Are the real and imaginary parts of Riemann zeta equal to each other by multiplication of a Riemann Siegel theta function expression?

Question

It appears that the real part of Riemann zeta is related to the imaginary part by this formula:

$$\Re\left(\zeta \left(\frac{1}{2}+i t\right)\right)=\frac{\Im\left(\zeta \left(\frac{1}{2}+i t\right)\right) (\cos (2 \vartheta (t))+1)}{-\sin (2 \vartheta (t))}$$

and vice versa:

$$\Im\left(\zeta \left(\frac{1}{2}+i t\right)\right)=\frac{\Re\left(\zeta \left(\frac{1}{2}+i t\right)\right) (-\sin (2 \vartheta (t)))}{\cos (2 \vartheta (t))+1}$$

where $\vartheta (t)$ is the Riemann Siegel theta function.

How does one prove this formula above?

I guess a starting point should be these formulas for the real and imaginary parts of the Riemann zeta function:

$$\Re\left(\zeta \left(a+ib\right)\right)=\frac{1}{2} \left(\zeta \left(a+ib\right)+\zeta \left(a-ib\right)\right)$$

$$\Im\left(\zeta \left(a+ib\right)\right)=\frac{1}{2 i}\left(\zeta \left(a+ib\right)-\zeta \left(a-ib\right)\right)$$

What I would do is to look at the series expansion of:

$$-\frac{\sin (2 \vartheta (t))}{\cos (2 \vartheta (t))+1}$$

Clear[t];
Normal[Series[(-Sin[RiemannSiegelTheta[t]*2])/(1 +
Cos[RiemannSiegelTheta[t]*2]), {t, 0, 1}]]
N[%, 20]

This gives: $$\frac{1}{2} t \left(\log (\pi )-\psi ^{(0)}\left(\frac{1}{4}\right)\right)$$ and numerically: $$2.6860917096128327911 t$$ where $\psi ^{(0)}\left(\frac{1}{4}\right)$ is the Polygamma function.

Then I would compare it to the series expansion of: $$\frac{\zeta \left(\frac{1}{2}+i t\right)-\zeta \left(\frac{1}{2}-i t\right)}{\frac{1}{2} (2 i) \left(\zeta \left(\frac{1}{2}+i t\right)+\zeta \left(\frac{1}{2}-i t\right)\right)}$$

which Mathematica:

Normal[Series[((Zeta[1/2 + I*t] - Zeta[1/2 - I*t])/(2*
       I))/((Zeta[1/2 + I*t] + Zeta[1/2 - I*t])/2), {t, 0, 1}]]
N[%, 20]

tells me is: $$\frac{t \zeta '\left(\frac{1}{2}\right)}{\zeta \left(\frac{1}{2}\right)}$$ and numerically: $$2.6860917096128327911 t$$

I then asked Mathematica to compare the two numerically equivalent expressions symbolically with the True Query command TrueQ:

TrueQ[1/2 t (Log[\[Pi]] - PolyGamma[0, 1/4]) == (
  t Derivative[1][Zeta][1/2])/Zeta[1/2]]

But the computer says no, Mathematica returns False although numerically the values so far are identical.

So is the real question to show that: $$\frac{1}{2} t \left(\log (\pi )-\psi ^{(0)}\left(\frac{1}{4}\right)\right)=\frac{t \zeta '\left(\frac{1}{2}\right)}{\zeta \left(\frac{1}{2}\right)}$$ ?

I know that: $$\log (\pi )=\frac{\zeta '(0)}{\zeta (0)}-\log (2)$$ but that won't take me far will it?

Update:

I misread the Mathematica manual at first and missed that this is an equality:

$$\frac{1}{2} t \left(\log (\pi )-\psi ^{(0)}\left(\frac{1}{4}\right)\right)=\frac{1}{2} t \left(\log (\pi )-\frac{\Gamma '\left(\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}\right)$$

Therefore the task is to show that: $$\frac{1}{2} t \left(\log (\pi )-\frac{\Gamma '\left(\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}\right)=\frac{t \zeta '\left(\frac{1}{2}\right)}{\zeta \left(\frac{1}{2}\right)}$$

TrueQ[1/2 t (Log[\[Pi]] - Gamma'[1/4]/Gamma[1/4]) == (
  t Derivative[1][Zeta][1/2])/Zeta[1/2]]

returns False.

N[1/2 t (Log[\[Pi]] - Gamma'[1/4]/Gamma[1/4]) == (
  t Derivative[1][Zeta][1/2])/Zeta[1/2], 20]

returns:
(2.6860917096128327911 t == 2.6860917096128327911 t)
True

Answer 1

Will you look at the definition of $Z(t)$ and $\vartheta(t)$ ?

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$$\text{For } t \in \mathbb{R}, \ Z(t) \in \mathbb{R}, \qquad \zeta(\tfrac{1}2+it) = Z(t) e^{-i \vartheta(t)} $$ thus $$\Re(\zeta(\tfrac{1}2+it)) = Z(t) \cos(\vartheta(t)), \qquad Z(t) = \frac{\Re(\zeta(\frac{1}2+it))}{\cos(\vartheta(t))}$$ $$\Im(\zeta(\tfrac{1}2+it)) = - Z(t) \sin(\vartheta(t)) = -\frac{\Re(\zeta(\frac{1}2+it))}{\cos(\vartheta(t))}\sin(\vartheta(t)) =\frac{ -\Re(\zeta(\frac{1}2+it))\sin(2\vartheta(t))}{1+\cos(2\vartheta(t))}$$ where I used $$\frac{\sin(x)}{\cos(x)} =\frac{\cos(x)\sin(x)}{\cos(x)^2}=\frac{\sin(2x)}{1+\cos(2x)}$$