# 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 $T=\cos(x+y)+\cos(x+z)+\cos(z+x)$

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.

• What happened when you set $$\frac{P}{Q} = \frac{\cos x + \cos y + \cos z}{\cos (x+y+z)}, ~\frac{R}{S} =\frac{\sin x + \sin y + \sin z}{\sin (x+y+z)}?$$ What did the constructed fraction $\frac{P+R}{Q+S}$ simplify to? Perhaps you could edit your question to show this work. With respect to influencing mathSE reviewers to react positively to your question, see this article. Oct 15, 2021 at 13:16

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, $$T = \cos (x+y) + \cos (y+z) + \cos (z+x)$$

• This solution is certainly more elegant than the other methods I have attempted. Now that you've mentioned it, the question does indeed lend itself very nicely into complex numbers. Oct 15, 2021 at 15:16