# Find modulus and argument of \omega=\frac{\sin(P+Q)+i(1-\cos(P+Q))}{(\cos P+\cos Q)+i(\sin P-\sin Q)}

Find modulus and argument of $\omega =\frac{\mathrm{sin}\left(P+Q\right)+i\left(1-\mathrm{cos}\left(P+Q\right)\right)}{\left(\mathrm{cos}P+\mathrm{cos}Q\right)+i\left(\mathrm{sin}P-\mathrm{sin}Q\right)}$
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coolbananas03ok
We have
$N\phantom{\rule{0.222em}{0ex}}={\mathrm{sin}}^{2}\left(P+Q\right)+{\left(1-\mathrm{cos}\left(P+Q\right)\right)}^{2}={\mathrm{sin}}^{2}\left(P+Q\right)+{\mathrm{cos}}^{2}\left(P+Q\right)+1-2\mathrm{cos}\left(P+Q\right)$
$=2\left(1-\mathrm{cos}\left(P+Q\right)\right)=2\cdot 2{\mathrm{sin}}^{2}\frac{P+Q}{2}=4{\mathrm{sin}}^{2}\frac{P+Q}{2}$
and
$D={\mathrm{cos}}^{2}P+{\mathrm{cos}}^{2}Q+{\mathrm{sin}}^{2}P+{\mathrm{sin}}^{2}Q+2\left(\mathrm{cos}P\mathrm{cos}Q-\mathrm{sin}P\mathrm{sin}Q\right)$
$=2+2\left(\mathrm{cos}\left(P+Q\right)\right)=2\left(1+\mathrm{cos}\left(P+Q\right)\right)=4{\mathrm{cos}}^{2}\frac{P+Q}{2}$
Now,
$|\omega |=\sqrt{\frac{N}{D}}=\mathrm{tan}\frac{P+Q}{2}$