Question

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Answer 1

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{\displaystyle[\dfrac{1+cos\frac{\pi}{8}+i\,sin\frac{\pi}{8}}{1+cos\frac{\pi}{8}-i\,sin\frac{\pi}{8}}]^8}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\displaystyle[\dfrac{1+cos\frac{\pi}{8}+i\,sin\frac{\pi}{8}}{1+cos\frac{\pi}{8}-i\,sin\frac{\pi}{8}}]^8}$

$\boxed{\begin{minipage}{3cm} \textsf{Take,}\\\\\mathsf{z=cos\frac{\pi}{8}+i\,sin\frac{\pi}{8}}\\\\\textsf{Then,}\\\\\mathsf{\dfrac{1}{z}=cos\frac{\pi}{8}-i\,sin\frac{\pi}{8}} \end{minipage}}$

$\mathsf{=\displaystyle[\dfrac{1+z}{1+\dfrac{1}{z}}]^8}$

$\mathsf{=\displaystyle[\dfrac{1+z}{\dfrac{1+z}{z}}]^8}$

$\mathsf{=\displaystyle[1+z{\times}\dfrac{z}{1+z}]^8}$

$\mathsf{=z^8}$

$\mathsf{=(cos\frac{\pi}{8}+i\,sin\frac{\pi}{8})^8}$

$\textsf{Applying Demovire's theorem}$

$\mathsf{=cos\frac{8\pi}{8}+i\,sin\frac{8\pi}{8}}$

$\mathsf{=cos\pi+i\,sin\pi}$

$\mathsf{=-1}$

$\implies\boxed{\mathsf{\displaystyle[\dfrac{1+cos\frac{\pi}{8}+i\,sin\frac{\pi}{8}}{1+cos\frac{\pi}{8}-i\,sin\frac{\pi}{8}}]^8=-1}}$

$\underline{\textsf{Answer:}}$

$\textsf{Option (2) is correct}$

Answer 2

Answer:

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Step-by-step explanation:

Answer. Answer: When computer programs store data in variables, each variable must be assigned a specific data type. Some common data types include integers, floating point numbers, characters, strings, and arrays. ...