I am having a hard time solving the following problem.
Question:
Show that if:
$$\frac{\cos x + \cos y + \cos z}{\cos (x+y+z)}=\frac{\sin x + \sin y + \sin z}{\sin (x+y+z)} = T$$
then $T=\cos (x+y) + \cos (x+z) + \cos (z+x)$
The question says to use the identity
if:$$\frac{P}Q = \frac{R}S$$
then: $$\frac{P}Q = \frac{P+R}{Q+S}$$
Other info: It is worth three marks. I am in my last year of high school. We are expected to know products to sums identities and how to expand compound trig expressions.
I've tried decomposing the compound angles and even squaring both fractions before merging them but even that hasn't really helped. (The reason why I considered $T^2$ was because then you would be able to use the Pythagorean identity a few times.) Any assistance would be greatly appreciated.
Best Answer
This solution uses complex numbers. We will use
$$\frac{P}{Q} = \frac{R}{S} \Rightarrow \frac{P}{Q} = \frac{P+iR}{Q+iS}$$
where $i^2=-1$. Since $T$ is a real number, we have
$$T = \Re{\Big( \frac{(\cos x + \cos y + \cos z) + i(\sin x + \sin y + \sin z)}{\cos (x+y+z) + i\sin(x+y+z)} \Big)}$$
where $\Re(w)$ denotes real part of complex number $w$.
Simplifying this using de Moivre's theorem : $$\cos \theta + i\sin \theta = e^{i\theta} = \exp(i\theta)$$
$$T = \Re{\Big( \frac{e^{ix} + e^{iy} + e^{iz}}{e^{i(x+y+z)}} \Big)} $$ $$T = \Re{\Big( (e^{ix} + e^{iy} + e^{iz})e^{-i(x+y+z)}\Big)}$$ $$T = \Re{\Big( e^{-i(y+z)} + e^{-i(z+x)} + e^{-i(x+y)}\Big)}$$
Hence, $$T = \cos (x+y) + \cos (y+z) + \cos (z+x)$$