If $$\tan\left(\frac{\pi}{12}-x\right) , \tan\left(\frac{\pi}{12}\right) , \tan \left(\frac{\pi}{12} +x\right)$$ in order are three consecutive terms of a GP then what is sum of all possible values of $x$.
I am not getting any start, can anybody provide me a hint?
Use :$$\Big(\tan\frac{\pi}{2}\Big)^2=\tan\Big(\frac{\pi}{12}-x\Big)\tan\Big(\frac{\pi}{12}+x\Big) \tag1$$ (Condition for G.P)
And :
$$\tan (A\pm B) = \frac{\tan A \pm \tan B}{1\mp \tan A\tan B}$$
Let $\tan \dfrac{\pi}{12} =c$
Put in $(1)$
$$c^2=\frac {c+\tan x}{1-c\tan x} \cdot \frac{c-\tan x}{1+c\tan x} \implies c^2-c^4\tan^2 x =c^2-\tan^2 x \implies \tan x =0$$
Sum of solutions in a bounded interval will be sum all angles in the form $k\pi ~;~ k \in \mathbb Z$
Clearly, $x=n\pi$ (where $n$ is any integer) is a trivial solution.
Otherwise
$$\dfrac{\tan\dfrac\pi{12}}{\tan\left(\dfrac\pi{12}-x\right)}=\dfrac{\tan\left(\dfrac\pi{12}+x\right)}{\tan\dfrac\pi{12}}$$
$$\iff\dfrac{\sin\dfrac\pi{12}\cos\left(\dfrac\pi{12}-x\right)}{\cos\dfrac\pi{12}\sin\left(\dfrac\pi{12}-x\right)}=\dfrac{\cos\dfrac\pi{12}\sin\left(\dfrac\pi{12}+x\right)}{\sin\dfrac\pi{12}\cos\left(\dfrac\pi{12}+x\right)}$$
Using Componendo & Dividendo and $\sin(A\pm B)$ formulae,
$$\dfrac{\sin\left(\dfrac\pi{12}+\dfrac\pi{12}-x\right)}{\sin\left\{\dfrac\pi{12}-\left(\dfrac\pi{12}-x\right\}\right)}=\dfrac{\sin\left(\dfrac\pi{12}+x+\dfrac\pi{12}\right)}{\sin\left\{\dfrac\pi{12}+x-\dfrac\pi{12}\right\}}$$
As $\sin x\ne0,$ $$\sin\left(\dfrac\pi6-x\right)=\sin\left(\dfrac\pi6+x\right)$$
Using Prosthaphaeresis Formulas, $$\sin\left(\dfrac\pi6+x\right)-\sin\left(\dfrac\pi6-x\right)=2\sin x\cos\dfrac\pi6$$
So, $\sin x$ has to be $0$ which is impossible as $x\ne n\pi$
You should be able to use this fact to set up an equation that can be solved for $x$.
– Paul Aljabar Apr 19 '17 at 17:18