MHB Therefore, $\sinh{(4-j3)} = \cos{(3)}\sinh{(4)} - \mathrm{i}\sin{(3)}\cosh{(4)}$

[paulmdrdo1]

Evaluate the following expressions, expressing answers in rectangular form.

1. $\cos(1+j)$
2.$\sinh(4-j3)$

can you help me on how to solve these problems.

thanks in advance!

 

[Prove It]

Evaluate the following expressions, expressing answers in rectangular form.

1. $\cos(1+j)$
2.$\sinh(4-j3)$

can you help me on how to solve these problems.

thanks in advance!

First of all, I refuse to use $\displaystyle \begin{align*} \mathrm{j} \end{align*}$ to represent the imaginary unit, I use $\displaystyle \begin{align*} \mathrm{i} \end{align*}$. Now...

$\displaystyle \begin{align*} \cos{ \left( \alpha + \beta \right) } &\equiv \cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } - \sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}$

so

$\displaystyle \begin{align*} \cos{ \left( 1 + \mathrm{i} \right) } &= \cos{ \left( 1 \right) } \cos{ \left( \mathrm{i} \right) } - \sin{ \left( 1 \right) } \sin{ \left( \mathrm{i} \right) } \end{align*}$

Playing around a bit with Euler's formula we have

$\displaystyle \begin{align*} \mathrm{e}^{\mathrm{i} \, \theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \\ \\ \mathrm{e}^{-\mathrm{i}\,\theta} &= \cos{ \left( -\theta \right) } + \mathrm{i}\sin{ \left( -\theta \right) } \\ &= \cos{ \left( \theta \right) } - \mathrm{i}\sin{ \left( \theta \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{-\mathrm{i}\,\theta} &= 2\cos{ \left( \theta \right) } \\ \\ \cos{ \left( \theta \right) } &= \frac{1}{2} \left( \mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{-\mathrm{i}\,\theta} \right) \\ \\ \cos{ \left( \mathrm{i} \right) } &= \frac{1}{2} \left( \mathrm{e}^{\mathrm{i}^2} + \mathrm{e}^{-\mathrm{i}^2} \right) \\ &= \frac{1}{2} \left( \mathrm{e}^{-1} + \mathrm{e}^{1} \right) \\ &= \cosh{ \left( 1 \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} &= 2\mathrm{i}\sin{ \left( \theta \right) } \\ \mathrm{i} \left( \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} \right) &= -2\sin{ \left( \theta \right) } \\ \sin{ \left( \theta \right) } &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} \right) \\ \\ \sin{ \left( \mathrm{i} \right) } &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{\mathrm{i}^2} - \mathrm{e}^{-\mathrm{i}^2} \right) \\ &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{-1} - \mathrm{e}^{1} \right) \\ &= \frac{\mathrm{i}}{2} \left( \mathrm{e}^{1} - \mathrm{e}^{-1} \right) \\ &= \mathrm{i}\sinh{ \left( 1 \right) } \end{align*}$

Thus:

$\displaystyle \begin{align*} \cos{ \left( 1 + \mathrm{i} \right) } &= \cos{ \left( 1 \right) } \cos{ \left( \mathrm{i} \right) } - \sin{ \left( 1 \right) } \sin{ \left( \mathrm{i} \right) } \\ &= \cos{ \left( 1 \right) } \cosh{ \left( 1 \right) } - \sin{ \left( 1 \right) } \left[ \mathrm{i}\sinh{ \left( 1 \right) } \right] \\ &= \cos{ \left( 1 \right) } \cosh{ \left( 1 \right) } - \mathrm{i} \sin{ \left( 1 \right) } \sinh{ \left( 1 \right) } \end{align*}$

 

[paulmdrdo1]

how would you express that in rectangular form?

 

[paulmdrdo1]

First of all, I refuse to use $\displaystyle \begin{align*} \mathrm{j} \end{align*}$ to represent the imaginary unit, I use $\displaystyle \begin{align*} \mathrm{i} \end{align*}$. Now...

$\displaystyle \begin{align*} \cos{ \left( \alpha + \beta \right) } &\equiv \cos{ \left( \alpha \right) } \cos{ \left( \beta \right) } - \sin{ \left( \alpha \right) } \sin{ \left( \beta \right) } \end{align*}$

so

$\displaystyle \begin{align*} \cos{ \left( 1 + \mathrm{i} \right) } &= \cos{ \left( 1 \right) } \cos{ \left( \mathrm{i} \right) } - \sin{ \left( 1 \right) } \sin{ \left( \mathrm{i} \right) } \end{align*}$

