How would I go about expressing a term like this in the form $a + ib$?
$$ \frac{(\cos \alpha+ i \ \sin \alpha)(cos \beta+ i \ \sin \beta)}{(\cos \gamma+ i \ \sin \gamma)(\cos\delta+ i \ \sin \delta)} $$
I tried rationalizing the denominator and got an expression as just a product of 4 terms but then I am unable to proceed as I am not aware of any formula for the cosine of the sum of more than 2 angles.
The expression I got after rationalizing is:
$$ (\cos \alpha+ i \ \sin \alpha)(\cos \beta + i \ \sin \beta)(\cos \gamma- i \ \sin \gamma)(\cos\delta- i \ \sin\delta) $$
Any help will be highly appreciated.
Best Answer
As pointed out in the comments. We can use the euler's formula to give us:
$\cos(\alpha)+i\sin(\alpha)=e^{i\alpha}$
$\cos(\beta)+i\sin(\beta)=e^{i\beta}$
$\cos(\gamma)+i\sin(\gamma)=e^{i\gamma}$
$\cos(\delta)+i\sin(\delta)=e^{i\delta}$
Which would give us-
$\textstyle\displaystyle{\frac{(\cos(\alpha)+i\sin(\alpha))(\cos(\beta)+i\sin(\beta))}{(\cos(\gamma)+i\sin(\gamma))(\cos(\delta)+i\sin(\delta)}}$
$\textstyle\displaystyle{=\frac{e^{i\alpha}e^{i\beta}}{e^{i\gamma}e^{i\delta}}}$
$\textstyle\displaystyle{=e^{i(\alpha+\beta-\gamma-\delta)}}$
$\textstyle\displaystyle{=\cos(\beta+\alpha-\gamma-\delta)+i\sin(\beta+\alpha-\gamma-\delta)}$