i \: sin\bigg(\dfrac{\pi}{16} \bigg)\bigg]} \bigg\}}^{8}$

We know,

By Eulers Formula

$\boxed{ \tt{ \: cosx + i \: sinx = {e}^{ix} \: }} \\ \\ \boxed{ \tt{ \: cosx - i \: sinx = {e}^{ - \: ix} \: }}$

So, using these Identities, we get

$\rm \: = \: {\bigg[\dfrac{ {e}^{i\dfrac{\pi}{16} } }{{e}^{ - \: i\dfrac{\pi}{16} }} \bigg]}^{8}$

$\rm \: = \: {\bigg[{e}^{i\dfrac{\pi}{16} + i\dfrac{\pi}{16} }\bigg]}^{8}$

$\rm \: = \: {\bigg[{e}^{i\dfrac{\pi}{8}}\bigg]}^{8}$

$\rm \: = \: {e}^{i \: \pi}$

$\rm \: = \: cos\pi \: + \: i \:sin \pi$

$\rm \: = \: - 1 + 0i$

$\rm \: = \: - 1$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \sf \: {\bigg[\dfrac{1 + cos\bigg(\dfrac{\pi}{8} \bigg) + isin\bigg(\dfrac{\pi}{8} \bigg) \: }{1 + cos\bigg(\dfrac{\pi}{8} \bigg) - i \: sin\bigg(\dfrac{\pi}{8} \bigg)} \bigg]}^{8} = - 1 \: }}}$

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More to know :-

$\boxed{ \tt{ \: {(cosx + i \: sinx)}^{n} = cos \: nx + i \: sin \: nx \: }}$

$\boxed{ \tt{ \: {(cosx + i \: sinx)}^{ - n} = cos \: nx - i \: sin \: nx \: }}$

$\boxed{ \tt{ \: \frac{1}{cosx - i \: sinx} = cosx + i \: sinx \: }}$