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Noting that $$\text{Re}[z_1z_2] = \text{Re}[z_1]\text{Re}[z_2]-\text{Im}[z_1]\text{Im}[z_2],$$ how can $$\cos(\alpha+\beta) = \text{Re}\left[e^{j(\alpha+\beta)}\right]$$ be expressed? Give the final answer in a simple form, without complex-valued functions.

Source.

and here is my attempt:

We know that $$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta).$$ So taking the real part of the cosine function gives $$\text{Re}\left[\cos(\alpha)\right]\text{Re}[\cos(\beta)]-\text{Im}[\sin(\alpha)]\text{Im}[\sin(\beta)].$$

Now we use Euler's formula to convert sin and cosine to complex exponentials \begin{align*} \cos x &= \text{Re}\{e^{ix}\} = \frac{e^{ix}+e^{-ix}}{2}\\ \sin x &= \text{Im}\{e^{ix}\} = \frac{e^{ix}-e^{-ix}}{2i} \end{align*} $$ = \left[\frac{e^{j\alpha}+e^{-j\alpha}}{2}\right]\left[\frac{e^{j\beta}+e^{-j\beta}}{2}\right]-\left[\frac{e^{j\alpha}-e^{-j\alpha}}{2j}\right]\left[\frac{e^{j\beta}-e^{-j\beta}}{2j}\right]$$ Now once I do this, I will cancel out some terms, but my final answer is not $e^{\alpha+\beta}$ as it should be.

Source.

Please let me know what it is I am doing wrong. Thanks

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  • $\begingroup$ why vote down? what is the problem buddy ? $\endgroup$ – user65652 Jan 22 '16 at 19:55
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    $\begingroup$ Why can't you enter the questions in MathJax like most others would? $\endgroup$ – JB King Jan 22 '16 at 19:57
  • $\begingroup$ how do I use MathJax ? $\endgroup$ – user65652 Jan 22 '16 at 19:58
  • $\begingroup$ I posted the question, and then my attempt, I think my attempt considered as the work I have done ! $\endgroup$ – user65652 Jan 22 '16 at 19:59
  • $\begingroup$ math.stackexchange.com/help/notation would be one link under that "help" option in the top row that may be useful. math.stackexchange.com/help is a link to the general help stuff. Posting the "enter image description here" for your images is a bit sloppy you do realize right? $\endgroup$ – JB King Jan 22 '16 at 19:59
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Hint:

You know that: $$ \cos (\alpha+\beta)=\mbox{Re}\left[e^{i(\alpha+\beta)}\right]=\mbox{Re}\left[e^{i\alpha}e^{i\beta}\right] $$ Now use the given rule for the real part of a product and you have done.


From your rule you have: $$ \mbox{Re}\left[e^{i\alpha}e^{i\beta}\right]=\mbox{Re}\left[e^{i\alpha}\right]\mbox{Re}\left[e^{i\beta}\right]-\mbox{Im}\left[e^{i\alpha}\right]\mbox{Im}\left[e^{i\beta}\right] $$ and: $\mbox{Re}\left[e^{i\alpha}\right]=\cos \alpha$, $\mbox{Im}\left[e^{i\alpha}\right]=\sin \alpha$, $\mbox{Re}\left[e^{i\beta}\right]=\cos \beta$, $\mbox{Im}\left[e^{i\beta}\right]=\sin \beta$

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  • $\begingroup$ you mean apply this to the last line I wrote ( long complex exponential equations) ? $\endgroup$ – user65652 Jan 22 '16 at 20:17
  • $\begingroup$ I've added to my answer. I hope it's useful :) $\endgroup$ – Emilio Novati Jan 22 '16 at 20:25

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