Question

Evaluate ${}^{\mathbf{cosh}\left(\gamma \right)}$and ${}^{\mathbf{tanh}\left(\gamma 1/2\right)}$for ${}^{\mathbf{\gamma }\mathbf{l}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{\angle }{\mathbf{85}}^{\mathbf{°}}}$per unit.

Short answer

The value of ${}^{\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma }\mathbf{\right)}}$and role="math" localid="1655358531240" ${}^{\mathbf{tanh}\mathbf{(}\mathbf{\gamma l}\mathbf{/}\mathbf{2}\mathbf{)}}$are ${}^{\mathbf{0}\mathbf{.}\mathbf{9223}\mathbf{\angle }\mathbf{0}\mathbf{.}{\mathbf{841}}^{\mathbf{°}}\mathbf{}\mathbf{pu}}$and${}^{\mathbf{0}\mathbf{.}\mathbf{2027}\mathbf{\angle }\mathbf{84}\mathbf{.}{\mathbf{86}}^{\mathbf{°}}\mathbf{}\mathbf{pu}}$

Step-by-step solution

Given data.

Assume the following data:

$\mathit{\gamma }\mathit{\iota }\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{\angle }{\mathbf{85}}^{\mathbf{°}}$

Here, the quantity role="math" localid="1655358666011" ${}^{\mathbf{\gamma }\mathbf{l}}$is dimensionless.

Determine the formula of, and.

Write the formula of ${}^{\mathbf{cosh}\left(\mathrm{\gamma l}\right)}$, ${}^{\mathbf{sinh}\left(\mathrm{\gamma l}\right)}$and ${}^{\mathbf{tanh}\left(\mathrm{\gamma l}/2\right)}$.

role="math" localid="1655359405332" $\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{\gamma l}}\mathbf{+}{\mathbf{e}}^{\mathbf{\gamma l}}}{\mathbf{2}}$ …… (1)

Here, ${}^{\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}}$is the hyperbolic function of dimensionless quantity.

role="math" localid="1655359392700" $\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{\gamma l}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{\gamma l}}}{\mathbf{2}}$ …… (2)

Here, ${}^{\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}}$is the hyperbolic function of dimensionless quantity.

$\mathbf{tanh}\left(\mathrm{\gamma l}/2\right)\mathbf{=}\frac{\mathbf{cos}\left(\mathrm{\gamma l}\right)\mathbf{-}\mathbf{1}}{\mathbf{sinh}\left(\mathrm{\gamma l}\right)}$ …… (3)

Here, ${}^{\mathbf{tanh}\mathbf{(}\mathbf{\gamma l}\mathbf{/}\mathbf{2}\mathbf{)}}$is the hyperbolic function of dimensionless quantity.

Determine the value of.

First, determine the value of ${}^{{\mathit{e}}^{\mathit{\gamma }\mathit{l}}}$.

$$\begin{gathered} {\mathbf{e}}^{\mathbf{\gamma l}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{034862}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{39848}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{034862}}\mathbf{\angle }\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{39848}}^{\mathbf{°}\mathbf{}}\mathbf{radians} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0965738}\mathbf{\angle }\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{39848}}^{\mathbf{°}\mathbf{}}\mathbf{radians} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89007}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{40178} \end{gathered}$$

Next, determine the value of localid="1655359729441" ${}^{{\mathit{e}}^{\mathbf{-}\mathit{\gamma }\mathit{l}}}$.

$$\begin{gathered} {\mathit{e}}^{\mathbf{-}\mathbf{\gamma }\mathbf{l}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{034862}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{39848}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{034862}}\mathbf{\angle }\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{39848}}^{\mathbf{°}\mathbf{}}\mathit{r}\mathit{a}\mathit{d}\mathit{i}\mathit{a}\mathit{n}\mathit{s} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0965738}\mathbf{\angle }\mathbf{-}\mathbf{0}\mathbf{.}{\mathbf{39848}}^{\mathbf{°}\mathbf{}}\mathit{r}\mathit{a}\mathit{d}\mathit{i}\mathit{a}\mathit{n}\mathit{s} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89007}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{40178} \end{gathered}$$

Write the formula of ${}^{\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}}$.

localid="1655361834790" $\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{\gamma l}}\mathbf{+}{\mathbf{e}}^{\mathbf{\gamma l}}}{\mathbf{2}}$ …… (4)

Here, ${}^{\mathbf{cosh}\mathit{\left(}\mathit{\gamma }\mathit{l}\mathit{\right)}}$is the hyperbolic function of dimensionless quantity.

