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

Question

A past examination paper had the following question that I found somewhat difficult. I tried having a go at it but haven't come around with any possible double angle identities. How would one go about tackling it?

Given:

$$\omega = {\frac {\sin (P + Q) + i (1 - \cos (P + Q))} {(\cos P + \cos Q) + i (\sin P - \sin Q) }} $$

To prove:

$$|\omega| = \tan \frac {P + Q} {2} \qquad\text{and}\qquad \arg(\omega) = Q $$

A guideline on how/ which identity to use would be greatly appreciated.

To give an idea how one would start it is by;

Proof:

$$|\omega| = {\frac {\sqrt{\sin^2 (P + Q) + (1 - \cos (P + Q))^2}} {\sqrt{(\cos P + \cos Q)^2 + (\sin P - \sin Q)^2 }}} $$

I'm still unsure about the above or how the square root come about

Answer 1

We have \begin{align} N & := \sin^2(P+Q) + (1-\cos(P+Q))^2 = \sin^2(P+Q) + \cos^2(P+Q) + 1 - 2\cos(P+Q) \\ & = 2 (1-\cos(P+Q)) = 2\cdot2\sin^2\frac{P+Q}{2} = 4\sin^2\frac{P+Q}{2} \end{align} and \begin{align} D & = \cos^2P +\cos^2Q + \sin^2P + \sin^2Q + 2(\cos P\cos Q - \sin P \sin Q) \\ &= 2 +2(\cos(P+Q)) = 2(1+\cos(P+Q)) = 4\cos^2\frac{P+Q}{2} \end{align}

Now, $$|\omega| = \sqrt{\frac{N}{D}} = \tan\frac{P+Q}{2}$$

Answer 2

You just need Pythagoras and compound angle formulae, and in particular identities for $\cos 2x$. The numerator surd simplifies to $$\sqrt{2\left(1-\cos(P+Q)\right)}=2\left|\sin\frac{P+Q}{2}\right|,$$while the denominator surd simplifies to $$\sqrt{2\left(1+\cos(P+Q)\right)}=2\left|\cos\frac{P+Q}{2}\right|.$$

Answer 3

Using Double Angle formula,

$$N_r=\sin(P+Q)+i(1-\cos(P+Q))$$

$$=2\sin\dfrac{P+Q}2\cos\dfrac{P+Q}2+2i\sin^2\dfrac{P+Q}2$$

$$=2\sin\dfrac{P+Q}2\left(\cos\dfrac{P+Q}2+i\sin\dfrac{P+Q}2\right)=2\sin\dfrac{P+Q}2 e^{i(P+Q)/2}$$ using Intuition behind euler's formula

Using Prosthaphaeresis Formulas, $$D_r=\cos P+\cos Q+i(\sin P-\sin Q)$$ $$=2\cos\dfrac{P+Q}2\left(\cos\dfrac{P-Q}2+i\sin\dfrac{P-Q}2\right)=2\cos\dfrac{P+Q}2e^{i(P-Q)/2}$$

$$\implies \omega=\tan \frac {P + Q}2 e^{i Q}$$

$\displaystyle|\omega| = \left|\tan \dfrac {P + Q}2\right|=\begin{cases} \tan \dfrac {P + Q}2&\mbox{if } \tan \dfrac {P + Q}2\ge 0 \\ -\tan \dfrac {P + Q}2 & \mbox{otherwise} \end{cases} $

arg$(\omega)$ can be computed atan2