It's quite elementary, my dear Watson. $ABCD$ is a square. More generally, it's a rhombus, and the following applies to that also: $AB ≅ AD$, $BC ≅ DC$, $CA ≅ CA$, thus $△ABC ≅ △ADC$; and from this, it follows $∠BAC = ∠DAC$.
This proves that the $90^∘$ right angle $∠BAD$ is bisected, thereby showing that both $∠BAC$ and $∠DAC$ are both $45^∘$ angles.
You would do better to consider the fact that since
$$\cos^2 x + \sin^2 x = 1, \quad \cos^2 x - \sin^2 x = \cos{2x},$$
then
$$2\cos^2 x = 1 + \cos{2x},$$
from which it follows that
$$\cos x = ±\sqrt{\frac{1 + \cos{2x}}{2}},$$
and likewise when $x$ is divided by $2$:
$$\cos \frac{x}{2} = ±\sqrt{\frac{1 + \cos{2x}}{2}},$$
For angles $x$ between $0^∘$ and $90^∘$, inclusive, $\cos x ≥ 0$, so that we always have
$$\cos \frac{x}{2} = +\sqrt{\frac{1 + \cos{2x}}{2}} = \frac{1}{2}\sqrt{2 + 2\cos{2x}}.$$
Therefore, we can iterate and interpolate, starting with
$$\cos 0^∘ = 1, \quad \cos 90^∘ = 0.$$
In the first round, we get:
$$\cos 45^∘ = \frac{1}{2}\sqrt{2 + 2·0} = \frac{1}{2}\sqrt{2}.$$
In the second round, making use of complements ($\cos{135^∘} = -\cos{45^∘}$), we get:
$$\cos 22½^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2}}, \quad \cos 67½^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2}}.$$
In the third round, again making use of complements ($\cos 112½^∘ = -\cos 67½^∘$ and $\cos 157½^∘ = -\cos 22½^∘$), we get:
$$
\cos 11¼^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2}}}, \quad \cos 33¾^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2}}},\\
\cos 56¼^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2}}}, \quad \cos 78¾^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2}}}.
$$
In the fourth round, once more making use of complements ($\cos 101¼^∘ = -\cos 78¾^∘$, $\cos 123¾^∘ = -\cos 56¼^∘$, $\cos 146¼^∘ = -\cos 33¾^∘$ and $\cos 168¾^∘ = -\cos 11¼^∘$), we get:
$$
\cos 5⅝^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}, \quad \cos 16⅞^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 - \sqrt{2}}}},\\
\cos 28⅛^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2}}}}, \quad \cos 39⅜^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2}}}},\\
\cos 50⅝^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2}}}}, \quad \cos 61⅞^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2}}}},\\
\cos 73⅛^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2}}}}, \quad \cos 84⅜^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}}.
$$
And so on...
It goes without saying that $0^∘$ and $90^∘$ fit within this scheme with:
$$
1 = \cos 0^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ⋯}}}}},\\
0 = \cos 90^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ⋯}}}}}.
$$