Substitute ${}^{\mathbf{0}\mathbf{.}\mathbf{9543}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{40178}}$for ${}^{{\mathit{e}}^{\mathit{\gamma }\mathit{l}}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{890007}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{37472}}$for ${}^{{\mathit{e}}^{\mathbf{-}\mathit{\gamma }\mathit{l}}}$into equation (4).

localid="1655361845164" $$\begin{gathered} \mathbf{cos}\mathrm{h}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{9543}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{40178}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{890007}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{37472}\mathbf{)}}{\mathbf{2}}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{92219}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{01353} \end{gathered}$$

Determine the value of.

Write the formula of ${}^{\mathbf{sinh}\mathbf{\left(}\mathit{\gamma }\mathbf{l}\mathbf{\right)}}$.

$\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{\gamma l}}\mathbf{+}{\mathbf{e}}^{\mathbf{\gamma l}}}{\mathbf{2}}$

…… (5)

Here, ${}^{\mathbf{sinh}\mathbf{\left(}\mathit{\gamma }\mathbf{l}\mathbf{\right)}}$is the hyperbolic function of dimensionless quantity.

Substitute ${}^{\mathbf{0}\mathbf{.}\mathbf{9543}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{40178}}$for ${}^{{\mathit{e}}^{\mathit{\gamma }\mathit{l}}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{89007}\mathbf{-}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{37472}}$for ${}^{{\mathit{e}}^{\mathbf{-}\mathit{\gamma }\mathit{l}}}$into equation (5).

$$\begin{gathered} \mathbf{sinh}\left(\mathit{\gamma }\mathit{l}\right)\mathbf{=}\frac{\left(0.9543+j0.40178\right)\mathbf{-}\left(0.89007-j0.37472\right)}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03212}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{38825} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{38958}\mathbf{\angle }\mathbf{85}\mathbf{.}{\mathbf{271}}^{\mathbf{°}}\mathbf{}\mathbf{pu} \end{gathered}$$

Determine the value of.

Write the formula of ${}^{\mathbf{tanh}\left(\gamma l/2\right)}$ .

$\mathbf{tan}\left(\mathit{\gamma }\mathit{l}/2\right)\mathbf{=}\frac{\mathbf{cos}\left(\gamma l\right)\mathbf{-}\mathbf{1}}{\mathbf{sinh}\left(\gamma l\right)}$ …… (6)

Here, ${}^{\mathbf{tanh}\mathbf{(}\mathbf{\gamma l}\mathbf{/}\mathbf{2}\mathbf{)}}$is the hyperbolic function of dimensionless quantity.

Now substitute ${}^{\mathbf{0}\mathbf{.}\mathbf{9223}\mathbf{\angle }\mathbf{0}\mathbf{.}{\mathbf{841}}^{\mathbf{°}}\mathbf{}\mathbf{pu}}$for ${}^{\mathbf{cosh}\mathbf{\left(}\mathbf{\gamma l}\mathbf{\right)}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{38958}\mathbf{\angle }\mathbf{85}\mathbf{.}{\mathbf{271}}^{\mathbf{°}}\mathbf{}\mathbf{pu}}$for ${}^{\mathbf{sinh}\left(\mathrm{\gamma l}\right)}$ in equation (6).

$$\begin{gathered} \mathbf{tanh}\frac{\mathbf{\gamma l}}{\mathbf{2}}\mathbf{=}\frac{\left(0.9222+j0.01353\right)}{\mathbf{0}\mathbf{.}\mathbf{38958}\mathbf{\angle }\mathbf{85}\mathbf{.}{\mathbf{271}}^{\mathbf{°}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{07897}\mathbf{\angle }\mathbf{170}\mathbf{.}{\mathbf{13}}^{\mathbf{°}}}{\mathbf{0}\mathbf{.}\mathbf{38958}\mathbf{\angle }\mathbf{85}\mathbf{.}{\mathbf{271}}^{\mathbf{°}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2027}\mathbf{\angle }\mathbf{84}\mathbf{.}{\mathbf{86}}^{\mathbf{°}}\mathbf{}\mathbf{pu} \end{gathered}$$