Suppose $a$ is a real number such that $\sin(\pi \cdot  \cos a) = \cos(\pi \cdot \sin a)$. Evaluate $35 \sin^2(2a) + 84 \cos^2(4a)$.
1. Given the equation $\sin(\pi \cdot \cos a) = \cos(\pi \cdot \sin a)$, we need to find the value of $35 \sin^2(2a) + 84 \cos^2(4a)$.

2. First, let's analyze the given trigonometric equation. We know that $\sin(x) = \cos(\pi/2 - x)$. Therefore, we can rewrite the given equation as:
   \[
   \sin(\pi \cdot \cos a) = \cos(\pi \cdot \sin a) \implies \sin(\pi \cdot \cos a) = \sin\left(\frac{\pi}{2} - \pi \cdot \sin a\right)
   \]

3. For the equality $\sin(\pi \cdot \cos a) = \sin\left(\frac{\pi}{2} - \pi \cdot \sin a\right)$ to hold, the arguments must be either equal or differ by an integer multiple of $\pi$. Thus, we have:
   \[
   \pi \cdot \cos a = \frac{\pi}{2} - \pi \cdot \sin a + 2k\pi \quad \text{or} \quad \pi \cdot \cos a = \pi - \left(\frac{\pi}{2} - \pi \cdot \sin a\right) + 2k\pi
   \]
   for some integer $k$.

4. Simplifying the first case:
   \[
   \pi \cdot \cos a = \frac{\pi}{2} - \pi \cdot \sin a + 2k\pi \implies \cos a + \sin a = \frac{1}{2} + 2k
   \]
   Since $\cos a$ and $\sin a$ are bounded between -1 and 1, the only feasible solution is when $k = 0$:
   \[
   \cos a + \sin a = \frac{1}{2}
   \]

5. Simplifying the second case:
   \[
   \pi \cdot \cos a = \pi - \left(\frac{\pi}{2} - \pi \cdot \sin a\right) + 2k\pi \implies \cos a = \frac{1}{2} + \sin a + 2k
   \]
   Again, considering the bounds of $\cos a$ and $\sin a$, the only feasible solution is when $k = 0$:
   \[
   \cos a = \frac{1}{2} + \sin a
   \]

6. From the above, we have two possible equations:
   \[
   \cos a + \sin a = \frac{1}{2} \quad \text{or} \quad \cos a = \frac{1}{2} + \sin a
   \]

7. Let's solve $\cos a + \sin a = \frac{1}{2}$:
   \[
   \sin 2a = 2 \sin a \cos a
   \]
   Squaring both sides:
   \[
   (\cos a + \sin a)^2 = \left(\frac{1}{2}\right)^2 \implies \cos^2 a + 2 \cos a \sin a + \sin^2 a = \frac{1}{4}
   \]
   Using $\cos^2 a + \sin^2 a = 1$:
   \[
   1 + 2 \cos a \sin a = \frac{1}{4} \implies 2 \cos a \sin a = -\frac{3}{4} \implies \sin 2a = -\frac{3}{4}
   \]
   Therefore:
   \[
   \sin^2 2a = \left(-\frac{3}{4}\right)^2 = \frac{9}{16}
   \]

8. Next, we need to find $\cos 4a$:
   \[
   \cos 4a = 2 \cos^2 2a - 1
   \]
   Since $\sin^2 2a = \frac{9}{16}$, we have:
   \[
   \cos^2 2a = 1 - \sin^2 2a = 1 - \frac{9}{16} = \frac{7}{16}
   \]
   Therefore:
   \[
   \cos 4a = 2 \cdot \frac{7}{16} - 1 = \frac{14}{16} - 1 = \frac{14}{16} - \frac{16}{16} = -\frac{1}{8}
   \]
   Thus:
   \[
   \cos^2 4a = \left(-\frac{1}{8}\right)^2 = \frac{1}{64}
   \]

9. Finally, we calculate the expression:
   \[
   35 \sin^2(2a) + 84 \cos^2(4a) = 35 \cdot \frac{9}{16} + 84 \cdot \frac{1}{64}
   \]
   Simplifying:
   \[
   35 \cdot \frac{9}{16} = \frac{315}{16}
   \]
   \[
   84 \cdot \frac{1}{64} = \frac{84}{64} = \frac{21}{16}
   \]
   Adding these:
   \[
   \frac{315}{16} + \frac{21}{16} = \frac{336}{16} = 21
   \]

The final answer is $\boxed{21}$