Playing around a bit with Euler's formula we have

$\displaystyle \begin{align*} \mathrm{e}^{\mathrm{i} \, \theta} &= \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \\ \\ \mathrm{e}^{-\mathrm{i}\,\theta} &= \cos{ \left( -\theta \right) } + \mathrm{i}\sin{ \left( -\theta \right) } \\ &= \cos{ \left( \theta \right) } - \mathrm{i}\sin{ \left( \theta \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{-\mathrm{i}\,\theta} &= 2\cos{ \left( \theta \right) } \\ \\ \cos{ \left( \theta \right) } &= \frac{1}{2} \left( \mathrm{e}^{\mathrm{i}\,\theta} + \mathrm{e}^{-\mathrm{i}\,\theta} \right) \\ \\ \cos{ \left( \mathrm{i} \right) } &= \frac{1}{2} \left( \mathrm{e}^{\mathrm{i}^2} + \mathrm{e}^{-\mathrm{i}^2} \right) \\ &= \frac{1}{2} \left( \mathrm{e}^{-1} + \mathrm{e}^{1} \right) \\ &= \cosh{ \left( 1 \right) } \\ \\ \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} &= 2\mathrm{i}\sin{ \left( \theta \right) } \\ \mathrm{i} \left( \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} \right) &= -2\sin{ \left( \theta \right) } \\ \sin{ \left( \theta \right) } &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{\mathrm{i}\,\theta} - \mathrm{e}^{-\mathrm{i}\,\theta} \right) \\ \\ \sin{ \left( \mathrm{i} \right) } &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{\mathrm{i}^2} - \mathrm{e}^{-\mathrm{i}^2} \right) \\ &= -\frac{\mathrm{i}}{2} \left( \mathrm{e}^{-1} - \mathrm{e}^{1} \right) \\ &= \frac{\mathrm{i}}{2} \left( \mathrm{e}^{1} - \mathrm{e}^{-1} \right) \\ &= \mathrm{i}\sinh{ \left( 1 \right) } \end{align*}$

Thus:

$\displaystyle \begin{align*} \cos{ \left( 1 + \mathrm{i} \right) } &= \cos{ \left( 1 \right) } \cos{ \left( \mathrm{i} \right) } - \sin{ \left( 1 \right) } \sin{ \left( \mathrm{i} \right) } \\ &= \cos{ \left( 1 \right) } \cosh{ \left( 1 \right) } - \sin{ \left( 1 \right) } \left[ \mathrm{i}\sinh{ \left( 1 \right) } \right] \\ &= \cos{ \left( 1 \right) } \cosh{ \left( 1 \right) } - \mathrm{i} \sin{ \left( 1 \right) } \sinh{ \left( 1 \right) } \end{align*}$

i don't understand why it became

$i(e^{i\theta}-e^{-i\theta})=2\sin\theta$

may be you mean

$\frac{(e^{i\theta}-e^{-i\theta})}{i}=2\sin\theta$

 

[topsquark]

how would you express that in rectangular form?

He did. The last line reads as a + i b. (cos(1)cosh(1)) - i (sin(1)sinh(1))

-Dan

 

[I like Serena]

i don't understand why it became

$i(e^{i\theta}-e^{-i\theta})=2\sin\theta$

may be you mean

$\frac{(e^{i\theta}-e^{-i\theta})}{i}=2\sin\theta$

It didn't.
It became:
$$i(e^{i\theta}-e^{-i\theta})=-2\sin\theta$$
This is the result of multiplying left and right with $i$.

Btw, note that $\frac 1 i = -i$, so this fits with what you were thinking.

 

[Prove It]

Evaluate the following expressions, expressing answers in rectangular form.

1. $\cos(1+j)$
2.$\sinh(4-j3)$

can you help me on how to solve these problems.

thanks in advance!

$\displaystyle \begin{align*} \sinh{ \left( 4 - 3\mathrm{i} \right) } &= \frac{1}{2} \left[ \mathrm{e}^{4 - 3\mathrm{i}} - \mathrm{e}^{- \left( 4 - 3\mathrm{i} \right) } \right] \\ &= \frac{1}{2} \left( \mathrm{e}^{4 - 3\mathrm{i}} - \mathrm{e}^{-4 + 3\mathrm{i}} \right) \\ &= \frac{1}{2} \left\{ \mathrm{e}^4 \left[ \cos{ \left( -3 \right) } + \mathrm{i}\sin{ \left( - 3 \right) } \right] - \mathrm{e}^{-4} \left[ \cos{ \left( 3 \right) } + \mathrm{i} \sin{ \left( 3 \right) } \right] \right\} \\ &= \frac{1}{2} \left\{ \mathrm{e}^4 \left[ \cos{ (3) } - \mathrm{i}\sin{(3)} \right] - \mathrm{e}^{-4} \left[ \cos{(3)} + \mathrm{i}\sin{(3)} \right] \right\} \\ &= \frac{1}{2} \left[ \mathrm{e}^4\cos{ \left( 3 \right) } - \mathrm{i}\,\mathrm{e}^4 \sin{ (3) } - \mathrm{e}^{-4}\cos{(3)} - \mathrm{i}\,\mathrm{e}^{-4}\sin{(3)} \right] \\ &= \frac{1}{2} \left[ \cos{(3)} \left( \mathrm{e}^4 - \mathrm{e}^{-4} \right) - \mathrm{i} \sin{(3)} \left( \mathrm{e}^4 + \mathrm{e}^{-4} \right) \right] \\ &= \cos{(3)} \left[ \frac{1}{2} \left( \mathrm{e}^{4} - \mathrm{e}^{-4} \right) \right] - \mathrm{i} \sin{(3)} \left[ \frac{1}{2} \left( \mathrm{e}^4 + \mathrm{e}^{-4} \right) \right] \\ &= \cos{(3)} \sinh{(4)} - \mathrm{i}\sin{(3)}\cosh{(4)} \end{align*}$