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Molodets [167]
3 years ago
8

Using Euler's relation, derive the following relationships:a. Cosθ=½(e^jθ+e^−jθ)b. Sinθ=½(e^jθ−e(^−jθ)

Mathematics
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

a. cosθ = ¹/₂[e^jθ + e^(-jθ)] b. sinθ = ¹/₂[e^jθ - e^(-jθ)]

Step-by-step explanation:

a.We know that

e^jθ = cosθ + jsinθ and

e^(-jθ) = cosθ - jsinθ

Adding both equations, we have

e^jθ = cosθ + jsinθ

+

e^(-jθ) = cosθ - jsinθ

e^jθ + e^(-jθ) = cosθ + cosθ + jsinθ - jsinθ

Simplifying, we have

e^jθ + e^(-jθ) = 2cosθ

dividing through by 2 we have

cosθ = ¹/₂[e^jθ + e^(-jθ)]

b. We know that

e^jθ = cosθ + jsinθ and

e^(-jθ) = cosθ - jsinθ

Subtracting both equations, we have

e^jθ = cosθ + jsinθ

-

e^(-jθ) = cosθ - jsinθ

e^jθ + e^(-jθ) = cosθ - cosθ + jsinθ - (-jsinθ)

Simplifying, we have

e^jθ - e^(-jθ) = 2jsinθ

dividing through by 2 we have

sinθ = ¹/₂[e^jθ - e^(-jθ)]

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        \large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}

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<h3>Way 1:</h3>

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We first expand the binomial squared, so we get

                         \large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}

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<h3>Way 2</h3>

Dividing numerator and denominator by the term of highest degree:

                            \large\displaystyle\text{$\begin{gathered}\sf L  = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4  }{x^{3}-5  }  \end{gathered}$}\\

                                \ \  = \lim_{x \to \infty\frac{\frac{x^{4}  }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} }    }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}}   }  }

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Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

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