Proving two functions are equal each other

Question

I have a problem when doing an approximation. The problem comes in the final results that I have to demonstrate two functions below equal each other

\begin{align} \frac{e^{j\sin{(\phi-\phi_0)}}+e^{-j\sin{(\phi-\phi_0)}}}{8\sqrt{2}\cos{\frac{\phi-\phi_0}{2}}(1-\sin{\frac{\phi-\phi_0}{2}})\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} = \frac{e^{j\sin{(\phi-\phi_0)}}+e^{-j\sin{(\phi-\phi_0)}}}{[1+\cos{(\phi-\phi_0)}]^2}. \end{align}

I have checked the two functions by numerical calculation to a graph and see that two functions give exactly the same shape with the $\phi\leq \pi$ as shown in the figure.

The condition for this is $0\leq|\phi-\phi_0|\leq\pi/2$.

We can see that two functions are the same until $\phi-\phi_0\leq\pi$. I dont understand why is that. Could please someone help me to prove that two functions are equal.

Thank you all very much! enter image description here

Answer 1

They are not always equal. Consider what happens when $\phi=\phi_0,$ for example.

To see that they won't always be equal, neglect their numerators, which are equal. Then one function has the form $$\frac{1}{K\cos(z/2)(1-\sin(z/2))\sqrt{1-\sin z}},$$ and the other $$\frac{1}{(1+\cos z)^2},$$ where $K=8\sqrt 2.$

One thing you spot immediately is that one always positive, while the other is not always.

Answer 2

Consider $$f=K \cos \left(\frac{t}{2}\right)\left(1-\sin \left(\frac{t}{2}\right)\right)^{3/2} $$ $$g=(1+\cos (t))^2$$ Using Taylor series built around $t=0$, we have $$f=K\left(1-\frac{3 t}{4}-\frac{t^2}{32}+\frac{17 t^3}{128}+O\left(t^4\right) \right)$$ $$g=4-2 t^2+O\left(t^4\right)$$

How could they be equal ?