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Diffraction grating formulawavelength

I understand how to derive the formula for diffraction gratings as you just have to compare the light rays approaching at a maxima point. The formula for diffraction grating formula is shown below : $$d \sin{\theta} = mλ$$ where $d =$ distance between slits/slit separation

Diffraction grating formulafor maxima

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The reason for the small angle is that the diffraction pattern satisfying the grating equation is the far-field (Fraunhofer) diffraction pattern. For which you would need the screen to be very far away from the slits/grating. Alternatively, you can have a lens, that brings the far-field pattern at its focal length $f$.

Diffraction gratingdefinition

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You can see that these two are the same in the small angle approximation, because for small $\Theta$ you have $\sin\Theta \approx \Theta \approx \tan\Theta$.

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From my understanding, the double-slit formula is derived from small-angle approximations but we can't assume small angles for a diffraction grating as there is a lot more interference and diffraction. Formula for double slits is shown below : $$w = \frac{mλD}s$$ Where $w =$ distance between fringes, $m =$ fringe order. $D =$ distance between slits, and screen $s =$ distance between slits/slit separation.

Diffraction grating formulaDerivation

Then, from geometry, $w = D \tan\Theta$, so you end up with: $$ d\tan\Theta = m\lambda $$ for the double slit diffraction pattern. Which, as you are saying, is different from the general diffraction grating formula: $$ d\sin\Theta = m\lambda.$$

Diffraction gratingexperiment

Writing your second equation using the same symbols of the first equation $(s\rightarrow d)$: $$ w = \frac{m \lambda D}{d}.$$

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What I can't seem to do is visualize this scenario as I do not fully understand why the angle, $\theta$ between the centre and a maxima point cannot be small for a diffraction grating as well? And so small angle approximations can't be used?