Prove: 1^° overlinesin z=sinbar z; 2^° overlinecos z=cosbar…
Prove: $\ 1^\circ\ \overline{\sin z}=\sin\bar z;\quad 2^\circ\ \overline{\cos z}=\cos\bar z;\quad 3^\circ\ \overline{\tan z}=\tan\bar z$.
Explanation of Method
Use $\overline{e^w}=e^{\bar w}$ (true since conjugation is a ring homomorphism commuting with the power series of $\exp$) together with $\bar i=-i$, applied to the exponential definitions.
Part $1^\circ$: $\overline{\sin z}=\sin\bar z$
$$ \overline{\sin z}=\overline{\left(\frac{e^{iz}-e^{-iz}}{2i}\right)}=\frac{\overline{e^{iz}}-\overline{e^{-iz}}}{\overline{2i}}=\frac{e^{\overline{iz}}-e^{\overline{-iz}}}{-2i} $$
Since $\overline{iz}=\bar i\bar z=-i\bar z$:
$$ =\frac{e^{-i\bar z}-e^{i\bar z}}{-2i}=\frac{e^{i\bar z}-e^{-i\bar z}}{2i}=\sin\bar z $$
Part $2^\circ$: $\overline{\cos z}=\cos\bar z$
$$ \overline{\cos z}=\overline{\left(\frac{e^{iz}+e^{-iz}}2\right)}=\frac{e^{-i\bar z}+e^{i\bar z}}2=\cos\bar z $$
Part $3^\circ$: $\overline{\tan z}=\tan\bar z$
Using $\tan z=\sin z/\cos z$ and the fact that conjugation distributes over division:
$$ \overline{\tan z}=\overline{\left(\frac{\sin z}{\cos z}\right)}=\frac{\overline{\sin z}}{\overline{\cos z}}=\frac{\sin\bar z}{\cos\bar z}=\tan\bar z $$
Cross-check via real/imaginary splits
With $z=x+iy$: $\ \sin z=\sin x\cosh y+i\cos x\sinh y$, so $\overline{\sin z}=\sin x\cosh y-i\cos x\sinh y$. Meanwhile $\bar z=x-iy$, and $\sin(x-iy)=\sin x\cosh(-y)+i\cos x\sinh(-y)=\sin x\cosh y-i\cos x\sinh y$ (using $\cosh$ even, $\sinh$ odd) — matching exactly.
General principle
Any function built from real constants, the four arithmetic operations, and $e^z$ automatically satisfies $\overline{f(z)}=f(\bar z)$; this covers every elementary trigonometric and hyperbolic function.
Result
All three follow directly from $\overline{e^{iz}}=e^{-i\bar z}$ — the single fact that conjugation commutes with the complex exponential.