Trigonometry and Inequality

Let's do a trivial but crucial rearrangement of the equation:

$\sqrt{\rho}\cos \theta = \sin(\theta + \phi)$

We first note that the addition of $\phi$ is a horizontal translation of the sine wave, and that the multiplication of $\sqrt{\rho}$ is a scaling of the (co)sine wave. Now, note that no $\rho \neq 1$ will permit a scaling of the (co)sine wave to be equal to a horizontal translation of the sine wave.

Thus, no $\phi$ can satisfy the equation.

Since the given equation is required to hold for all $\theta$ , by putting $\theta=\frac{\pi}{2}$ , we must have $\sqrt \rho\cos(\phi) =0$ Since $\rho \neq 0$ , we have $\cos(\phi)=0$ . Hence $\sin^2(\phi)= 1$ . Now put $\theta=0$ in the above identity and square both sides to obtain $\rho=\sin^2(\phi) = 1,$ which is a contradiction.

Note that $\rho = (-\infty,1)\cup(1, +\infty) \Rightarrow \rho \neq 1$ .

Consider the (expression 1) ---> $\rho\cos\theta$ . We need to transform expression 1 to (expression 2) ---> $\sqrt{\rho}\sin(\theta + \phi)$ .

Now, By expression 1

$\rho\cos\theta = \sqrt{\rho}(\sqrt{\rho}\cos\theta + 0\sin\theta)$

Let $\sin\phi = \sqrt{\rho}$ and $\cos\phi = 0$

So, the expression 1 will become...

$\sqrt{\rho}(\sin\phi \cos\theta + \cos\phi \sin \theta)$

$\Rightarrow \sqrt{\rho}(\sin(\theta + \phi)$ --> expression 2

Note that, $\cos\phi = 0 \Rightarrow \sin\phi = 1$

Hence, $\sqrt{\rho} = 1 \Rightarrow \rho = 1$ .

But $\rho \neq 1$ .

So, If $\rho \neq 1$ then $\phi$ has no solution.